Count occurrences of a substring: Difference between revisions

It should return an integer count.

3

// do not count substrings that overlap with previously-counted substrings:

print countSubstring("ababababab","abab")

The matching should yield the highest number of non-overlapping matches.

=={{header|11l}}==

{{out}}

=={{header|360 Assembly}}==

The program uses two ASSIST macros (XDECO,XPRNT) to keep the code as short as possible.

COUNTSTR CSECT

USING COUNTSTR,R13 base register

* ---- -------------------------------------------------------

YREGS

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<pre>

<code>DE</code>.

jmp demo

;;; Count non-overlapping substrings (BC) in string (HL)

sub2: db 'abab',0 ; result should be 2

str3: db 'cat',0

{{out}}

=={{header|8086 Assembly}}==

org 100h

section .text

.sub2: db 'abab',0 ; result should be 2

.str3: db 'cat',0

{{out}}

=={{header|Action!}}==

BYTE i,j,res,found

Test("11111111","11")

Test("abcdefg","123")

{{out}}

[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Count_occurrences_of_a_substring.png Screenshot from Atari 8-bit computer]

=={{header|Ada}}==

procedure Substrings is

Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source => "ababababab",

Pattern => "abab"));

{{out}}

{{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d] - due to extensive use of '''format'''[ted] ''transput''.}}

Algol68 has no build in function to do this task, hence the next to create a ''count string in string'' routine.

PROC count string in string = (STRING needle, haystack)INT: (

count string in string("a*b", "abaabba*bbaba*bbab"), # expect 2 #

$l$

<pre>

=={{header|Apex}}==

Apex example for 'Count occurrences of a substring'.

String substr = 'ABC';

String str = 'ABCZZZABCYABCABCXXABC';

}

System.debug('Count String : '+count);

<pre>

{{works with|Dyalog APL}}

{{out}}

Here we use a generic ''evalOSA(language, code)'' function to apply a JavaScript for Automation regex to a pair of AppleScript strings, using OSAKit.

on run

return oError's NSLocalizedDescription as text

{{out}}

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.

set astid to AppleScript's text item delimiters

set AppleScript's text item delimiters to theSubstring

end countSubstring

{{Out}}

=={{header|Arturo}}==

loop [["the three truths" "th"] ["ababababab" "abab"]] 'pair ->

~"occurrences of '|last pair|' in '|first pair|':"

countOccurrences first pair last pair

{{out}}

AutoHotkey has a rather unconventional method which outperforms this.

StringReplace sets the number of replaced strings to ErrorLevel.

MsgBox % countSubstring("ababababab","abab") ; 2

StringReplace, junk, fullstring, %substring%, , UseErrorLevel

return errorlevel

=={{header|AWK}}==

# countsubstring(string, pattern)

# Returns number of occurrences of pattern in string

print countsubstring_regex("[do&d~run?d!run&>run&]", "run[&]")

print countsubstring("the three truths","th")

{{out}}

<pre>$ awk -f countsubstring.awk

=={{header|BaCon}}==

LOCAL x

WHILE TALLY(text$, part$)

PRINT "the three truths - th: ", Uniq_Tally("the three truths", "th")

{{out}}

<pre>

In FreeBASIC, this needs to be compiled with <code>-lang qb</code> or <code>-lang fblite</code>.

PRINT "the three truths, th:", countSubstring&("the three truths", "th")

LOOP

countSubstring = c

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==={{header|Applesoft BASIC}}===

20 S$ = "THE THREE TRUTHS"

30 GOSUB 100"COUNT SUBSTRING

170 IF F$ = MID$(S$, I, F) THEN R = R + 1 : I = I + F - 1

180 NEXT I

==={{header|IS-BASIC}}===

110 INPUT PROMPT "Substring: ":SUB$

120 PRINT COUNT(LCASE$(TXT$),LCASE$(SUB$))

180 LOOP UNTIL PO=0

190 LET COUNT=N

==={{header|Sinclair ZX81 BASIC}}===

Works with 1k of RAM.

20 LET U$="TH"

30 GOSUB 100

150 LET N=N+1

160 LET I=I+LEN U$

==={{header|True BASIC}}===

{{trans|QBasic}}

LET c = 0

LET s = 1-LEN(what$)

PRINT "the three truths, th:", countSubstring("the three truths", "th")

PRINT "ababababab, abab:", countSubstring("ababababab", "abab")

==={{header|BASIC256}}===

{{trans|Run BASIC}}

print countSubstring("ababababab","abab")

end

i = instr(s$,find$,i) + length(find$)

end while

{{out}}

<pre>Igual que la entrada de Run BASIC.</pre>

==={{header|Yabasic}}===

{{trans|Run BASIC}}

print countSubstring("ababababab","abab")

end

end while

return countSubstring

{{out}}

<pre>Igual que la entrada de Run BASIC.</pre>

=={{header|Batch File}}==

setlocal enabledelayedexpansion

set input=!trimmed!

set /a cnt+=1

{{Out}}

<pre>3

=={{header|BBC BASIC}}==

sub$ = "th"

PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"

UNTIL I% = 0

= N%

{{out}}

<pre>3 "th" in "the three truths"

=={{header|BCPL}}==

let countsubstr(str, match) = valof

show("ababababab", "abab")

show("cat", "dog")

{{out}}

<pre>"th" in "the three truths": 3

=={{header|BQN}}==

<code>/𝕨⍷𝕩</code> finds locations of substrings, rest of the function suppresses overlapping substrings.

•Show "abab" Find "ababababab"

Using strings.bqn from bqn-libs, another solution is <code>Find←+´Locate</code>, since <code>Locate</code> performs a non-overlapping search.

=={{header|Bracmat}}==

= n S s p

. 0:?n:?p

& out$(count-substring$("the three truths".th))

& out$(count-substring$(ababababab.abab))

{{out}}

<pre>3

=={{header|C}}==

#include <string.h>

printf("not: %d\n", match("abababababa", "aba", 0));

return 0;

Alternate version:

#include <string.h>

return 0;

{{out}}

<pre>

=={{header|C sharp|C#}}==

class SubStringTestClass

return count;

}

Using C# 6.0's expression-bodied member, null-conditional operator, and coalesce operator features:

class SubStringTestClass

{

public static int CountSubStrings(this string testString, string testSubstring) =>

testString?.Split(new [] { testSubstring }, StringSplitOptions.None)?.Length - 1 ?? 0;

=={{header|C++}}==

#include <string>

return 0;

{{out}}

<pre>

=={{header|Clojure}}==

Use a sequence of regexp matches to count occurrences.

(defn re-quote

"Produces a string that can be used to create a Pattern

(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

(.results)

(.count)))

=={{header|COBOL}}==

<code>INSPECT</code> can be used for this task without having to create a function.

PROGRAM-ID. testing.

GOBACK

{{out}}

=={{header|CoffeeScript}}==

countSubstring = (str, substr) ->

n = 0

console.log countSubstring "the three truths", "th"

console.log countSubstring "ababababab", "abab"

=={{header|Common Lisp}}==

(loop with z = 0 with s = 0 while s do

(when (setf s (search pat str :start2 s))

(count-sub "ababa" "ab") ; 2

=={{header|Cowgol}}==

sub countSubstring(str: [uint8], match: [uint8]): (count: uint8) is

print_nl();

print_i8(countSubstring("cat","dog")); # should print 0

{{out}}

=={{header|D}}==

import std.stdio, std.algorithm;

"the three truths".count("th").writeln;

"ababababab".count("abab").writeln;

{{out}}

<pre>3

=={{header|Delphi}}==

{$APPTYPE CONSOLE}

Writeln(CountSubstring('the three truths', 'th'));

Writeln(CountSubstring('ababababab', 'abab'));

=={{header|Dyalect}}==

var idx = 0

var count = 0

print(countSubstring("the three truths", "th"))

{{out}}

=={{header|Déjà Vu}}==

{{out}}

<pre>3

=={{header|EchoLisp}}==

;; from Racket

(define count-substring

(count-substring "/ .e/" "Longtemps je me suis couché de bonne heure") ;; regexp

→ 4

=={{header|EGL}}==

{{works with|EDT}}

The "remove and count the difference" and "manual loop" methods. Implementation includes protection from empty source and search strings.

function main()

end

{{out}}

<pre>Remove and Count:

=={{header|Eiffel}}==

class

APPLICATION

search_for:STRING = "abab"

end

=={{header|Elixir}}==

(str, sub) -> length(String.split(str, sub)) - 1 end

Enum.each(data, fn{str, sub} ->

IO.puts countSubstring.(str, sub)

{{out}}

=={{header|Erlang}}==

%% Count non-overlapping substrings in Erlang for the rosetta code wiki.

%% Implemented by J.W. Luiten

main(String, Sub) ->

{{out}}

<pre>

Alternative using built in functions:

main( String, Sub ) -> erlang:length( binary:split(binary:list_to_bin(String), binary:list_to_bin(Sub), [global]) ) - 1.

=={{header|Euphoria}}==

integer from,count

count = 0

? countSubstring("the three truths","th")

{{out}}

=={{header|F_Sharp|F#}}==

"Remove and count the difference" method, as shown by J, Java, ...

let countSubstring (where :string) (what : string) =

show "ababababab" "abab"

show "abc" ""

<pre>countSubstring("the three truths", "th") = 3

countSubstring("ababababab", "abab") = 2

=={{header|Factor}}==

=={{header|Forth}}==

2swap 0 >r

begin 2over search

s" the three truths" s" th" str-count . \ 3

=={{header|Fortran}}==

{{works with|Fortran|90 and later}}

implicit none

integer :: n

end do

end function

{{out}}

<pre>3

=={{header|FreeBASIC}}==

Function countSubstring(s As String, search As String) As Integer

Print

Print "Press any key to quit"

{{out}}

=={{header|FunL}}==

def countSubstring( str, substr ) = Regex( substr ).findAllMatchIn( str ).length()

println( countSubstring("the three truths", "th") )

{{out}}

=={{header|Go}}==

Using strings.Count() method:

import (

"fmt"

fmt.Println(strings.Count("the three truths", "th")) // says: 3

fmt.Println(strings.Count("ababababab", "abab")) // says: 2

=={{header|Groovy}}==

Solution, uses the Groovy "find" operator (=~), and the Groovy-extended Matcher property "count":

println (('ababababab' =~ /abab/).count)

println (('abaabba*bbaba*bbab' =~ /a*b/).count)

{{out}}

=={{header|Haskell}}==

=== Text-based solution ===

-- Return the number of non-overlapping occurrences of sub in str.

print $ countSubStrs "the three truths" "th"

print $ countSubStrs "ababababab" "abab"

{{out}}

<pre>

Alternatively, in a language built around currying, it might make more sense to reverse the suggested order of arguments.

import Data.Text hiding (length)

"abelian absurdity",

"babel kebab"

{{Out}}

<pre>[5,2,2]</pre>

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 [] = error "empty substring"

count sub = go

scan [] xs = 1 + go xs

scan (x:xs) (y:ys) | x == y = scan xs ys

{{Out}}

<pre>λ> count "th" "the three truths"

The following solution is almost two times faster than the previous one.

import Data.Maybe (catMaybes)

count :: Eq a => [a] -> [a] -> Int

=={{header|Icon}} and {{header|Unicon}}==

every A := ![ ["the three truths","th"], ["ababababab","abab"] ] do

write("The string ",image(A[2])," occurs as a non-overlapping substring ",

}

return c

{{out}}

=={{header|J}}==

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:

3

'ababababab' countss 'abab'

=={{header|Java}}==

{{works with|Java|1.5+}}

The "remove and count the difference" method:

public static int countSubstring(String subStr, String str){

return (str.length() - str.replace(subStr, "").length()) / subStr.length();

System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));

}

{{out}}

<pre>3

{{works with|Java|1.5+}}

The "split and count" method:

public class CountSubstring {

System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));

}

{{out}}

<pre>3

Manual looping

public static int countSubstring(String subStr, String str){

int count = 0;

System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));

}

{{out}}

<pre>3

=={{header|JavaScript}}==

Using regexes:

var matches = str.match(new RegExp(subStr, "g"));

return matches ? matches.length : 0;

Using 'split' and ES6 notation:

=={{header|jq}}==

Using regexes (available in jq versions after June 19, 2014):

def countSubstring(sub):

=={{header|Julia}}==

'''Main'''

ts = ["the three truths", "ababababab"]

tsub = ["th", "abab"]

println(length(matchall(Regex(tsub[i]), ts[i], true)))

end

{{out}}

=={{header|K}}==

The dyadic verb _ss gives the positions of substring y in string x.

0 4 13

#"ababababab" _ss "abab"

2

=={{header|Klingphix}}==

:count %s !s

"ababababab" "abab" count ?

Other solution

:count "- " convert "-" 2 tolist split len nip ;

"ababababab" "abab" count ?

{{out}}

<pre>3

=={{header|Kotlin}}==

fun countSubstring(s: String, sub: String): Int = s.split(sub).size - 1

println(countSubstring("ababababab","abab"))

println(countSubstring("",""))

{{out}}

=={{header|Lambdatalk}}==

{def countSubstring

{def countSubstring.r

{countSubstring aba ababa}

-> 1

=={{header|langur}}==

{{out}}

=={{header|Lasso}}==

local(i = 1, foundpos = -1, found = 0)

while(#i < #str->size && #foundpos != 0) => {

//3

countSubstring_bothways('ababababab','abab')

=={{header|Liberty BASIC}}==

print countSubstring( "the three truths", "th")

print countSubstring( "ababababab", "abab")

countSubstring =c

end function

=={{header|Logtalk}}==

Using atoms for string representation:

:- object(counting).

:- end_object.

{{out}}

| ?- counting::count('the three truths', th, N).

N = 3

N = 2

yes

=={{header|Lua}}==

Solution 1:

return select(2, s1:gsub(s2, ""))

end

print(countSubstring("the three truths", "th"))

<pre>3

2</pre>

Solution 2:

local count = 0

for eachMatch in s1:gmatch(s2) do

print(countSubstring("the three truths", "th"))

<pre>3

2</pre>

=={{header|Maple}}==

f:=proc(s::string,c::string,count::nonnegint) local n;

n:=StringTools:-Search(c,s);

f("ababababab","abab",0);

{{out}}

<pre>

=={{header|Mathematica}} / {{header|Wolfram Language}}==

3

StringPosition["ababababab","abab",Overlaps->False]//Length

=={{header|MATLAB}} / {{header|Octave}}==

length(findstr("ababababab","abab",0))

length(findstr("the three truths","th",0))

% Count occurrences of a substring with overlap

{{out}}

=={{header|Maxima}}==

[n: 0, k: 1],

while integerp(k: ssearch(e, s, k)) do (n: n + 1, k: k + 1),

scount("na", "banana");

=={{header|MiniScript}}==

return self.split(s).len - 1

end function

print "the three truths".count("th")

{{out}}

<pre>

=={{header|Mirah}}==

import java.util.regex.Matcher

puts count_substring3("abab", "ababababab") # ==> 2

puts count_substring3("a*b", "abaabba*bbaba*bbab") # ==> 2

=={{header|Nanoquery}}==

{{trans|Java}}

return int((len(str) - len(str.replace(subStr, ""))) / len(subStr))

=={{header|Nemerle}}==

{{trans|F#}}

module CountSubStrings

WriteLine($"$target2 occurs $(text2.CountSubStrings(target2)) times in $text2");

}

{{out}}

<pre>th occurs 3 times in the three truths

NetRexx provides the <tt>''string''.countstr(''needle'')</tt> built-in function:

options replace format comments java crossref symbols nobinary

return

{{out}}

<pre>

=={{header|NewLISP}}==

; url: http://rosettacode.org/wiki/Count_occurrences_of_a_substring

; author: oofoe 2012-01-29

)

{{out}}

=={{header|Nim}}==

proc count(s, sub: string): int =

echo count("the three truths","th")

{{out}}

<pre>3

=={{header|Objective-C}}==

The "split and count" method:

- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;

@end

}

return 0;

{{out}}

<pre>3

The "remove and count the difference" method:

- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;

@end

}

return 0;

{{out}}

<pre>3

Manual looping:

- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;

@end

}

return 0;

{{out}}

<pre>3

=={{header|OCaml}}==

let sub_len = String.length sub in

let len_diff = (String.length str) - sub_len

Printf.printf "count 1: %d\n" (count_substring "the three truth" "th");

Printf.printf "count 2: %d\n" (count_substring "ababababab" "abab");

=={{header|Oforth}}==

: countSubString(s, sub)

0 1 while(sub swap s indexOfAllFrom dup notNull) [ sub size + 1 under+ ]

{{out}}

=={{header|ooRexx}}==

bag="the three truths"

x="th"

x="abab"

say left(bag,30) left(x,15) 'found' bag~caselesscountstr(x)

{{out}}

<pre style="height:10ex;overflow:scroll">

=={{header|PARI/GP}}==

my(i=1,s);

while(i+#u<=#v,

substr(s1,s2)=subvec(Vec(s1),Vec(s2));

substr("the three truths","th")

{{out}}

<pre>%1 = 3

=={{header|Perl}}==

my $str = shift;

my $sub = quotemeta(shift);

print countSubstring("the three truths","th"), "\n"; # prints "3"

=={{header|Phix}}==

<span style="color: #004080;">sequence</span> <span style="color: #000000;">tests</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{{</span><span style="color: #008000;">"the three truths"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"th"</span><span style="color: #0000FF;">},</span>

<span style="color: #0000FF;">{</span><span style="color: #008000;">"ababababab"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"abab"</span><span style="color: #0000FF;">},</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"The string \"%s\" occurs as a non-overlapping substring %d times in \"%s\"\n"</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">substring</span><span style="color: #0000FF;">,</span><span style="color: #000000;">count</span><span style="color: #0000FF;">,</span><span style="color: #000000;">test</span><span style="color: #0000FF;">})</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>

{{out}}

<pre>

=={{header|PHP}}==

echo substr_count("the three truths", "th"), PHP_EOL; // prints "3"

echo substr_count("ababababab", "abab"), PHP_EOL; // prints "2"

=={{header|Picat}}==

===Recursion===

count_rec(S,SB,0,C).

count_rec([T|Rest],SB,Count0,Count) :-

T != SB, % this character is not a substring

===Iterative===

Iterative version using find/4 (wrap with once/1 to avoid backtracking).

{{trans|Euphoria}}