Count occurrences of a substring
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Create a function, or show a built-in function, to count the number of non-overlapping occurrences of a substring inside a string.
The function should take two arguments:
- the first argument being the string to search, and
- the second a substring to be searched for.
It should return an integer count.
print countSubstring("the three truths","th")
3
// do not count substrings that overlap with previously-counted substrings:
print countSubstring("ababababab","abab")
2The matching should yield the highest number of non-overlapping matches.
In general, this essentially means matching from left-to-right or right-to-left (see proof on talk page).
Other tasks related to string operations:
- Metrics
- Counting
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contain the most consonants
- Find words which contains more than 3 vowels
- Find words whose first and last three letters are equal
- Find words with alternating vowels and consonants
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 bottles of beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
11l
print(‘the three truths’.count(‘th’))
print(‘ababababab’.count(‘abab’))- Output:
3 2
360 Assembly
The program uses two ASSIST macros (XDECO,XPRNT) to keep the code as short as possible.
* Count occurrences of a substring 05/07/2016
COUNTSTR CSECT
USING COUNTSTR,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
MVC HAYSTACK,=CL32'the three truths'
MVC LENH,=F'17' lh=17
MVC NEEDLE,=CL8'th' needle='th'
MVC LENN,=F'2' ln=2
BAL R14,SHOW call show
MVC HAYSTACK,=CL32'ababababab'
MVC LENH,=F'11' lh=11
MVC NEEDLE,=CL8'abab' needle='abab'
MVC LENN,=F'4' ln=4
BAL R14,SHOW call show
L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
HAYSTACK DS CL32 haystack
NEEDLE DS CL8 needle
LENH DS F length(haystack)
LENN DS F length(needle)
*------- ---- show---------------------------------------------------
SHOW ST R14,SAVESHOW save return address
BAL R14,COUNT count(haystack,needle)
LR R11,R0 ic=count(haystack,needle)
MVC PG(20),HAYSTACK output haystack
MVC PG+20(5),NEEDLE output needle
XDECO R11,PG+25 output ic
XPRNT PG,80 print buffer
L R14,SAVESHOW restore return address
BR R14 return to caller
SAVESHOW DS A return address of caller
PG DC CL80' ' buffer
*------- ---- count--------------------------------------------------
COUNT ST R14,SAVECOUN save return address
SR R7,R7 n=0
LA R6,1 istart=1
L R10,LENH lh
S R10,LENN ln
LA R10,1(R10) lh-ln+1
LOOPI CR R6,R10 do istart=1 to lh-ln+1
BH ELOOPI
LA R8,NEEDLE @needle
L R9,LENN ln
LA R4,HAYSTACK-1 @haystack[0]
AR R4,R6 +istart
LR R5,R9 ln
CLCL R4,R8 if substr(haystack,istart,ln)=needle
BNE NOTEQ
LA R7,1(R7) n=n+1
A R6,LENN istart=istart+ln
NOTEQ LA R6,1(R6) istart=istart+1
B LOOPI
ELOOPI LR R0,R7 return(n)
L R14,SAVECOUN restore return address
BR R14 return to caller
SAVECOUN DS A return address of caller
* ---- -------------------------------------------------------
YREGS
END COUNTSTR- Output:
the three truths th 3 ababababab abab 2
8080 Assembly
The routine subcnt takes the string pointer in HL
and the substring pointer in BC, and returns a 16-bit count in
DE.
org 100h
jmp demo
;;; Count non-overlapping substrings (BC) in string (HL)
;;; Returns amount of matches in DE
subcnt: lxi d,0 ; Amount of matches
s_scan: mov a,m ; Get current character
ana a ; End of string?
rz ; Then stop
push b ; Keep start of substring search
push h ; Keep current location in string
s_cmp: ldax b ; Get character from substring
cmp m ; Compare to curent charracter of search string
inx b ; Advance pointers
inx h
jz s_cmp ; Keep going if they were equal
ana a ; Did we reach the end of the substring?
jz s_find ; If so, we found a match
pop h ; Otherwise, no match - restore search position
pop b ; Restore start of substring
inx h ; Try next position
jmp s_scan
s_find: inx d ; We found a match
pop b ; Discard start of the search, keep going after match
pop b ; Restore start of substring
dcx h ; The comparison routine overshoots by one
jmp s_scan
;;; Test on a few strings
demo: lxi h,pairs
loop: mov e,m ; Load string pointer
inx h
mov d,m
inx h
mov a,d ; If 0, stop
ora e
rz
mov c,m ; Load substring pointer
inx h
mov b,m
inx h
push h ; Save example pointer
xchg ; Put string pointer in HL
call subcnt ; Count substrings
mvi a,'0' ; Assuming output is <10, print output
add e ; (This is true for all examples, and a proper numeric
mov e,a ; output routine is big and not relevant.)
mvi c,2 ; CP/M character output
call 5
pop h ; Restore example pointer
jmp loop
pairs: dw str1,sub1,str2,sub2,str3,sub3,0
str1: db 'the three truths',0
sub1: db 'th',0 ; result should be 3
str2: db 'ababababab',0
sub2: db 'abab',0 ; result should be 2
str3: db 'cat',0
sub3: db 'dog',0 ; result should be 0- Output:
320
8086 Assembly
cpu 8086
org 100h
section .text
jmp demo
;;; Count non-overlapping substrings [ES:DI] in [DS:SI]
;;; Return count in AX
subcnt: xor ax,ax ; Set count to 0
xor dl,dl ; Zero to compare to
mov bp,di ; Keep copy of substring pointer
.scan: cmp dl,[si] ; End of string?
je .out ; Then we're done
mov bx,si ; Keep copy of search position
mov di,bp ; Start at beginning of substring
.cmp: xor cx,cx
dec cx
repe cmpsb ; Scan until no match
dec si ; Go to first non-match
dec di
cmp dl,[es:di] ; Reached end of substring?
je .match ; Then we found a match
mov si,bx ; If not, continue searching one
inc si ; position further
jmp .scan
.match: inc ax ; Found a match - increment count
jmp .scan
.out: ret
;;; Test the routine on a few examples
demo: mov si,pairs
.loop: lodsw ; Load string pointer
test ax,ax ; If 0, stop
jz .out
xchg dx,ax
lodsw ; Load substring pointer
xchg di,ax
push si ; Keep example pointer
xchg si,dx
call subcnt ; Count substrings
call prax ; Print amount of substrings
pop si ; Restore example pointer
jmp .loop
.out: ret
;;; Print AX as number
prax: mov bx,num ; Pointer to end of number string
mov cx,10 ; Divisor
.dgt: xor dx,dx ; Divide by 10
div cx
add dl,'0' ; Add ASCII 0 to remainder
dec bx ; Store digit
mov [bx],dl
test ax,ax ; If number not zero yet
jnz .dgt ; Find rest of digits
mov dx,bx ; Print number string
mov ah,9
int 21h
ret
section .data
db '*****' ; Output number placeholder
num: db ' $'
;;; Examples
pairs: dw .str1,.sub1,.str2,.sub2,.str3,.sub3,0
.str1: db 'the three truths',0
.sub1: db 'th',0 ; result should be 3
.str2: db 'ababababab',0
.sub2: db 'abab',0 ; result should be 2
.str3: db 'cat',0
.sub3: db 'dog',0 ; result should be 0
- Output:
3 2 0
AArch64 Assembly
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program strcptsub64.s */
/************************************/
/* Constantes */
/************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
/************************************/
/* Initialized data */
/************************************/
.data
szMessResult: .asciz "Result: "
szString: .asciz "the three truths"
szSubString: .asciz "th"
szString1: .asciz "ababababab"
szSubString1: .asciz "abab"
szCarriageReturn: .asciz "\n"
szMessStart: .asciz "Program 64 bits start.\n"
/************************************/
/* UnInitialized data */
/************************************/
.bss
sZoneConv: .skip 24
/************************************/
/* code section */
/************************************/
.text
.global main
main: // entry of program
ldr x0,qAdrszMessStart
bl affichageMess
ldr x0,qAdrszString
ldr x1,qAdrszSubString
bl countSubString
ldr x0,qAdrszString1
ldr x1,qAdrszSubString1
bl countSubString
100: // standard end of the program
mov x0, #0 // return code
mov x8, #EXIT // request to exit program
svc 0 // perform the system call
qAdrszString: .quad szString
qAdrszSubString: .quad szSubString
qAdrszString1: .quad szString1
qAdrszSubString1: .quad szSubString1
qAdrsZoneConv: .quad sZoneConv
qAdrszMessResult: .quad szMessResult
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrszMessStart: .quad szMessStart
/***************************************************/
/* count sub string of string */
/***************************************************/
/* r0 contains a string */
/* r1 contains a substring */
/* r0 return substring count */
countSubString:
stp x1,lr,[sp,-16]!
stp x2,x3,[sp,-16]!
stp x4,x5,[sp,-16]!
stp x6,x7,[sp,-16]!
mov x4,#0 // counter
mov x3,#0 // index string
mov x5,#0 // index substring
1:
ldrb w6,[x0,x5] // load byte of string
ldrb w7,[x1,x3] // load byte of substring
cmp x6,x7 // compare byte
bne 2f // not equal
cmp x6,#0 // string end ?
beq 3f // yes
add x5,x5,#1 // else increment index
add x3,x3,#1
b 1b // and loop
2: // characters not equal
cmp x6,#0 // end string ?
beq 4f
cmp x7,#0 // end substring ?
add x6,x4,1
csel x4,x6,x4,eq // yes -> increment counter
mov x3,#0 // raz index substring
add x5,x5,#1 // increment string index
b 1b // and loop
3: // end string and end substring
add x4,x4,#1 // increment counter
4: // result display
mov x0,x4
ldr x1,qAdrsZoneConv
bl conversion10
ldr x0,qAdrszMessResult
bl affichageMess
ldr x0,qAdrsZoneConv
bl affichageMess
ldr x0,qAdrszCarriageReturn
bl affichageMess
mov x0,x4
100:
ldp x6,x7,[sp],16
ldp x4,x5,[sp],16
ldp x2,x3,[sp],16
ldp x1,lr,[sp],16
ret
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeARM64.inc"- Output:
Program 64 bits start. Result: 3 Result: 2
Action!
BYTE FUNC CountSubstring(CHAR ARRAY s,sub)
BYTE i,j,res,found
i=1 res=0
WHILE i-1+sub(0)<=s(0)
DO
found=1
FOR j=1 TO sub(0)
DO
IF s(j+i-1)#sub(j) THEN
found=0
EXIT
FI
OD
IF found=1 THEN
i==+sub(0)
res==+1
ELSE
i==+1
FI
OD
RETURN (res)
PROC Test(CHAR ARRAY s,sub)
BYTE c
c=CountSubstring(s,sub)
PrintF("%B ""%S"" in ""%S""%E",c,sub,s)
RETURN
PROC Main()
Test("the three truths","th")
Test("ababababab","abab")
Test("11111111","11")
Test("abcdefg","123")
RETURN- Output:
Screenshot from Atari 8-bit computer
3 "th" in "the three truths" 2 "abab" in "ababababab" 4 "11" in "11111111" 0 "123" in "abcdefg"
Ada
with Ada.Strings.Fixed, Ada.Integer_Text_IO;
procedure Substrings is
begin
Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source => "the three truths",
Pattern => "th"));
Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source => "ababababab",
Pattern => "abab"));
end Substrings;
- Output:
3 2
ALGOL 68
Algol68 has no build in function to do this task, hence the need to create a count string in string routine.
If your Algol 68 compiler/interpreter does not have string in string, there is an implementation on Rosetta Code here.
PROC count string in string = (STRING needle, haystack)INT: (
INT start:=LWB haystack, next, out:=0;
FOR count WHILE string in string(needle, next, haystack[start:]) DO
start+:=next+UPB needle-LWB needle;
out:=count
OD;
out
);
print((
whole( count string in string("th", "the three truths"), 0 ) # expect 3 #, " ",
whole( count string in string("abab", "ababababab"), 0 ) # expect 2 #, " ",
whole( count string in string("a*b", "abaabba*bbaba*bbab"), 0 ) # expect 2 #, newline
))- Output:
3 2 2
Apex
Apex example for 'Count occurrences of a substring'.
String substr = 'ABC';
String str = 'ABCZZZABCYABCABCXXABC';
Integer substrLen = substr.length();
Integer count = 0;
Integer index = str.indexOf(substr);
while (index >= 0) {
count++;
str = str.substring(index+substrLen);
index = str.indexOf(substr);
}
System.debug('Count String : '+count);Count String : 5
APL
csubs←{0=x←⊃⍸⍺⍷⍵:0 ⋄ 1+⍺∇(¯1+x+⍴⍺)↓⍵}
- Output:
'th' 'abab' 'dog' csubs¨ 'the three truths' 'ababababab' 'cat' 3 2 0
AppleScript
This is a good example of the kind of problem to which standard libraries (a regex library in this case) would offer most languages a simple and immediate solution. AppleScript, however, for want of various basics like regex and math library functions, can require scripters to draw on the supplementary resources of Bash, using the built-in do shell script function.
A slightly faster approach to seeking outside library help has, however, become possible since OS X 10.10 added JavaScript as an osalang sibling for AppleScript. Rather than using the resources of Bash, we can more quickly use AppleScript's built in ObjC interface to pass a subproblem over to the more richly endowed JavaScript core of JavaScript for Automation.
Here we use a generic evalOSA(language, code) function to apply a JavaScript for Automation regex to a pair of AppleScript strings, using OSAKit.
use framework "OSAKit"
on run
{countSubstring("the three truths", "th"), ¬
countSubstring("ababababab", "abab")}
end run
on countSubstring(str, subStr)
return evalOSA("JavaScript", "var matches = '" & str & "'" & ¬
".match(new RegExp('" & subStr & "', 'g'));" & ¬
"matches ? matches.length : 0") as integer
end countSubstring
-- evalOSA :: ("JavaScript" | "AppleScript") -> String -> String
on evalOSA(strLang, strCode)
set ca to current application
set oScript to ca's OSAScript's alloc's initWithSource:strCode ¬
|language|:(ca's OSALanguage's languageForName:(strLang))
set {blnCompiled, oError} to oScript's compileAndReturnError:(reference)
if blnCompiled then
set {oDesc, oError} to oScript's executeAndReturnError:(reference)
if (oError is missing value) then return oDesc's stringValue as text
end if
return oError's NSLocalizedDescription as text
end evalOSA
- Output:
{3, 2}
The above assertions notwithstanding, it's always been possible to use AppleScript's text item delimiters for this purpose with its native strings, except that the TIDs have only observed the current considering/ignoring state with the 'unicode text' class, which was introduced around Mac OS X 10.4 and became AppleScript's only native text class with the introduction of AppleScript 2.0 in Mac OS X 10.5.
on countSubstring(theString, theSubstring)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to theSubstring
set substringCount to (count theString's text items) - 1
set AppleScript's text item delimiters to astid
return substringCount
end countSubstring
{countSubstring("the three truths", "th"), countSubstring("ababababab", "abab")}
- Output:
{3, 2}
ARM Assembly
/* ARM assembly Raspberry PI */
/* program strcptsub.s */
/************************************/
/* Constantes */
/************************************/
/* for this file see task include a file in language ARM assembly*/
.include "../constantes.inc"
/************************************/
/* Initialized data */
/************************************/
.data
szMessResult: .asciz "Result: "
szString: .asciz "the three truths"
szSubString: .asciz "th"
szString1: .asciz "ababababab"
szSubString1: .asciz "abab"
szCarriageReturn: .asciz "\n"
szMessStart: .asciz "Program 32 bits start.\n"
/************************************/
/* UnInitialized data */
/************************************/
.bss
sZoneConv: .skip 24
/************************************/
/* code section */
/************************************/
.text
.global main
main: @ entry of program
ldr r0,iAdrszMessStart
bl affichageMess
ldr r0,iAdrszString
ldr r1,iAdrszSubString
bl countSubString
ldr r0,iAdrszString1
ldr r1,iAdrszSubString1
bl countSubString
100: @ standard end of the program
mov r0, #0 @ return code
mov r7, #EXIT @ request to exit program
svc 0 @ perform the system call
iAdrszString: .int szString
iAdrszSubString: .int szSubString
iAdrszString1: .int szString1
iAdrszSubString1: .int szSubString1
iAdrsZoneConv: .int sZoneConv
iAdrszMessResult: .int szMessResult
iAdrszCarriageReturn: .int szCarriageReturn
iAdrszMessStart: .int szMessStart
/***************************************************/
/* count sub string of string */
/***************************************************/
/* r0 contains a string */
/* r1 contains a substring */
/* r0 return substring count */
countSubString:
push {r1-r7,lr} @ save registers
mov r4,#0 @ counter
mov r3,#0 @ index string
Mov r5,#0 @ index substring
1:
ldrb r6,[r0,r5] @ load byte of string
ldrb r7,[r1,r3] @ load byte of substring
cmp r6,r7 @ compare byte
bne 2f @ not equal
cmp r6,#0 @ string end ?
beq 3f @ yes
add r5,r5,#1 @ else increment index
add r3,r3,#1
b 1b @ and loop
2: @ characters not equal
cmp r6,#0 @ end string ?
beq 4f
cmp r7,#0 @ end substring ?
addeq r4,r4,#1 @ yes -> increment counter
mov r3,#0 @ raz index substring
add r5,r5,#1 @ increment string index
b 1b @ and loop
3: @ end string and end substring
add r4,r4,#1 @ increment counter
4: @ result display
mov r0,r4
ldr r1,iAdrsZoneConv
bl conversion10
ldr r0,iAdrszMessResult
bl affichageMess
ldr r0,iAdrsZoneConv
bl affichageMess
ldr r0,iAdrszCarriageReturn
bl affichageMess
mov r0,r4
100:
pop {r1-r7,pc}
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
/* for this file see task include a file in language ARM assembly*/
.include "../affichage.inc"- Output:
Program 32 bits start. Result: 3 Result: 2
Arturo
countOccurrences: function [str, substr]-> size match str substr
loop [["the three truths" "th"] ["ababababab" "abab"]] 'pair ->
print [
~"occurrences of '|last pair|' in '|first pair|':"
countOccurrences first pair last pair
]
- Output:
occurrences of 'th' in 'the three truths': 3 occurrences of 'abab' in 'ababababab': 2
AutoHotkey
While it is simple enough to parse the string, AutoHotkey has a rather unconventional method which outperforms this. StringReplace sets the number of replaced strings to ErrorLevel.
MsgBox % countSubstring("the three truths","th") ; 3
MsgBox % countSubstring("ababababab","abab") ; 2
CountSubstring(fullstring, substring){
StringReplace, junk, fullstring, %substring%, , UseErrorLevel
return errorlevel
}
AWK
#
# countsubstring(string, pattern)
# Returns number of occurrences of pattern in string
# Pattern treated as a literal string (regex characters not expanded)
#
function countsubstring(str, pat, len, i, c) {
c = 0
if( ! (len = length(pat) ) )
return 0
while(i = index(str, pat)) {
str = substr(str, i + len)
c++
}
return c
}
#
# countsubstring_regex(string, regex_pattern)
# Returns number of occurrences of pattern in string
# Pattern treated as regex
#
function countsubstring_regex(str, pat, c) {
c = 0
c += gsub(pat, "", str)
return c
}
BEGIN {
print countsubstring("[do&d~run?d!run&>run&]", "run&")
print countsubstring_regex("[do&d~run?d!run&>run&]", "run[&]")
print countsubstring("the three truths","th")
}
- Output:
$ awk -f countsubstring.awk 2 2 3
BaCon
FUNCTION Uniq_Tally(text$, part$)
LOCAL x
WHILE TALLY(text$, part$)
INCR x
text$ = MID$(text$, INSTR(text$, part$)+LEN(part$))
WEND
RETURN x
END FUNCTION
PRINT "the three truths - th: ", Uniq_Tally("the three truths", "th")
PRINT "ababababab - abab: ", Uniq_Tally("ababababab", "abab")
- Output:
the three truths - th: 3 ababababab - abab: 2
BASIC
In FreeBASIC, this needs to be compiled with -lang qb or -lang fblite.
DECLARE FUNCTION countSubstring& (where AS STRING, what AS STRING)
PRINT "the three truths, th:", countSubstring&("the three truths", "th")
PRINT "ababababab, abab:", countSubstring&("ababababab", "abab")
FUNCTION countSubstring& (where AS STRING, what AS STRING)
DIM c AS LONG, s AS LONG
s = 1 - LEN(what)
DO
s = INSTR(s + LEN(what), where, what)
IF 0 = s THEN EXIT DO
c = c + 1
LOOP
countSubstring = c
END FUNCTION
- Output:
the three truths, th: 3 ababababab, abab: 2
See also: Liberty BASIC, PowerBASIC, PureBasic.
Applesoft BASIC
10 F$ = "TH"
20 S$ = "THE THREE TRUTHS"
30 GOSUB 100"COUNT SUBSTRING
40 PRINT R
50 F$ = "ABAB"
60 S$ = "ABABABABAB"
70 GOSUB 100"COUNT SUBSTRING
80 PRINT R
90 END
100 R = 0
110 F = LEN(F$)
120 S = LEN(S$)
130 IF F > S THEN RETURN
140 IF F = 0 THEN RETURN
150 IF F = S AND F$ = S$ THEN R = 1 : RETURN
160 FOR I = 1 TO S - F
170 IF F$ = MID$(S$, I, F) THEN R = R + 1 : I = I + F - 1
180 NEXT I
190 RETURNBASIC256
print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")
end
function countSubstring(s$,find$)
i = 1
while instr(s$,find$,i) <> 0
countSubstring += 1
i = instr(s$,find$,i) + length(find$)
end while
end function
- Output:
Igual que la entrada de Run BASIC.
Chipmunk Basic
10 CLS : REM 10 HOME for Applesoft BASIC
20 F$ = "TH"
30 S$ = "THE THREE TRUTHS"
40 GOSUB 110: REM COUNT SUBSTRING
50 PRINT R
60 F$ = "ABAB"
70 S$ = "ABABABABAB"
80 GOSUB 110: REM COUNT SUBSTRING
90 PRINT R
100 END
110 R = 0
120 F = LEN(F$)
130 S = LEN(S$)
140 IF F > S THEN RETURN
150 IF F = 0 THEN RETURN
160 IF F = S AND F$ = S$ THEN R = 1: RETURN
170 FOR I = 1 TO S - F
180 IF F$ = MID$(S$, I, F) THEN R = R + 1: I = I + F - 1
190 NEXT I
200 RETURN
GW-BASIC
The Chipmunk Basic solution works without any changes.
IS-BASIC
100 INPUT PROMPT "String: ":TXT$
110 INPUT PROMPT "Substring: ":SUB$
120 PRINT COUNT(LCASE$(TXT$),LCASE$(SUB$))
130 DEF COUNT(TXT$,SUB$)
140 LET N=0:LET PO=1
150 DO
160 LET PO=POS(TXT$,SUB$,PO)
170 IF PO THEN LET N=N+1:LET PO=PO+LEN(SUB$)
180 LOOP UNTIL PO=0
190 LET COUNT=N
200 END DEFMSX Basic
The Chipmunk Basic solution works without any changes.
Sinclair ZX81 BASIC
Works with 1k of RAM.
10 LET S$="THE THREE TRUTHS"
20 LET U$="TH"
30 GOSUB 100
40 PRINT N
50 LET S$="ABABABABAB"
60 LET U$="ABAB"
70 GOSUB 100
80 PRINT N
90 STOP
100 LET N=0
110 LET I=0
120 LET I=I+1
130 IF I+LEN U$>LEN S$ THEN RETURN
140 IF S$(I TO I+LEN U$-1)<>U$ THEN GOTO 120
150 LET N=N+1
160 LET I=I+LEN U$
170 GOTO 130
True BASIC
FUNCTION countsubstring(where$, what$)
LET c = 0
LET s = 1-LEN(what$)
DO
LET s = POS(where$,what$,s+LEN(what$))
IF 0 = s THEN EXIT DO
LET c = c+1
LOOP
LET countsubstring = c
END FUNCTION
PRINT "the three truths, th:", countSubstring("the three truths", "th")
PRINT "ababababab, abab:", countSubstring("ababababab", "abab")
END
Yabasic
print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")
end
sub countSubstring(s$,find$)
countSubstring = 0
i = 1
while instr(s$,find$,i) <> 0
countSubstring = countSubstring + 1
i = instr(s$,find$,i) + len(find$)
end while
return countSubstring
end sub- Output:
Igual que la entrada de Run BASIC.
Batch File
@echo off
setlocal enabledelayedexpansion
::Main
call :countString "the three truths","th"
call :countString "ababababab","abab"
pause>nul
exit /b
::/Main
::Procedure
:countString
set input=%~1
set cnt=0
:count_loop
set trimmed=!input:*%~2=!
if "!trimmed!"=="!input!" (echo.!cnt!&goto :EOF)
set input=!trimmed!
set /a cnt+=1
goto count_loop
- Output:
3 2
BBC BASIC
tst$ = "the three truths"
sub$ = "th"
PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"
tst$ = "ababababab"
sub$ = "abab"
PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"
END
DEF FNcountSubstring(A$, B$)
LOCAL I%, N%
I% = 1 : N% = 0
REPEAT
I% = INSTR(A$, B$, I%)
IF I% THEN N% += 1 : I% += LEN(B$)
UNTIL I% = 0
= N%
- Output:
3 "th" in "the three truths" 2 "abab" in "ababababab"
BCPL
get "libhdr"
let countsubstr(str, match) = valof
$( let i, count = 1, 0
while i <= str%0 do
test valof
$( for j = 1 to match%0
unless match%j = str%(i+j-1)
resultis false
resultis true
$)
then
$( count := count + 1
i := i + match%0
$)
else
i := i + 1
resultis count
$)
let show(str, match) be
writef("*"%S*" in *"%S*": %N*N",
match, str, countsubstr(str, match))
let start() be
$( show("the three truths", "th")
show("ababababab", "abab")
show("cat", "dog")
$)- Output:
"th" in "the three truths": 3 "abab" in "ababababab": 2 "dog" in "cat": 0
BQN
/𝕨⍷𝕩 finds locations of substrings, rest of the function suppresses overlapping substrings.
Find←{i←/𝕨⍷𝕩, i/˜i≥»0≤◶⟨⊣,(≠𝕨)+⊢⟩`i}
•Show "abab" Find "ababababab"
•Show "th" Find "the three truths"
2
3
Using strings.bqn from bqn-libs, another solution is Find←+´Locate, since Locate performs a non-overlapping search.
Bracmat
( count-substring
= n S s p
. 0:?n:?p
& !arg:(?S.?s)
& @( !S
: ?
( [!p ? !s [?p ?
& !n+1:?n
& ~
)
)
| !n
)
& out$(count-substring$("the three truths".th))
& out$(count-substring$(ababababab.abab))
& ;- Output:
3 2
C
#include <stdio.h>
#include <string.h>
int match(const char *s, const char *p, int overlap)
{
int c = 0, l = strlen(p);
while (*s != '\0') {
if (strncmp(s++, p, l)) continue;
if (!overlap) s += l - 1;
c++;
}
return c;
}
int main()
{
printf("%d\n", match("the three truths", "th", 0));
printf("overlap:%d\n", match("abababababa", "aba", 1));
printf("not: %d\n", match("abababababa", "aba", 0));
return 0;
}
Alternate version:
#include <stdio.h>
#include <string.h>
// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const char *str, const char *sub)
{
int length = strlen(sub);
if (length == 0) return 0;
int count = 0;
for (str = strstr(str, sub); str; str = strstr(str + length, sub))
++count;
return count;
}
int main()
{
printf("%d\n", countSubstring("the three truths", "th"));
printf("%d\n", countSubstring("ababababab", "abab"));
printf("%d\n", countSubstring("abaabba*bbaba*bbab", "a*b"));
return 0;
}
- Output:
3 2 2
C#
using System;
class SubStringTestClass
{
public static int CountSubStrings(this string testString, string testSubstring)
{
int count = 0;
if (testString.Contains(testSubstring))
{
for (int i = 0; i < testString.Length; i++)
{
if (testString.Substring(i).Length >= testSubstring.Length)
{
bool equals = testString.Substring(i, testSubstring.Length).Equals(testSubstring);
if (equals)
{
count++;
i += testSubstring.Length - 1; // Fix: Don't count overlapping matches
}
}
}
}
return count;
}
}Using C# 6.0's expression-bodied member, null-conditional operator, and coalesce operator features:
using System;
class SubStringTestClass
{
public static int CountSubStrings(this string testString, string testSubstring) =>
testString?.Split(new [] { testSubstring }, StringSplitOptions.None)?.Length - 1 ?? 0;
}C++
#include <iostream>
#include <string>
// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const std::string& str, const std::string& sub)
{
if (sub.length() == 0) return 0;
int count = 0;
for (size_t offset = str.find(sub); offset != std::string::npos;
offset = str.find(sub, offset + sub.length()))
{
++count;
}
return count;
}
int main()
{
std::cout << countSubstring("the three truths", "th") << '\n';
std::cout << countSubstring("ababababab", "abab") << '\n';
std::cout << countSubstring("abaabba*bbaba*bbab", "a*b") << '\n';
return 0;
}
- Output:
3 2 2
Clojure
Use a sequence of regexp matches to count occurrences.
(defn re-quote
"Produces a string that can be used to create a Pattern
that would match the string text as if it were a literal pattern.
Metacharacters or escape sequences in text will be given no special
meaning"
[text]
(java.util.regex.Pattern/quote text))
(defn count-substring [txt sub]
(count (re-seq (re-pattern (re-quote sub)) txt)))
Use the trick of blank replacement and maths to count occurrences.
(defn count-substring1 [txt sub]
(/ (- (count txt) (count (.replaceAll txt sub "")))
(count sub)))
A Java 8 stream-based solution, which should avoid creation of temporary strings (though it will produce temporary MatchResult instances).
(defn count-substring2 [txt sub]
(-> sub
(re-quote)
(re-pattern)
(re-matcher txt)
(.results)
(.count)))
COBOL
INSPECT can be used for this task without having to create a function.
IDENTIFICATION DIVISION.
PROGRAM-ID. testing.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 occurrences PIC 99.
PROCEDURE DIVISION.
INSPECT "the three truths" TALLYING occurrences FOR ALL "th"
DISPLAY occurrences
MOVE 0 TO occurrences
INSPECT "ababababab" TALLYING occurrences FOR ALL "abab"
DISPLAY occurrences
MOVE 0 TO occurrences
INSPECT "abaabba*bbaba*bbab" TALLYING occurrences
FOR ALL "a*b"
DISPLAY occurrences
GOBACK
.
- Output:
03 02 02
CoffeeScript
countSubstring = (str, substr) ->
n = 0
i = 0
while (pos = str.indexOf(substr, i)) != -1
n += 1
i = pos + substr.length
n
console.log countSubstring "the three truths", "th"
console.log countSubstring "ababababab", "abab"
Common Lisp
(defun count-sub (str pat)
(loop with z = 0 with s = 0 while s do
(when (setf s (search pat str :start2 s))
(incf z) (incf s (length pat)))
finally (return z)))
(count-sub "ababa" "ab") ; 2
(count-sub "ababa" "aba") ; 1
Cowgol
include "cowgol.coh";
sub countSubstring(str: [uint8], match: [uint8]): (count: uint8) is
count := 0;
while [str] != 0 loop
var find := match;
var loc := str;
while [loc] == [find] loop
find := @next find;
loc := @next loc;
end loop;
if [find] == 0 then
str := loc;
count := count + 1;
else
str := @next str;
end if;
end loop;
end sub;
print_i8(countSubstring("the three truths","th")); # should print 3
print_nl();
print_i8(countSubstring("ababababab","abab")); # should print 2
print_nl();
print_i8(countSubstring("cat","dog")); # should print 0
print_nl();- Output:
3 2 0
D
void main() {
import std.stdio, std.algorithm;
"the three truths".count("th").writeln;
"ababababab".count("abab").writeln;
}
- Output:
3 2
Delphi
program OccurrencesOfASubstring;
{$APPTYPE CONSOLE}
uses StrUtils;
function CountSubstring(const aString, aSubstring: string): Integer;
var
lPosition: Integer;
begin
Result := 0;
lPosition := PosEx(aSubstring, aString);
while lPosition <> 0 do
begin
Inc(Result);
lPosition := PosEx(aSubstring, aString, lPosition + Length(aSubstring));
end;
end;
begin
Writeln(CountSubstring('the three truths', 'th'));
Writeln(CountSubstring('ababababab', 'abab'));
end.
Draco
proc countSubstring(*char string, substring) word:
word count;
*char pos, loc;
count := 0;
while string* /= '\e' do
pos := substring;
loc := string;
while loc* = pos* do
loc := loc + 1;
pos := pos + 1
od;
if pos* = '\e' then
string := loc;
count := count + 1
else
string := string + 1
fi
od;
count
corp
proc main() void:
writeln(countSubstring("the three truths", "th"));
writeln(countSubstring("ababababab", "abab"))
corp- Output:
3 2
Dyalect
func countSubstring(str, val) {
var idx = 0
var count = 0
while true {
idx = str.IndexOf(val, idx)
if idx == -1 {
break
}
idx += val.Length()
count += 1
}
return count
}
print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))- Output:
3 2
Déjà Vu
!. count "the three truths" "th"
!. count "ababababab" "abab"- Output:
3 2
EasyLang
func count str$ pat$ .
ind = 1
while ind + len pat$ - 1 <= len str$
if substr str$ ind len pat$ = pat$
cnt += 1
ind += len pat$
else
ind += 1
.
.
return cnt
.
print count "the three truths" "th"
print count "ababababab" "abab"
print count "11111111" "11"
print count "11111111" "12"
print count "12" "12"EchoLisp
;; from Racket
(define count-substring
(compose length regexp-match*))
(count-substring "aab" "graabaabdfaabgh") ;; substring
→ 3
(count-substring "/ .e/" "Longtemps je me suis couché de bonne heure") ;; regexp
→ 4
ed
H
,p
# Change the substring here to search it
g/th/s//^/g
g/[^^]/s///g
g/\^/s//i/g
,j
# unary -> decimal (up to 10^10)
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/^\(i*\)\1\1\1\1\1\1\1\1\1\(i\{0,9\}\)/\1 \2/
g/i/s/[ ]$/ 0 /g
g/i/s/[ ][ ]/ 0 /g
g/i/s/[ ][ ]/ 0 /g
g/i/s/[ ]iiiiiiiii\>/ 9 /g
g/i/s/[ ]iiiiiiii\>/ 8 /g
g/i/s/[ ]iiiiiii\>/ 7 /g
g/i/s/[ ]iiiiii\>/ 6 /g
g/i/s/[ ]iiiii\>/ 5 /g
g/i/s/[ ]iiii\>/ 4 /g
g/i/s/[ ]iii\>/ 3 /g
g/i/s/[ ]ii\>/ 2 /g
g/i/s/[ ]i\>/ 1 /g
g/[ ]/s///g
g/^0\{1,\}\([0-9]\)/s//\1/
,p
Q
- Output:
$ ed -s count-substrings.input < count-substrings.ed Newline appended the three truths 3
EGL
The "remove and count the difference" and "manual loop" methods. Implementation includes protection from empty source and search strings.
program CountStrings
function main()
SysLib.writeStdout("Remove and Count:");
SysLib.writeStdout(countSubstring("th", "the three truths"));
SysLib.writeStdout(countSubstring("abab", "ababababab"));
SysLib.writeStdout(countSubstring("a*b", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring("a", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring(" ", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring("", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring("a", ""));
SysLib.writeStdout(countSubstring("", ""));
SysLib.writeStdout("Manual Loop:");
SysLib.writeStdout(countSubstringWithLoop("th", "the three truths"));
SysLib.writeStdout(countSubstringWithLoop("abab", "ababababab"));
SysLib.writeStdout(countSubstringWithLoop("a*b", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop("a", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop(" ", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop("", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop("a", ""));
SysLib.writeStdout(countSubstringWithLoop("", ""));
end
function countSubstring(substr string in, str string in) returns(int)
if(str.length() > 0 and substr.length() > 0)
return (str.length() - str.replaceStr(subStr, "").length()) / subStr.length();
else
return 0;
end
end
function countSubstringWithLoop(substr string in, str string in) returns(int)
count int = 0;
loc, index int = 1;
strlen int = str.length();
substrlen int = substr.length();
if(strlen > 0 and substrlen > 0)
while(loc != 0 and index <= strlen)
loc = str.indexOf(substr, index);
if(loc > 0)
count += 1;
index = loc + substrlen;
end
end
end
return count;
end
end- Output:
Remove and Count: 3 2 2 7 0 0 0 0 Manual Loop: 3 2 2 7 0 0 0 0
Eiffel
class
APPLICATION
inherit
ARGUMENTS
create
make
feature {NONE} -- Initialization
make
-- Run application.
do
occurance := 0
from
index := 1
until
index > text.count
loop
temp := text.fuzzy_index(search_for, index, 0)
if
temp /= 0
then
index := temp + search_for.count
occurance := occurance + 1
else
index := text.count + 1
end
end
print(occurance)
end
index:INTEGER
temp:INTEGER
occurance:INTEGER
text:STRING = "ababababab"
search_for:STRING = "abab"
end
Elixir
countSubstring = fn(_, "") -> 0
(str, sub) -> length(String.split(str, sub)) - 1 end
data = [ {"the three truths", "th"},
{"ababababab", "abab"},
{"abaabba*bbaba*bbab", "a*b"},
{"abaabba*bbaba*bbab", "a"},
{"abaabba*bbaba*bbab", " "},
{"abaabba*bbaba*bbab", ""},
{"", "a"},
{"", ""} ]
Enum.each(data, fn{str, sub} ->
IO.puts countSubstring.(str, sub)
end)
- Output:
3 2 2 7 0 0 0 0
Emacs Lisp
Two Emacs Lisp solutions are shown below
;; version 1, which takes advantage of the how-many function,
;; which runs only in a buffer
(defun count-substrings (text substring)
"Count non-overlapping occurrences of SUBSTRING in TEXT."
(with-temp-buffer ; create a temporary buffer, which will be deleted when function finishes
(insert text) ; insert TEXT into the empty buffer
(goto-char (point-min)) ; go to the beginning of the buffer
(how-many substring))) ; count how many occurrences of SUBSTRING
;; version 2, which operates on a string
(defun count-substrings (text substring)
"Count non-overlapping occurences of SUBSTRING in TEXT."
(let ((substrings)) ; empty list to add occurrences of SUBSTRING as we find them
(while (string-match substring text) ; as long as we can find SUBSTRING in TEXT
(push (match-string 0 text) substrings) ; add the SUBSTRING we found to the list of substrings
(setq text (replace-match "" nil nil text))) ; remove SUBSTRING from text, and repeat while loop
(length substrings))) ; count number of items in substrings list
- Output:
(count-substrings "the three truths" "th") 3 (count-substrings "ababababab" "abab") 2
Erlang
%% Count non-overlapping substrings in Erlang for the rosetta code wiki.
%% Implemented by J.W. Luiten
-module(substrings).
-export([main/2]).
%% String and Sub exhausted, count a match and present result
match([], [], _OrigSub, Acc) ->
Acc+1;
%% String exhausted, present result
match([], _Sub, _OrigSub, Acc) ->
Acc;
%% Sub exhausted, count a match
match(String, [], Sub, Acc) ->
match(String, Sub, Sub, Acc+1);
%% First character matches, advance
match([X|MainTail], [X|SubTail], Sub, Acc) ->
match(MainTail, SubTail, Sub, Acc);
%% First characters do not match. Keep scanning for sub in remainder of string
match([_X|MainTail], [_Y|_SubTail], Sub, Acc)->
match(MainTail, Sub, Sub, Acc).
main(String, Sub) ->
match(String, Sub, Sub, 0).
Command:
substrings:main("ababababab","abab").
- Output:
2
Alternative using built in functions:
main( String, Sub ) -> erlang:length( binary:split(binary:list_to_bin(String), binary:list_to_bin(Sub), [global]) ) - 1.
Euphoria
function countSubstring(sequence s, sequence sub)
integer from,count
count = 0
from = 1
while 1 do
from = match_from(sub,s,from)
if not from then
exit
end if
from += length(sub)
count += 1
end while
return count
end function
? countSubstring("the three truths","th")
? countSubstring("ababababab","abab")- Output:
3 2
F#
"Remove and count the difference" method, as shown by J, Java, ...
open System
let countSubstring (where :string) (what : string) =
match what with
| "" -> 0 // just a definition; infinity is not an int
| _ -> (where.Length - where.Replace(what, @"").Length) / what.Length
[<EntryPoint>]
let main argv =
let show where what =
printfn @"countSubstring(""%s"", ""%s"") = %d" where what (countSubstring where what)
show "the three truths" "th"
show "ababababab" "abab"
show "abc" ""
0
countSubstring("the three truths", "th") = 3
countSubstring("ababababab", "abab") = 2
countSubstring("abc", "") = 0
Factor
USING: math sequences splitting ;
: occurences ( seq subseq -- n ) split-subseq length 1 - ;
Forth
: str-count ( s1 len s2 len -- n )
2swap 0 >r
begin 2over search
while 2over nip /string
r> 1+ >r
repeat 2drop 2drop r> ;
s" the three truths" s" th" str-count . \ 3
s" ababababab" s" abab" str-count . \ 2
Fortran
program Example
implicit none
integer :: n
n = countsubstring("the three truths", "th")
write(*,*) n
n = countsubstring("ababababab", "abab")
write(*,*) n
n = countsubstring("abaabba*bbaba*bbab", "a*b")
write(*,*) n
contains
function countsubstring(s1, s2) result(c)
character(*), intent(in) :: s1, s2
integer :: c, p, posn
c = 0
if(len(s2) == 0) return
p = 1
do
posn = index(s1(p:), s2)
if(posn == 0) return
c = c + 1
p = p + posn + len(s2) - 1
end do
end function
end program
- Output:
3 2 2
FreeBASIC
' FB 1.05.0 Win64
Function countSubstring(s As String, search As String) As Integer
If s = "" OrElse search = "" Then Return 0
Dim As Integer count = 0, length = Len(search)
For i As Integer = 1 To Len(s)
If Mid(s, i, length) = Search Then
count += 1
i += length - 1
End If
Next
Return count
End Function
Print countSubstring("the three truths","th")
Print countSubstring("ababababab","abab")
Print countSubString("zzzzzzzzzzzzzzz", "z")
Print
Print "Press any key to quit"
Sleep
- Output:
3 2 15
FunL
import util.Regex
def countSubstring( str, substr ) = Regex( substr ).findAllMatchIn( str ).length()
println( countSubstring("the three truths", "th") )
println( countSubstring("ababababab", "abab") )- Output:
3 2
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
FutureBasic
window 1
local fn CountSubstring( string as CFStringRef, substring as CFStringRef ) as long
end fn = fn ArrayCount( fn StringComponentsSeparatedByString( string, substring ) ) - 1
print fn CountSubstring( @"the three truths", @"th" )
print fn CountSubstring( @"ababababab", @"abab" )
HandleEvents- Output:
3 2
Gambas
Public Sub Main()
Print countSubstring("the three truths", "th")
Print countSubstring("ababababab", "abab")
Print countSubString("zzzzzzzzzzzzzzz", "z")
End
Function countSubstring(s As String, search As String) As Integer
If s = "" Or search = "" Then Return 0
Dim count As Integer = 0, length As Integer = Len(search)
For i As Integer = 1 To Len(s)
If Mid(s, i, length) = Search Then
count += 1
i += length - 1
End If
Next
Return count
End Function
- Output:
Same as FreeBASIC entry.
Go
Using strings.Count() method:
package main
import (
"fmt"
"strings"
)
func main() {
fmt.Println(strings.Count("the three truths", "th")) // says: 3
fmt.Println(strings.Count("ababababab", "abab")) // says: 2
}
Groovy
Solution, uses the Groovy "find" operator (=~), and the Groovy-extended Matcher property "count":
println (('the three truths' =~ /th/).count)
println (('ababababab' =~ /abab/).count)
println (('abaabba*bbaba*bbab' =~ /a*b/).count)
println (('abaabba*bbaba*bbab' =~ /a\*b/).count)
- Output:
3 2 9 2
Haskell
Text-based solution
import Data.Text hiding (length)
-- Return the number of non-overlapping occurrences of sub in str.
countSubStrs str sub = length $ breakOnAll (pack sub) (pack str)
main = do
print $ countSubStrs "the three truths" "th"
print $ countSubStrs "ababababab" "abab"
- Output:
3 2
Alternatively, in a language built around currying, it might make more sense to reverse the suggested order of arguments.
{-# LANGUAGE OverloadedStrings #-}
import Data.Text hiding (length)
--------- COUNT OF SUBSTRING INSTANCES IN A STRING -------
countAll :: Text -> Text -> Int
countAll needle haystack =
length
(breakOnAll needle haystack)
--------------------------- TEST -------------------------
main :: IO ()
main =
print $
countAll "ab"
<$> [ "ababababab",
"abelian absurdity",
"babel kebab"
]
- Output:
[5,2,2]
List-based solution
Even though list-based strings are not "the right" way of representing texts, the problem of counting subsequences in a list is generally useful.
count :: Eq a => [a] -> [a] -> Int
count [] = error "empty substring"
count sub = go
where
go = scan sub . dropWhile (/= head sub)
scan _ [] = 0
scan [] xs = 1 + go xs
scan (x:xs) (y:ys) | x == y = scan xs ys
| otherwise = go ys
- Output:
λ> count "th" "the three truths" 3 λ> count "abab" "ababababab" 2 λ> count [2,3] [1,2,1,2,3,4,3,2,3,4,3,2] 2 λ> count "123456" $ foldMap show [1..1000000] 7
List-based solution using Data.List
The following solution is almost two times faster than the previous one.
import Data.List (tails, stripPrefix)
import Data.Maybe (catMaybes)
count :: Eq a => [a] -> [a] -> Int
count sub = length . catMaybes . map (stripPrefix sub) . tails
Icon and Unicon
- Output:
The string "th" occurs as a non-overlapping substring 3 times in "the three truths" The string "abab" occurs as a non-overlapping substring 2 times in "ababababab"
J
require'strings'
countss=: #@] %~ #@[ - [ #@rplc '';~]
In other words: find length of original string, replace the string to be counted with the empty string, find the difference in lengths and divide by the length of the string to be counted.
Example use:
'the three truths' countss 'th'
3
'ababababab' countss 'abab'
2
Java
Using regular expression pattern matching
int countSubstring(String string, String substring) {
substring = Pattern.quote(substring);
Pattern pattern = Pattern.compile(substring);
Matcher matcher = pattern.matcher(string);
int count = 0;
while (matcher.find())
count++;
return count;
}
The "remove and count the difference" method:
public class CountSubstring {
public static int countSubstring(String subStr, String str){
return (str.length() - str.replace(subStr, "").length()) / subStr.length();
}
public static void main(String[] args){
System.out.println(countSubstring("th", "the three truths"));
System.out.println(countSubstring("abab", "ababababab"));
System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
}
}
- Output:
3 2 2
The "split and count" method:
import java.util.regex.Pattern;
public class CountSubstring {
public static int countSubstring(String subStr, String str){
// the result of split() will contain one more element than the delimiter
// the "-1" second argument makes it not discard trailing empty strings
return str.split(Pattern.quote(subStr), -1).length - 1;
}
public static void main(String[] args){
System.out.println(countSubstring("th", "the three truths"));
System.out.println(countSubstring("abab", "ababababab"));
System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
}
}
- Output:
3 2 2
Manual looping
public class CountSubstring {
public static int countSubstring(String subStr, String str){
int count = 0;
for (int loc = str.indexOf(subStr); loc != -1;
loc = str.indexOf(subStr, loc + subStr.length()))
count++;
return count;
}
public static void main(String[] args){
System.out.println(countSubstring("th", "the three truths"));
System.out.println(countSubstring("abab", "ababababab"));
System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
}
}
- Output:
3 2 2
JavaScript
Using regexes:
function countSubstring(str, subStr) {
var matches = str.match(new RegExp(subStr, "g"));
return matches ? matches.length : 0;
}
Using 'split' and ES6 notation:
const countSubString = (str, subStr) => str.split(subStr).length - 1;
jq
Using regexes (available in jq versions after June 19, 2014):
def countSubstring(sub):
[match(sub; "g")] | length;Example:
"the three truths" | countSubstring("th")Julia
Built-in Function
matchall(r::Regex, s::String[, overlap::Bool=false]) -> Vector{String}
Return a vector of the matching substrings from eachmatch.
Main
ts = ["the three truths", "ababababab"]
tsub = ["th", "abab"]
println("Test of non-overlapping substring counts.")
for i in 1:length(ts)
print(ts[i], " (", tsub[i], ") => ")
println(length(matchall(Regex(tsub[i]), ts[i])))
end
println()
println("Test of overlapping substring counts.")
for i in 1:length(ts)
print(ts[i], " (", tsub[i], ") => ")
println(length(matchall(Regex(tsub[i]), ts[i], true)))
end
- Output:
Test of non-overlapping substring counts. the three truths (th) => 3 ababababab (abab) => 2 Test of overlapping substring counts. the three truths (th) => 3 ababababab (abab) => 4
K
The dyadic verb _ss gives the positions of substring y in string x.
"the three truths" _ss "th"
0 4 13
#"the three truths" _ss "th"
3
"ababababab" _ss "abab"
0 4
#"ababababab" _ss "abab"
2
Klingphix
include ..\Utilitys.tlhy
:count %s !s
0 >ps
[ps> 1 + >ps
$s len nip + snip nip] [$s find dup] while
drop drop ps>
;
"the three truths" "th" count ?
"ababababab" "abab" count ?
" " inputOther solution
include ..\Utilitys.tlhy
:count "- " convert "-" 2 tolist split len nip ;
"the three truths" "th" count ?
"ababababab" "abab" count ?
" " input- Output:
3 2
Kotlin
// version 1.0.6
fun countSubstring(s: String, sub: String): Int = s.split(sub).size - 1
fun main(args: Array<String>) {
println(countSubstring("the three truths","th"))
println(countSubstring("ababababab","abab"))
println(countSubstring("",""))
}
- Output:
3 2 1
Lambdatalk
{def countSubstring
{def countSubstring.r
{lambda {:n :i :acc :s}
{if {>= :i :n}
then :acc
else {countSubstring.r :n
{+ :i 1}
{if {W.equal? {W.get :i :s} ⫖}
then {+ :acc 1}
else :acc}
:s} }}}
{lambda {:w :s}
{countSubstring.r {W.length :s} 0 0
{S.replace \s by ⫕ in
{S.replace :w by ⫖ in :s}}}}}
-> countSubstring
{countSubstring th the three truths}
-> 3
{countSubstring ab ababa}
-> 2
{countSubstring aba ababa}
-> 1
langur
writeln len(indices("th", "the three truths"))
writeln len(indices("abab", "ababababab"))- Output:
3 2
Lasso
define countSubstring(str::string, substr::string)::integer => {
local(i = 1, foundpos = -1, found = 0)
while(#i < #str->size && #foundpos != 0) => {
protect => {
handle_error => { #foundpos = 0 }
#foundpos = #str->find(#substr, -offset=#i)
}
if(#foundpos > 0) => {
#found += 1
#i = #foundpos + #substr->size
else
#i++
}
}
return #found
}
define countSubstring_bothways(str::string, substr::string)::integer => {
local(found = countSubstring(#str,#substr))
#str->reverse
local(found2 = countSubstring(#str,#substr))
#found > #found2 ? return #found | return #found2
}
countSubstring_bothways('the three truths','th')
//3
countSubstring_bothways('ababababab','abab')
//2
Liberty BASIC
print countSubstring( "the three truths", "th")
print countSubstring( "ababababab", "abab")
end
function countSubstring( a$, s$)
c =0
la =len( a$)
ls =len( s$)
for i =1 to la -ls
if mid$( a$, i, ls) =s$ then c =c +1: i =i +ls -1
next i
countSubstring =c
end functionLogtalk
Using atoms for string representation:
:- object(counting).
:- public(count/3).
count(String, SubString, Count) :-
count(String, SubString, 0, Count).
count(String, SubString, Count0, Count) :-
( sub_atom(String, Before, Length, After, SubString) ->
Count1 is Count0 + 1,
Start is Before + Length,
sub_atom(String, Start, After, 0, Rest),
count(Rest, SubString, Count1, Count)
; Count is Count0
).
:- end_object.
- Output:
| ?- counting::count('the three truths', th, N).
N = 3
yes
| ?- counting::count(ababababab, abab, N).
N = 2
yes
Lua
Solution 1:
function countSubstring(s1, s2)
return select(2, s1:gsub(s2, ""))
end
print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3 2
Solution 2:
function countSubstring(s1, s2)
local count = 0
for eachMatch in s1:gmatch(s2) do
count = count + 1
end
return count
end
print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3 2
MACRO-11
.TITLE CSUBS
.MCALL .TTYOUT,.EXIT
CSUBS:: JMP DEMO
; COUNT SUBSTRINGS R1 IN R0
COUNT: CLR R2
BR 4$
1$: MOV R0,R3
MOV R1,R4
2$: CMPB (R3)+,(R4)+
BEQ 2$
TSTB -(R4)
BNE 3$
INC R2
DEC R3
MOV R3,R0
BR 4$
3$: INC R0
4$: TSTB (R0)
BNE 1$
RTS PC
; TEST EXAMPLES
DEMO: MOV #ST1,R0
MOV #SU1,R1
JSR PC,1$
MOV #ST2,R0
MOV #SU2,R1
JSR PC,1$
.EXIT
1$: JSR PC,COUNT
ADD #60,R2
MOVB R2,3$
MOV #3$,R2
2$: MOVB (R2)+,R0
.TTYOUT
BNE 2$
RTS PC
3$: .BYTE 0,15,12,0
ST1: .ASCIZ /THE THREE TRUTHS/
SU1: .ASCIZ /TH/
ST2: .ASCIZ /ABABABABAB/
SU2: .ASCIZ /ABAB/
.END CSUBS- Output:
3 2
Maple
f:=proc(s::string,c::string,count::nonnegint) local n;
n:=StringTools:-Search(c,s);
if n>0 then 1+procname(s[n+length(c)..],c,count);
else 0; end if;
end proc:
f("the three truths","th",0);
f("ababababab","abab",0);- Output:
3
2
Mathematica / Wolfram Language
StringPosition["the three truths","th",Overlaps->False]//Length
3
StringPosition["ababababab","abab",Overlaps->False]//Length
2
MATLAB / Octave
% Count occurrences of a substring without overlap
length(findstr("ababababab","abab",0))
length(findstr("the three truths","th",0))
% Count occurrences of a substring with overlap
length(findstr("ababababab","abab",1))
- Output:
>> % Count occurrences of a substring without overlap
>> length(findstr("ababababab","abab",0))
ans = 2
>> length(findstr("the three truths","th",0))
ans = 3
>> % Count occurrences of a substring with overlap
>> length(findstr("ababababab","abab",1))
ans = 4
>>
Maxima
scount(e, s) := block(
[n: 0, k: 1],
while integerp(k: ssearch(e, s, k)) do (n: n + 1, k: k + 1),
n
)$
scount("na", "banana");
2
MiniScript
string.count = function(s)
return self.split(s).len - 1
end function
print "the three truths".count("th")
print "ababababab".count("abab")
- Output:
3 2
Mirah
import java.util.regex.Pattern
import java.util.regex.Matcher
#The "remove and count the difference" method
def count_substring(pattern:string, source:string)
(source.length() - source.replace(pattern, "").length()) / pattern.length()
end
puts count_substring("th", "the three truths") # ==> 3
puts count_substring("abab", "ababababab") # ==> 2
puts count_substring("a*b", "abaabba*bbaba*bbab") # ==> 2
# The "split and count" method
def count_substring2(pattern:string, source:string)
# the result of split() will contain one more element than the delimiter
# the "-1" second argument makes it not discard trailing empty strings
source.split(Pattern.quote(pattern), -1).length - 1
end
puts count_substring2("th", "the three truths") # ==> 3
puts count_substring2("abab", "ababababab") # ==> 2
puts count_substring2("a*b", "abaabba*bbaba*bbab") # ==> 2
# This method does a match and counts how many times it matches
def count_substring3(pattern:string, source:string)
result = 0
Matcher m = Pattern.compile(Pattern.quote(pattern)).matcher(source);
while (m.find())
result = result + 1
end
result
end
puts count_substring3("th", "the three truths") # ==> 3
puts count_substring3("abab", "ababababab") # ==> 2
puts count_substring3("a*b", "abaabba*bbaba*bbab") # ==> 2Miranda
main :: [sys_message]
main = [Stdout (show (countSubstring "the three truths" "th") ++ "\n"),
Stdout (show (countSubstring "ababababab" "abab") ++ "\n")]
countSubstring :: [*]->[*]->num
countSubstring str ss
= 0, if str = []
= 1 + countSubstring (drop len str) ss, if match
= countSubstring (tl str) ss, otherwise
where len = #ss
match = take len str = ss- Output:
3 2
Nanoquery
def countSubstring(str, subStr)
return int((len(str) - len(str.replace(subStr, ""))) / len(subStr))
endNemerle
using System.Console;
module CountSubStrings
{
CountSubStrings(this text : string, target : string) : int
{
match (target) {
|"" => 0
|_ => (text.Length - text.Replace(target, "").Length) / target.Length
}
}
Main() : void
{
def text1 = "the three truths";
def target1 = "th";
def text2 = "ababababab";
def target2 = "abab";
WriteLine($"$target1 occurs $(text1.CountSubStrings(target1)) times in $text1");
WriteLine($"$target2 occurs $(text2.CountSubStrings(target2)) times in $text2");
}
}
- Output:
th occurs 3 times in the three truths abab occurs 2 times in ababababab
NetRexx
NetRexx provides the string.countstr(needle) built-in function:
/* NetRexx */
options replace format comments java crossref symbols nobinary
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method countSubstring(inStr, findStr) public static
return inStr.countstr(findStr)
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method main(args = String[]) public static
strings = ''
find = 'FIND'
ix = 0
ix = ix + 1; strings[0] = ix; find[0] = ix; strings[ix] = 'the three truths'; strings[ix, find] = 'th'
ix = ix + 1; strings[0] = ix; find[0] = ix; strings[ix] = 'ababababab'; strings[ix, find] = 'abab'
loop ix = 1 to strings[0]
str = strings[ix]
fnd = strings[ix, find]
say 'there are' countSubstring(str, fnd) 'occurences of "'fnd'" in "'str'"'
end ix
return
- Output:
there are 3 occurences of "th" in "the three truths" there are 2 occurences of "abab" in "ababababab"
NewLISP
; file: stringcount.lsp
; url: http://rosettacode.org/wiki/Count_occurrences_of_a_substring
; author: oofoe 2012-01-29
; Obvious (and non-destructive...)
; Note that NewLISP performs an /implicit/ slice on a string or list
; with this form "(start# end# stringorlist)". If the end# is omitted,
; the slice will go to the end of the string. This is handy here to
; keep removing the front part of the string as it gets matched.
(define (scount needle haystack)
(let ((h (copy haystack)) ; Copy of haystack string.
(i 0) ; Cursor.
(c 0)) ; Count of occurences.
(while (setq i (find needle h))
(inc c)
(setq h ((+ i (length needle)) h)))
c)) ; Return count.
; Tricky -- Uses functionality from replace function to find all
; non-overlapping occurrences, replace them, and return the count of
; items replaced in system variable $0.
(define (rcount needle haystack)
(replace needle haystack "X") $0)
; Test
(define (test f needle haystack)
(println "Found " (f needle haystack)
" occurences of '" needle "' in '" haystack "'."))
(dolist (f (list scount rcount))
(test f "glart" "hinkerpop")
(test f "abab" "ababababab")
(test f "th" "the three truths")
(println)
)
(exit)
- Output:
Found 0 occurences of 'glart' in 'hinkerpop'. Found 2 occurences of 'abab' in 'ababababab'. Found 3 occurences of 'th' in 'the three truths'. Found 0 occurences of 'glart' in 'hinkerpop'. Found 2 occurences of 'abab' in 'ababababab'. Found 3 occurences of 'th' in 'the three truths'.
Solution 2.
(begin (find-all "th" "the three truths") $count)
3
(begin (find-all "abab" "ababababab") $count)
2
(begin (find-all "ab" "ababababab") $count)
5
(begin (find-all "11" "11111111") $count)
4
Nim
import strutils
proc count(s, sub: string): int =
var i = 0
while true:
i = s.find(sub, i)
if i < 0:
break
i += sub.len # i += 1 for overlapping substrings
inc result
echo count("the three truths","th")
echo count("ababababab","abab")
- Output:
3 2
Nu
def 'str count' [s: string] {
split row $s | length | $in - 1
}
{th: 'the three truths' abab: 'ababababab'} | items {|k v| $v | str count $k }- Output:
╭───┬───╮ │ 0 │ 3 │ │ 1 │ 2 │ ╰───┴───╯
Objective-C
The "split and count" method:
@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end
@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
return [[self componentsSeparatedByString:subStr] count] - 1;
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
}
return 0;
}
- Output:
3 2 2
The "remove and count the difference" method:
@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end
@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
return ([self length] - [[self stringByReplacingOccurrencesOfString:subStr withString:@""] length]) / [subStr length];
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
}
return 0;
}
- Output:
3 2 2
Manual looping:
@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end
@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
NSUInteger count = 0;
for (NSRange range = [self rangeOfString:subStr]; range.location != NSNotFound;
range.location += range.length,
range = [self rangeOfString:subStr options:0
range:NSMakeRange(range.location, [self length] - range.location)])
count++;
return count;
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
}
return 0;
}
- Output:
3 2 2
OCaml
let count_substring str sub =
let sub_len = String.length sub in
let len_diff = (String.length str) - sub_len
and reg = Str.regexp_string sub in
let rec aux i n =
if i > len_diff then n else
try
let pos = Str.search_forward reg str i in
aux (pos + sub_len) (succ n)
with Not_found -> n
in
aux 0 0
let () =
Printf.printf "count 1: %d\n" (count_substring "the three truth" "th");
Printf.printf "count 2: %d\n" (count_substring "ababababab" "abab");
;;
Oforth
: countSubString(s, sub)
0 1 while(sub swap s indexOfAllFrom dup notNull) [ sub size + 1 under+ ]
drop ;- Output:
countSubString("the three truths", "th") println
3
countSubString("ababababab", "abab") println
2
ooRexx
bag="the three truths"
x="th"
say left(bag,30) left(x,15) 'found' bag~countstr(x)
bag="ababababab"
x="abab"
say left(bag,30) left(x,15) 'found' bag~countstr(x)
-- can be done caselessly too
bag="abABAbaBab"
x="abab"
say left(bag,30) left(x,15) 'found' bag~caselesscountstr(x)
- Output:
the three truths th found 3 ababababab abab found 2 abABAbaBab abab found 2
PARI/GP
subvec(v,u)={
my(i=1,s);
while(i+#u<=#v,
for(j=1,#u,
if(v[i+j-1]!=u[j], i++; next(2))
);
s++;
i+=#u
);
s
};
substr(s1,s2)=subvec(Vec(s1),Vec(s2));
substr("the three truths","th")
substr("ababababab","abab")- Output:
%1 = 3 %2 = 2
Pascal
See Delphi
PascalABC.NET
function CountSubstrings(s,subs: string): integer := s.IndicesOf(subs).Count;
begin
Print(CountSubstrings('the three truths','th'));
Print(CountSubstrings('ababababab','abab'));
end.
- Output:
3 2
Perl
sub countSubstring {
my $str = shift;
my $sub = quotemeta(shift);
my $count = () = $str =~ /$sub/g;
return $count;
# or return scalar( () = $str =~ /$sub/g );
}
print countSubstring("the three truths","th"), "\n"; # prints "3"
print countSubstring("ababababab","abab"), "\n"; # prints "2"
Phix
sequence tests = {{"the three truths","th"}, {"ababababab","abab"}, {"ababababab","aba"}, {"ababababab","ab"}, {"ababababab","a"}, {"ababababab",""}} integer start, count string test, substring for i=1 to length(tests) do start = 1 count = 0 {test, substring} = tests[i] while 1 do start = match(substring,test,start) if start=0 then exit end if start += length(substring) count += 1 end while printf(1,"The string \"%s\" occurs as a non-overlapping substring %d times in \"%s\"\n",{substring,count,test}) end for
- Output:
The string "th" occurs as a non-overlapping substring 3 times in "the three truths" The string "abab" occurs as a non-overlapping substring 2 times in "ababababab" The string "aba" occurs as a non-overlapping substring 2 times in "ababababab" The string "ab" occurs as a non-overlapping substring 5 times in "ababababab" The string "a" occurs as a non-overlapping substring 5 times in "ababababab" The string "" occurs as a non-overlapping substring 0 times in "ababababab"
PHP
<?php
echo substr_count("the three truths", "th"), PHP_EOL; // prints "3"
echo substr_count("ababababab", "abab"), PHP_EOL; // prints "2"
Picat
Picat has a predicate for searching for substrings (find/4) but on backtracking (e.g. in a findall/2 wrapper) it yields all overlapping substrings which is not correct in this task.
So we have to roll our own. Here are two versions.
Recursion
count_substrings_rec(S, SB) = C =>
count_rec(S,SB,0,C).
count_rec([],_SB,Count,Count).
count_rec(SBRest,SB,Count0,Count) :-
SBRest = SB ++ Rest, % "split" into substring and the rest of the string
count_rec(Rest,SB,Count0+1,Count).
count_rec([T|Rest],SB,Count0,Count) :-
T != SB, % this character is not a substring
count_rec(Rest,SB,Count0,Count).Iterative
Iterative version using find/4 (wrap with once/1 to avoid backtracking).
count_substrings_find(S, SB) = C =>
SLen = S.len,
Count = 0,
From = 1,
while (From <= SLen)
(
once(find(slice(S,From),SB,_From2,To)) ->
Count := Count + 1,
From := From + To
;
From := From + 1
)
end,
C = Count.The time differences between these two versions are quite large which is shown in a benchmark of searching the substring "ab" in a string of 100 000 random characters from the set of "abcde":
- count_substrings_rec/2: 0.009s
- count_substrings_find: 0.501s
PicoLisp
(de countSubstring (Str Sub)
(let (Cnt 0 H (chop Sub))
(for (S (chop Str) S (cdr S))
(when (head H S)
(inc 'Cnt)
(setq S (map prog2 H S)) ) )
Cnt ) )Test:
: (countSubstring "the three truths" "th") -> 3 : (countSubstring "ababababab" "abab") -> 2
Pike
write("%d %d\n",
String.count("the three truths", "th"),
String.count("ababababab", "abab"));
- Output:
3 2
PL/I
cnt: procedure options (main);
declare (i, tally) fixed binary;
declare (text, key) character (100) varying;
get edit (text) (L); put skip data (text);
get edit (key) (L); put skip data (key);
tally = 0; i = 1;
do until (i = 0);
i = index(text, key, i);
if i > 0 then do; tally = tally + 1; i = i + length(key); end;
end;
put skip list (tally);
end cnt;Output for the two specified strings is as expected.
- Output:
for the following data
TEXT='AAAAAAAAAAAAAAA';
KEY='AA';
7
PL/M
100H:
/* CP/M CALLS */
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0, 0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9, S); END PRINT;
/* PRINT A NUMBER */
PRINT$NUMBER: PROCEDURE (N);
DECLARE S (8) BYTE INITIAL ('.....',13,10,'$');
DECLARE (P, N) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1;
C = N MOD 10 + '0';
N = N / 10;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT$NUMBER;
/* COUNT OCCURRENCES OF SUBSTRING IN STRING */
COUNT$SUBSTRING: PROCEDURE (STR, MATCH) ADDRESS;
DECLARE (STR, MATCH) ADDRESS, C BASED STR BYTE;
DECLARE (SP, MP) ADDRESS, (SC BASED SP, MC BASED MP) BYTE;
DECLARE COUNT ADDRESS;
COUNT = 0;
DO WHILE C <> '$';
SP = STR;
MP = MATCH;
DO WHILE SC = MC;
SP = SP + 1;
MP = MP + 1;
END;
IF MC = '$' THEN DO;
STR = SP;
COUNT = COUNT + 1;
END;
ELSE DO;
STR = STR + 1;
END;
END;
RETURN COUNT;
END COUNT$SUBSTRING;
CALL PRINT$NUMBER(COUNT$SUBSTRING(.'THE THREE TRUTHS$', .'TH$')); /* PRINTS 3 */
CALL PRINT$NUMBER(COUNT$SUBSTRING(.'ABABABABAB$', .'ABAB$')); /* PRINTS 2 */
CALL PRINT$NUMBER(COUNT$SUBSTRING(.'CAT$', .'DOG$')); /* PRINTS 0 */
CALL EXIT;
EOF- Output:
3 2 0
PowerBASIC
Windows versions of PowerBASIC (at least since PB/Win 7, and possibly earlier) provide the TALLY function, which does exactly what this task requires (count non-overlapping substrings).
PB/DOS can use the example under BASIC, above.
Note that while this example is marked as working with PB/Win, the PRINT statement would need to be replaced with MSGBOX, or output to a file. (PB/Win does not support console output.)
FUNCTION PBMAIN () AS LONG
PRINT "the three truths, th:", TALLY("the three truths", "th")
PRINT "ababababab, abab:", TALLY("ababababab", "abab")
END FUNCTION- Output:
the three truths, th: 3 ababababab, abab: 2
PowerShell
[regex]::Matches("the three truths", "th").count
Output:
3
[regex]::Matches("ababababab","abab").count
Output:
2
Prolog
Using SWI-Prolog's string facilities (this solution is very similar to the Logtalk solution that uses sub_atom/5):
count_substring(String, Sub, Total) :-
count_substring(String, Sub, 0, Total).
count_substring(String, Sub, Count, Total) :-
( substring_rest(String, Sub, Rest)
->
succ(Count, NextCount),
count_substring(Rest, Sub, NextCount, Total)
;
Total = Count
).
substring_rest(String, Sub, Rest) :-
sub_string(String, Before, Length, Remain, Sub),
DropN is Before + Length,
sub_string(String, DropN, Remain, 0, Rest).
Usage:
?- count_substring("the three truths","th",X).
X = 3.
?- count_substring("ababababab","abab",X).
X = 2.
version using DCG
:- system:set_prolog_flag(double_quotes,chars) .
occurrences(TARGETz0,SUBSTRINGz0,COUNT)
:-
prolog:phrase(occurrences(SUBSTRINGz0,0,COUNT),TARGETz0)
.
occurrences("",_,_)
-->
! ,
{ false }
.
occurrences(SUBSTRINGz0,COUNT0,COUNT)
-->
SUBSTRINGz0 ,
! ,
{ COUNT1 is COUNT0 + 1 } ,
occurrences(SUBSTRINGz0,COUNT1,COUNT)
.
occurrences(SUBSTRINGz0,COUNT0,COUNT)
-->
[_] ,
! ,
occurrences(SUBSTRINGz0,COUNT0,COUNT)
.
occurrences(_SUBSTRINGz0_,COUNT,COUNT)
-->
!
.
- Output:
/*
?- occurrences("the three truths","th",COUNT) .
COUNT = 3.
?- occurrences("the three truths","",COUNT) .
false .
?- occurrences("ababababab","abab",COUNT) .
COUNT = 2.
?- occurrences("ababababab","",COUNT) .
false .
?-
*/
PureBasic
a = CountString("the three truths","th")
b = CountString("ababababab","abab")
; a = 3
; b = 2
Python
>>> "the three truths".count("th")
3
>>> "ababababab".count("abab")
2
Quackery
Quackery does not come equipped with a "find substring m within string n" function, but one is defined in The Book of Quackery, as a demonstration of creating a finite state machine in Quackery. It is reproduced here with permission.
[ [] 95 times
[ i^ space +
join ] ] constant is alphabet ( --> $ )
[ [ 2dup != while
-1 split drop
swap 1 split
unrot drop again ]
drop size ] is overlap ( [ [ --> n )
[ temp put [] swap
alphabet witheach
[ over -1 poke
over overlap
dup temp share
= if negate
swap dip join ]
drop temp release ] is eachend ( [ n --> [ )
[ [] swap
dup temp put
size times
[ temp share
i 1+ split drop
temp share size eachend
nested swap join ]
temp release ] is buildfsm ( $ --> [ )
[ dup [] = iff -1
else
[ behead
dup carriage = if
[ drop space ]
space - ]
swap ] is nextcharn ( $ --> n $ )
[ swap dup size
swap temp put
swap 0
[ over swap peek
temp take
nextcharn
temp put
dup 0 < iff
[ 2drop 0 ] done
peek
dup 0 < until ]
nip
temp take size - + ] is usefsm ( $ [ --> n )
[ over size 0 = iff
[ 2drop 0 ]
else
[ swap buildfsm
usefsm ] ] is find$ ( $ $ --> n )find$ builds a finite state machine to search for m, (an O(m³) operation), then uses it to search in n with O(n). Rather than use find$, and repeatedly build the same fsm, we will define a word findall$ which returns a nest (i.e. list) of positions of m within n. (It actually returns the positions of the end of the substring, relative to (for the first instance) the start of the string, or (for subsequent instances) the end of the previous instance of the substring.)
[ over size 0 = iff
[ 2drop [] ] done
[] unrot
swap buildfsm
[ 2dup usefsm
rot 2dup found while
dip [ over size + ]
dip
[ rot over join
unrot ]
swap split
nip swap again ]
2drop drop ] is findall$ ( $ $ --> [ )- Output:
Testing findall$ in the Quackery shell.
O> $ "th" $ "the three truths" tuck findall$ ... witheach [ split swap echo$ cr ] ... echo$ ... th e th ree truth s Stack empty.
Finally we can use findall$ to fulfil the task requirements.
[ swap findall$ size ] is occurences ( $ $ --> n )
$ "the three truths" $ "th" occurences echo cr
$ "ababababab" $ "abab" occurences echo cr- Output:
3 2
R
The fixed parameter (and, in stringr, the function of the same name) is used to specify a search for a fixed string. Otherwise, the search pattern is interpreted as a POSIX regular expression. PCRE is also an option: use the perl parameter or function.
count = function(haystack, needle)
{v = attr(gregexpr(needle, haystack, fixed = T)[[1]], "match.length")
if (identical(v, -1L)) 0 else length(v)}
print(count("hello", "l"))
library(stringr)
print(str_count("hello", fixed("l")))
Racket
(define count-substring
(compose length regexp-match*))
> (count-substring "th" "the three truths")
3
> (count-substring "abab" "ababababab")
2
Raku
(formerly Perl 6)
sub count-substring($big, $little) { +$big.comb: / :r $little / }
say count-substring("the three truths", "th"); # 3
say count-substring("ababababab", "abab"); # 2
say count-substring(123123123, 12); # 3
The :r adverb makes the regex "ratchet forward" and skip any overlapping matches. .comb - when given a Regex as an argument - returns instances of that substring. Also, prefix + forces numeric context in Raku (it's a no-op in Perl 5). For the built in listy types that is the same as calling .elems method. One other style point: we now tend to prefer hyphenated names over camelCase.
Red
Red []
count-occurrences: function [string substring] [
length? parse string [collect [some [keep substring to substring]]]
]
test-case-1: "the three truths"
test-case-2: "ababababab"
print [test-case-1 "-" count-occurrences test-case-1 "th"]
print [test-case-2 "-" count-occurrences test-case-2 "abab"]
- Output:
the three truths - 3 ababababab - 2
Refal
$ENTRY Go {
= <Prout <Count ('th') 'the three truths'>>
<Prout <Count ('abab') 'abababab'>>;
};
Count {
(e.item) e.item e.rest = <+ 1 <Count (e.item) e.rest>>;
(e.item) s.x e.rest = <Count (e.item) e.rest>;
(e.item) = 0;
};- Output:
3 2
REXX
Some older REXXes don't have the built-in function countstr, so one is included within the REXX program as a function.
The countstr subroutine (below) mimics the BIF in newer REXXes (except for error checking).
Either of the first two strings may be null.
The third argument is optional and is the start position to start counting (the default is 1, meaning the first character).
If specified, it must be a positive integer (and it may exceed the length of the 1st string).
The third argument was added here to be compatible with the newer REXXes BIF.
No checks are made (in the countstr subroutine) for:
- missing arguments
- too many arguments
- if start is a positive integer (when specified)
/*REXX program counts the occurrences of a (non─overlapping) substring in a string. */
w=. /*max. width so far.*/
bag= 'the three truths' ; x= "th" ; call showResult
bag= 'ababababab' ; x= "abab" ; call showResult
bag= 'aaaabacad' ; x= "aa" ; call showResult
bag= 'abaabba*bbaba*bbab' ; x= "a*b" ; call showResult
bag= 'abaabba*bbaba*bbab' ; x= " " ; call showResult
bag= ; x= "a" ; call showResult
bag= ; x= ; call showResult
bag= 'catapultcatalog' ; x= "cat" ; call showResult
bag= 'aaaaaaaaaaaaaa' ; x= "aa" ; call showResult
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
countstr: procedure; parse arg haystack,needle,start; if start=='' then start=1
width=length(needle)
do $=0 until p==0; p=pos(needle,haystack,start)
start=width + p /*prevent overlaps.*/
end /*$*/
return $ /*return the count.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
showResult: if w==. then do; w=30 /*W: largest haystack width.*/
say center('haystack',w) center('needle',w%2) center('count',5)
say left('', w, "═") left('', w%2, "═") left('', 5, "═")
end
if bag=='' then bag= " (null)" /*handle displaying of nulls.*/
if x=='' then x= " (null)" /* " " " " */
say left(bag, w) left(x, w%2) center(countstr(bag, x), 5)
returnoutput when using the default (internal) inputs:
haystack needle count ══════════════════════════════ ═══════════════ ═════ the three truths th 3 ababababab abab 2 aaaabacad aa 2 abaabba*bbaba*bbab a*b 2 abaabba*bbaba*bbab 0 (null) a 0 (null) (null) 1 catapultcatalog cat 2 aaaaaaaaaaaaaa aa 7
Ring
aString = "Ring Welcome Ring to the Ring Ring Programming Ring Language Ring"
bString = "Ring"
see count(aString,bString)
func count cString,dString
sum = 0
while substr(cString,dString) > 0
sum++
cString = substr(cString,substr(cString,dString)+len(string(sum)))
end
return sumOutput:
6
RPL
≪ DUP SIZE 1 - → str substr subsize
≪ 0
1 str SIZE subsize + FOR j
str j DUP subsize + SUB
IF substr == THEN
1 +
j subsize + 'j' STO
END
NEXT
≫ ≫ 'CNTSUB' STO
"the three truths" CNTSUB "ababababab" CNTSUB
- Output:
2: 3 1: 2
Ruby
def countSubstrings str, subStr
str.scan(subStr).length
end
p countSubstrings "the three truths", "th" #=> 3
p countSubstrings "ababababab", "abab" #=> 2String#scan returns an array of substrings, and Array#length (or Array#size) counts them.
Run BASIC
print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")
FUNCTION countSubstring(s$,find$)
WHILE instr(s$,find$,i) <> 0
countSubstring = countSubstring + 1
i = instr(s$,find$,i) + len(find$)
WEND
END FUNCTION- Output:
3 2
Rust
fn main() {
println!("{}","the three truths".matches("th").count());
println!("{}","ababababab".matches("abab").count());
}- Output:
3 2
Scala
Using Recursion
import scala.annotation.tailrec
def countSubstring(str1:String, str2:String):Int={
@tailrec def count(pos:Int, c:Int):Int={
val idx=str1 indexOf(str2, pos)
if(idx == -1) c else count(idx+str2.size, c+1)
}
count(0,0)
}Using Sliding
def countSubstring(str: String, sub: String): Int =
str.sliding(sub.length).count(_ == sub)
Using Regular Expressions
def countSubstring( str:String, substr:String ) = substr.r.findAllMatchIn(str).length
println(countSubstring("ababababab", "abab"))
println(countSubstring("the three truths", "th"))- Output:
2 3
Scheme
gosh> (use gauche.lazy)
#<undef>
gosh> (length (lrxmatch "th" "the three truths"))
3
gosh> (length (lrxmatch "abab" "ababababab"))
2Seed7
$ include "seed7_05.s7i";
const func integer: countSubstring (in string: stri, in string: searched) is func
result
var integer: count is 0;
local
var integer: offset is 0;
begin
offset := pos(stri, searched);
while offset <> 0 do
incr(count);
offset := pos(stri, searched, offset + length(searched));
end while;
end func;
const proc: main is func
begin
writeln(countSubstring("the three truths", "th"));
writeln(countSubstring("ababababab", "abab"));
end func;- Output:
3 2
SETL
program count_overlapping_substrings;
tests := [["the three truths", "th"], ["ababababab", "abab"]];
loop for [s, subs] in tests do
print("'" + subs + "' in '" + s + "': "
+ str countSubs(s, subs));
end loop;
proc countSubs(s, subs);
count := 0;
loop while s(subs) /= om do
s(subs) := "";
count +:= 1;
end loop;
return count;
end proc;
end program;- Output:
'th' in 'the three truths': 3 'abab' in 'ababababab': 2
SenseTalk
Simply stated:
put the number of occurrences of "th" in "the three truths" --> 3
put the number of occurrences of "abab" in "ababababab" -- > 2User-created function:
put countSubstring("aaaaa","a") // 5
put countSubstring("abababa","aba") // 2
function countSubstring mainString, subString
return number of occurrences of subString in mainString
end countSubstringSidef
Built-in:
say "the three truths".count("th");
say "ababababab".count("abab");User-created function:
func countSubstring(s, ss) {
var re = Regex.new(ss.escape, 'g'); # 'g' for global
var counter = 0;
while (s =~ re) { ++counter };
return counter;
}
say countSubstring("the three truths","th");
say countSubstring("ababababab","abab");- Output:
3 2
Simula
BEGIN
INTEGER PROCEDURE COUNTSUBSTRING(T,TSUB); TEXT T,TSUB;
BEGIN
INTEGER N,I;
I := 1;
WHILE I+TSUB.LENGTH-1 <= T.LENGTH DO
IF T.SUB(I,TSUB.LENGTH) = TSUB THEN
BEGIN
N := N+1;
I := I+MAX(TSUB.LENGTH,1);
END ELSE I := I+1;
COUNTSUBSTRING:= N;
END COUNTSUBSTRING;
OUTINT(COUNTSUBSTRING("THE THREE TRUTHS", "TH"),0);
OUTIMAGE;
OUTINT(COUNTSUBSTRING("ABABABABAB", "ABAB"),0);
OUTIMAGE;
END.- Output:
3 2
Smalltalk
Transcript showCR:('the three truths' occurrencesOfString:'th').
Transcript showCR:('ababababab' occurrencesOfString:'abab')- Output:
3 2
SNOBOL4
DEFINE("countSubstring(t,s)")
OUTPUT = countSubstring("the three truths","th")
OUTPUT = countSubstring("ababababab","abab")
:(END)
countSubstring t ARB s = :F(RETURN)
countSubstring = countSubstring + 1 :(countSubstring)
END
3
2Standard ML
fun count_substrings (str, sub) =
let
fun aux (str', count) =
let
val suff = #2 (Substring.position sub str')
in
if Substring.isEmpty suff then
count
else
aux (Substring.triml (size sub) suff, count + 1)
end
in
aux (Substring.full str, 0)
end;
print (Int.toString (count_substrings ("the three truths", "th")) ^ "\n");
print (Int.toString (count_substrings ("ababababab", "abab")) ^ "\n");
print (Int.toString (count_substrings ("abaabba*bbaba*bbab", "a*b")) ^ "\n");Stata
function strcount(s, x) {
n = 0
k = 1-(i=strlen(x))
do {
if (k = ustrpos(s, x, k+i)) n++
} while(k)
return(n)
}
strcount("peter piper picked a peck of pickled peppers", "pe")
5
strcount("ababababab","abab")
2Swift
import Foundation
func countSubstring(str: String, substring: String) -> Int {
return str.components(separatedBy: substring).count - 1
}
print(countSubstring(str: "the three truths", substring: "th"))
print(countSubstring(str: "ababababab", substring: "abab"))- Output:
3 2
Tcl
The regular expression engine is ideal for this task, especially as the ***= prefix makes it interpret the rest of the argument as a literal string to match:
proc countSubstrings {haystack needle} {
regexp -all ***=$needle $haystack
}
puts [countSubstrings "the three truths" "th"]
puts [countSubstrings "ababababab" "abab"]
puts [countSubstrings "abaabba*bbaba*bbab" "a*b"]- Output:
3 2 2
Transd
#lang transd
MainModule: {
countSubstring: (λ s String() sub String()
(with n 0 pl 0
(while (> (= pl (find s sub pl)) -1)
(+= pl (size sub)) (+= n 1))
(lout n))
),
_start: (λ
(countSubstring "the three truths" "th")
(countSubstring "ababababab" "abab")
)
}- Output:
3 2
TUSCRIPT
$$ MODE TUSCRIPT, {}
occurences=COUNT ("the three truths", ":th:")
occurences=COUNT ("ababababab", ":abab:")
occurences=COUNT ("abaabba*bbaba*bbab",":a\*b:")- Output:
3 2 2
TXR
@(next :args)
@(do (defun count-occurrences (haystack needle)
(for* ((occurrences 0)
(old-pos 0)
(new-pos (search-str haystack needle old-pos nil)))
(new-pos occurrences)
((inc occurrences)
(set old-pos (+ new-pos (length needle)))
(set new-pos (search-str haystack needle old-pos nil))))))
@ndl
@hay
@(output)
@(count-occurrences hay ndl) occurrences(s) of @ndl inside @hay
@(end)$ ./txr count-occurrences.txr "baba" "babababa" 2 occurence(s) of baba inside babababa $ ./txr count-occurrences.txr "cat" "catapultcatalog" 2 occurence(s) of cat inside catapultcatalog
Uiua
F ← /↥⦷:
F "the three truths" "th"
F "ababababab" "abab"- Output:
3 2
UNIX Shell
#!/bin/bash
function countString(){
input=$1
cnt=0
until [ "${input/$2/}" == "$input" ]; do
input=${input/$2/}
let cnt+=1
done
echo $cnt
}
countString "the three truths" "th"
countString "ababababab" "abab"- Output:
3 2
VBA
Function CountStringInString(stLookIn As String, stLookFor As String)
CountStringInString = UBound(Split(stLookIn, stLookFor))
End FunctionVBScript
user created function
Function CountSubstring(str,substr)
CountSubstring = 0
For i = 1 To Len(str)
If Len(str) >= Len(substr) Then
If InStr(i,str,substr) Then
CountSubstring = CountSubstring + 1
i = InStr(i,str,substr) + Len(substr) - 1
End If
Else
Exit For
End If
Next
End Function
WScript.StdOut.Write CountSubstring("the three truths","th") & vbCrLf
WScript.StdOut.Write CountSubstring("ababababab","abab") & vbCrLfUsing built-in Regexp
Run it with CScript.
function CountSubstring(str,substr)
with new regexp
.pattern=substr
.global=true
set m=.execute(str)
end with
CountSubstring =m.count
end function
WScript.StdOut.Writeline CountSubstring("the three truths","th")
WScript.StdOut.Writeline CountSubstring("ababababab","abab")- Output:
3 2
Visual Basic .NET
Module Count_Occurrences_of_a_Substring
Sub Main()
Console.WriteLine(CountSubstring("the three truths", "th"))
Console.WriteLine(CountSubstring("ababababab", "abab"))
Console.WriteLine(CountSubstring("abaabba*bbaba*bbab", "a*b"))
Console.WriteLine(CountSubstring("abc", ""))
End Sub
Function CountSubstring(str As String, substr As String) As Integer
Dim count As Integer = 0
If (Len(str) > 0) And (Len(substr) > 0) Then
Dim p As Integer = InStr(str, substr)
Do While p <> 0
p = InStr(p + Len(substr), str, substr)
count += 1
Loop
End If
Return count
End Function
End Module- Output:
3 2 2 0
V (Vlang)
fn main(){
println('the three truths'.count('th'))
println('ababababab'.count('abab'))
}- Output:
3 2
Wortel
@let {
c &[s t] #!s.match &(t)g
[[
!!c "the three truths" "th"
!!c "ababababab" "abab"
]]
}Returns:
[3 2]
Wren
import "./pattern" for Pattern
import "./fmt" for Fmt
var countSubstring = Fn.new { |str, sub|
var p = Pattern.new(sub)
return p.findAll(str).count
}
var tests = [
["the three truths", "th"],
["ababababab", "abab"],
["abaabba*bbaba*bbab", "a*b"],
["aaaaaaaaaaaaaa", "aa"],
["aaaaaaaaaaaaaa", "b"],
]
for (test in tests) {
var count = countSubstring.call(test[0], test[1])
Fmt.print("$6s occurs $d times in $q.", Fmt.q(test[1]), count, test[0])
}- Output:
"th" occurs 3 times in "the three truths". "abab" occurs 2 times in "ababababab". "a*b" occurs 2 times in "abaabba*bbaba*bbab". "aa" occurs 7 times in "aaaaaaaaaaaaaa". "b" occurs 0 times in "aaaaaaaaaaaaaa".
Alternatively, using a library method (output same as before):
import "./str" for Str
import "./fmt" for Fmt
var tests = [
["the three truths", "th"],
["ababababab", "abab"],
["abaabba*bbaba*bbab", "a*b"],
["aaaaaaaaaaaaaa", "aa"],
["aaaaaaaaaaaaaa", "b"],
]
for (test in tests) {
var count = Str.occurs(test[0], test[1])
Fmt.print("$6s occurs $d times in $q.", Fmt.q(test[1]), count, test[0])
}XPL0
include c:\cxpl\codes; \intrinsic 'code' declarations
string 0; \use zero-terminated strings, instead of MSb terminated
func StrNCmp(A, B, N); \Compare string A to string B up to N bytes long
\This returns:
\ >0 if A > B
\ =0 if A = B
\ <0 if A < B
char A, B; \strings to be compared
int N; \number of bytes to compare
int I;
[for I:= 0 to N-1 do
if A(I) # B(I) then
return A(I) - B(I);
return 0; \they're equal
]; \StrNCmp
func StrLen(Str); \Return the number of characters in an ASCIIZ string
char Str;
int I;
for I:= 0 to -1>>1-1 do
if Str(I) = 0 then return I;
func SubStr(A, B); \Count number of times string B occurs in A
char A, B;
int LA, LB, C, I;
[LA:= StrLen(A); LB:= StrLen(B);
C:= 0; I:= 0;
while I < LA do
if StrNCmp(B, A+I, LB) = 0 then [C:= C+1; I:= I+LB]
else I:= I+1;
return C;
];
[IntOut(0, SubStr("the three truths", "th")); CrLf(0);
IntOut(0, SubStr("ababababab", "abab")); CrLf(0);
]- Output:
3 2
>> % Count occurrences of a substring without overlap
>> length(findstr("ababababab","abab",0))
ans = 2
>> length(findstr("the three truths","th",0))
ans = 3
>> % Count occurrences of a substring with overlap
>> length(findstr("ababababab","abab",1))
ans = 4
>>
zkl
Two solutions:
fcn countSubstring(s,p){ pn:=p.len(); cnt:=n:=0;
while(Void!=(n:=s.find(p,n))){cnt+=1; n+=pn}
cnt
}fcn countSubstring(s,p){ (pl:=p.len()) and (s.len()-(s-p).len())/pl }- Output:
zkl: println(countSubstring("the three truths","th"))
3
zkl: println(countSubstring("ababababab","abab"))
2
zkl: println(countSubstring("ababababab","v"))
0
ZX Spectrum Basic
10 LET t$="ABABABABAB": LET p$="ABAB": GO SUB 1000
20 LET t$="THE THREE TRUTHS": LET p$="TH": GO SUB 1000
30 STOP
1000 PRINT t$: LET c=0
1010 LET lp=LEN p$
1020 FOR i=1 TO LEN t$-lp+1
1030 IF (t$(i TO i+lp-1)=p$) THEN LET c=c+1: LET i=i+lp-1
1040 NEXT i
1050 PRINT p$;"=";c''
1060 RETURNCategories:
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