# Entropy

Entropy
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Calculate the Shannon entropy   H   of a given input string.

Given the discrete random variable ${\displaystyle X}$ that is a string of ${\displaystyle N}$ "symbols" (total characters) consisting of ${\displaystyle n}$ different characters (n=2 for binary), the Shannon entropy of X in bits/symbol is :

${\displaystyle H_{2}(X)=-\sum _{i=1}^{n}{\frac {count_{i}}{N}}\log _{2}\left({\frac {count_{i}}{N}}\right)}$

where ${\displaystyle count_{i}}$ is the count of character ${\displaystyle n_{i}}$.

For this task, use X="1223334444" as an example. The result should be 1.84644... bits/symbol. This assumes X was a random variable, which may not be the case, or it may depend on the observer.

This coding problem calculates the "specific" or "intensive" entropy that finds its parallel in physics with "specific entropy" S0 which is entropy per kg or per mole, not like physical entropy S and therefore not the "information" content of a file. It comes from Boltzmann's H-theorem where ${\displaystyle S=k_{B}NH}$ where N=number of molecules. Boltzmann's H is the same equation as Shannon's H, and it gives the specific entropy H on a "per molecule" basis.

The "total", "absolute", or "extensive" information entropy is

${\displaystyle S=H_{2}N}$ bits

This is not the entropy being coded here, but it is the closest to physical entropy and a measure of the information content of a string. But it does not look for any patterns that might be available for compression, so it is a very restricted, basic, and certain measure of "information". Every binary file with an equal number of 1's and 0's will have S=N bits. All hex files with equal symbol frequencies will have ${\displaystyle S=N\log _{2}(16)}$ bits of entropy. The total entropy in bits of the example above is S= 10*18.4644 = 18.4644 bits.

The H function does not look for any patterns in data or check if X was a random variable. For example, X=000000111111 gives the same calculated entropy in all senses as Y=010011100101. For most purposes it is usually more relevant to divide the gzip length by the length of the original data to get an informal measure of how much "order" was in the data.

Two other "entropies" are useful:

Normalized specific entropy:

${\displaystyle H_{n}={\frac {H_{2}*\log(2)}{\log(n)}}}$

which varies from 0 to 1 and it has units of "entropy/symbol" or just 1/symbol. For this example, Hn<\sub>= 0.923.

Normalized total (extensive) entropy:

${\displaystyle S_{n}={\frac {H_{2}N*\log(2)}{\log(n)}}}$

which varies from 0 to N and does not have units. It is simply the "entropy", but it needs to be called "total normalized extensive entropy" so that it is not confused with Shannon's (specific) entropy or physical entropy. For this example, Sn<\sub>= 9.23.

Shannon himself is the reason his "entropy/symbol" H function is very confusingly called "entropy". That's like calling a function that returns a speed a "meter". See section 1.7 of his classic A Mathematical Theory of Communication and search on "per symbol" and "units" to see he always stated his entropy H has units of "bits/symbol" or "entropy/symbol" or "information/symbol". So it is legitimate to say entropy NH is "information".

In keeping with Landauer's limit, the physics entropy generated from erasing N bits is ${\displaystyle S=H_{2}Nk_{B}\ln(2)}$ if the bit storage device is perfectly efficient. This can be solved for H2*N to (arguably) get the number of bits of information that a physical entropy represents.

Related tasks

## 11l

F entropy(source)
DefaultDict[Char, Int] hist
L(c) source
hist[c]++
V r = 0.0
L(v) hist.values()
V c = Float(v) / source.len
r -= c * log2(c)
R r

print(entropy(‘1223334444’))
Output:
1.84644


## Ada

Uses Ada 2012.

with Ada.Text_IO, Ada.Float_Text_IO, Ada.Numerics.Elementary_Functions;

procedure Count_Entropy is

package TIO renames Ada.Text_IO;

Count: array(Character) of Natural := (others => 0);
Sum:   Natural := 0;
Line: String := "1223334444";

begin
for I in Line'Range loop   -- count the characters
Count(Line(I)) := Count(Line(I))+1;
Sum := Sum + 1;
end loop;

declare   -- compute the entropy and print it
function P(C: Character) return Float is (Float(Count(C)) / Float(Sum));
use Ada.Numerics.Elementary_Functions, Ada.Float_Text_IO;
Result: Float := 0.0;
begin
for Ch in Character loop
Result := Result -
(if P(Ch)=0.0 then 0.0 else P(Ch) * Log(P(Ch), Base => 2.0));
end loop;
Put(Result, Fore => 1, Aft => 5, Exp => 0);
end;
end Count_Entropy;


## Aime

integer c;
real h, v;
index x;
data s;

for (, c in (s = argv(1))) {
x[c] += 1r;
}

h = 0;
for (, v in x) {
v /= ~s;
h -= v * log2(v);
}

o_form("/d6/\n", h);

Examples:

$aime -a tmp/entr 1223334444 1.846439$ aime -a tmp/entr 'Rosetta Code is the best site in the world!'
3.646513
$aime -a tmp/entr 1234567890abcdefghijklmnopqrstuvwxyz 5.169925 ## ALGOL 68 BEGIN # calculate the shannon entropy of a string # PROC shannon entropy = ( STRING s )REAL: BEGIN INT string length = ( UPB s - LWB s ) + 1; # count the occurences of each character # [ 0 : max abs char ]INT char count; FOR char pos FROM LWB char count TO UPB char count DO char count[ char pos ] := 0 OD; FOR char pos FROM LWB s TO UPB s DO char count[ ABS s[ char pos ] ] +:= 1 OD; # calculate the entropy, we use log base 10 and then convert # # to log base 2 after calculating the sum # REAL entropy := 0; FOR char pos FROM LWB char count TO UPB char count DO IF char count[ char pos ] /= 0 THEN # have a character that occurs in the string # REAL probability = char count[ char pos ] / string length; entropy -:= probability * log( probability ) FI OD; entropy / log( 2 ) END; # shannon entropy # # test the shannon entropy routine # print( ( shannon entropy( "1223334444" ), newline ) ) END Output: +1.84643934467102e +0  ## ALGOL W Translation of: ALGOL 68 begin % calculates the shannon entropy of a string % % strings are fixed length in algol W and the length is part of the % % type, so we declare the string parameter to be the longest possible % % string length (256 characters) and have a second parameter to % % specify how much is actually used % real procedure shannon_entropy ( string(256) value s ; integer value stringLength ); begin real probability, entropy; % algol W assumes there are 256 possible characters % integer MAX_CHAR; MAX_CHAR := 256; % declarations must preceed statements, so we start a new % % block here so we can use MAX_CHAR as an array bound % begin % increment an integer variable % procedure incI ( integer value result a ) ; a := a + 1; integer array charCount( 1 :: MAX_CHAR ); % count the occurances of each character in s % for charPos := 1 until MAX_CHAR do charCount( charPos ) := 0; for sPos := 0 until stringLength - 1 do incI( charCount( decode( s( sPos | 1 ) ) ) ); % calculate the entropy, we use log base 10 and then convert % % to log base 2 after calculating the sum % entropy := 0.0; for charPos := 1 until MAX_CHAR do begin if charCount( charPos ) not = 0 then begin % have a character that occurs in the string % probability := charCount( charPos ) / stringLength; entropy := entropy - ( probability * log( probability ) ) end end charPos end; entropy / log( 2 ) end shannon_entropy ; % test the shannon entropy routine % r_format := "A"; r_w := 12; r_d := 6; % set output to fixed format % write( shannon_entropy( "1223334444", 10 ) ) end. Output:  1.846439  ## APL  ENTROPY←{-+/R×2⍟R←(+⌿⍵∘.=∪⍵)÷⍴⍵} ⍝ How it works: ⎕←UNIQUE←∪X←'1223334444' 1234 ⎕←TABLE_OF_OCCURENCES←X∘.=UNIQUE 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 ⎕←COUNT←+⌿TABLE_OF_OCCURENCES 1 2 3 4 ⎕←N←⍴X 10 ⎕←RATIO←COUNT÷N 0.1 0.2 0.3 0.4 -+/RATIO×2⍟RATIO 1.846439345  Output:  ENTROPY X 1.846439345  ## Arturo entropy: function [s][ t: #[] loop s 'c [ unless key? t c -> t\[c]: 0 t\[c]: t\[c] + 1 ] result: new 0 loop values t 'x -> 'result - (x//(size s)) * log x//(size s) 2 return result ] print entropy "1223334444"  Output: 1.846439344671015 ## AutoHotkey MsgBox, % Entropy(1223334444) Entropy(n) { a := [], len := StrLen(n), m := n while StrLen(m) { s := SubStr(m, 1, 1) m := RegExReplace(m, s, "", c) a[s] := c } for key, val in a { m := Log(p := val / len) e -= p * m / Log(2) } return, e }  Output: 1.846440 ## AWK #!/usr/bin/awk -f { N = length for (i = 1; i <= N; ++i) ++H[substr($0, i, 1)]
}

END {
for (i in H)
S += H[i] * log(H[i])
print (log(N) - S / N) / log(2)
}

Usage:
 echo 1223334444 |./entropy.awk
1.84644


## BASIC

Works with older (unstructured) Microsoft-style BASIC.

10 DEF FN L(X)=LOG(X)/LOG(2)
20 S$="1223334444" 30 U$=""
40 FOR I=1 TO LEN(S$) 50 K=0 60 FOR J=1 TO LEN(U$)
70 IF MID$(U$,J,1)=MID$(S$,I,1) THEN K=1
80 NEXT J
90 IF K=0 THEN U$=U$+MID$(S$,I,1)
100 NEXT I
110 DIM R(LEN(U$)-1) 120 FOR I=1 TO LEN(U$)
130 C=0
140 FOR J=1 TO LEN(S$) 150 IF MID$(U$,I,1)=MID$(S$,J,1) THEN C=C+1 160 NEXT J 170 R(I-1)=(C/LEN(S$))*FN L(C/LEN(S$)) 180 NEXT I 190 E=0 200 FOR I=0 TO LEN(U$)-1
210 E=E-R(I)
220 NEXT I
230 PRINT E

Output:
1.84643935

### QBasic

FUNCTION L (X)
L = LOG(X) / LOG(2)
END FUNCTION

S$= "1223334444" U$ = ""
FOR I = 1 TO LEN(S$) K = 0 FOR J = 1 TO LEN(U$)
IF MID$(U$, J, 1) = MID$(S$, I, 1) THEN K = 1
NEXT J
IF K = 0 THEN U$= U$ + MID$(S$, I, 1)
NEXT I
DIM R(LEN(U$) - 1) FOR I = 1 TO LEN(U$)
C = 0
FOR J = 1 TO LEN(S$) IF MID$(U$, I, 1) = MID$(S$, J, 1) THEN C = C + 1 NEXT J R(I - 1) = (C / LEN(S$)) * L(C / LEN(S$)) NEXT I E = 0 FOR I = 0 TO LEN(U$) - 1
E = E - R(I)
NEXT I
PRINT E
END


### Sinclair ZX81 BASIC

Works with 1k of RAM.

 10 LET X$="1223334444" 20 LET U$=""
30 FOR I=1 TO LEN X$40 LET K=0 50 FOR J=1 TO LEN U$
60 IF U$(J)=X$(I) THEN LET K=K+1
70 NEXT J
80 IF K=0 THEN LET U$=U$+X$(I) 90 NEXT I 100 DIM R(LEN U$)
110 FOR I=1 TO LEN U$120 LET C=0 130 FOR J=1 TO LEN X$
140 IF U$(I)=X$(J) THEN LET C=C+1
150 NEXT J
160 LET R(I)=C/LEN X$*LN (C/LEN X$)/LN 2
170 NEXT I
180 LET E=0
190 FOR I=1 TO LEN U$200 LET E=E-R(I) 210 NEXT I 220 PRINT E  Output: 1.8464393 ### uBasic/4tH Translation of: QBasic uBasic/4tH is an integer BASIC only. So, fixed point arithmetic is required go fulfill this task. Some loss of precision is unavoidable. If Info("wordsize") < 64 Then Print "This program requires a 64-bit uBasic" : End s := "1223334444" u := "" x := FUNC(_Fln(FUNC(_Ntof(2)))) ' calculate LN(2) For i = 0 TO Len(s)-1 k = 0 For j = 0 TO Len(u)-1 If Peek(u, j) = Peek(s, i) Then k = 1 Next If k = 0 THEN u = Join(u, Char (Peek (s, i))) Next Dim @r(Len(u)-1) For i = 0 TO Len(u)-1 c = 0 For J = 0 TO Len(s)-1 If Peek(u, i) = Peek (s, j) Then c = c + 1 Next q = FUNC(_Fdiv(c, Len(s))) @r(i) = FUNC(_Fmul(q, FUNC(_Fdiv(FUNC(_Fln(q)), x)))) Next e = 0 For i = 0 To Len(u) - 1 e = e - @r(i) Next Print Using "+?.####"; FUNC(_Ftoi(e)) End _Fln Param (1) : Return (FUNC(_Ln(a@*4))/4) _Fmul Param (2) : Return ((a@*b@)/16384) _Fdiv Param (2) : Return ((a@*16384)/b@) _Ntof Param (1) : Return (a@*16384) _Ftoi Param (1) : Return ((10000*a@)/16384) _Ln Param (1) Local (2) c@=681391 If (a@<32768) Then a@=SHL(a@, 16) : c@=c@-726817 If (a@<8388608) Then a@=SHL(a@, 8) : c@=c@-363409 If (a@<134217728) Then a@=SHL(a@, 4) : c@=c@-181704 If (a@<536870912) Then a@=SHL(a@, 2) : c@=c@-90852 If (a@<1073741824) Then a@=SHL(a@, 1) : c@=c@-45426 b@=a@+SHL(a@, -1) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-26573 b@=a@+SHL(a@, -2) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-14624 b@=a@+SHL(a@, -3) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-7719 b@=a@+SHL(a@, -4) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-3973 b@=a@+SHL(a@, -5) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-2017 b@=a@+SHL(a@, -6) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-1016 b@=a@+SHL(a@, -7) : If (AND(b@, 2147483648)) = 0 Then a@=b@ : c@=c@-510 a@=2147483648-a@; c@=c@-SHL(a@, -15) Return (c@)  Output: 1.8461 0 OK, 0:638 ## BBC BASIC Translation of: APL REM >entropy PRINT FNentropy("1223334444") END : DEF FNentropy(x$)
LOCAL unique$, count%, n%, ratio(), u%, i%, j% unique$ = ""
n% = LEN x$FOR i% = 1 TO n% IF INSTR(unique$, MID$(x$, i%, 1)) = 0 THEN unique$+= MID$(x$, i%, 1) NEXT u% = LEN unique$
DIM ratio(u% - 1)
FOR i% = 1 TO u%
count% = 0
FOR j% = 1 TO n%
IF MID$(unique$, i%, 1) = MID$(x$, j%, 1) THEN count% += 1
NEXT
ratio(i% - 1) = (count% / n%) * FNlogtwo(count% / n%)
NEXT
= -SUM(ratio())
:
DEF FNlogtwo(n)
= LN n / LN 2

Output:
1.84643934

## BQN

H ← -∘(+´⊢×2⋆⁼⊢)∘((+˝⊢=⌜⍷)÷≠)

H "1223334444"

Output:
1.8464393446710154

## Burlesque

blsq ) "1223334444"F:u[vv^^{1\/?/2\/LG}m[?*++
1.8464393446710157

## C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <math.h>

#define MAXLEN 100 //maximum string length

int makehist(unsigned char *S,int *hist,int len){
int wherechar[256];
int i,histlen;
histlen=0;
for(i=0;i<256;i++)wherechar[i]=-1;
for(i=0;i<len;i++){
if(wherechar[(int)S[i]]==-1){
wherechar[(int)S[i]]=histlen;
histlen++;
}
hist[wherechar[(int)S[i]]]++;
}
return histlen;
}

double entropy(int *hist,int histlen,int len){
int i;
double H;
H=0;
for(i=0;i<histlen;i++){
H-=(double)hist[i]/len*log2((double)hist[i]/len);
}
return H;
}

int main(void){
unsigned char S[MAXLEN];
int len,*hist,histlen;
double H;
scanf("%[^\n]",S);
len=strlen(S);
hist=(int*)calloc(len,sizeof(int));
histlen=makehist(S,hist,len);
//hist now has no order (known to the program) but that doesn't matter
H=entropy(hist,histlen,len);
printf("%lf\n",H);
return 0;
}


Examples:

$./entropy 1223334444 1.846439$ ./entropy
Rosetta Code is the best site in the world!
3.646513


## C#

Translation of C++.

using System;
using System.Collections.Generic;
namespace Entropy
{
class Program
{
public static double logtwo(double num)
{
return Math.Log(num)/Math.Log(2);
}
public static void Main(string[] args)
{
label1:
string input = Console.ReadLine();
double infoC=0;
Dictionary<char,double> table = new Dictionary<char, double>();

foreach (char c in input)
{
if (table.ContainsKey(c))
table[c]++;
else
table.Add(c,1);

}
double freq;
foreach (KeyValuePair<char,double> letter in table)
{
freq=letter.Value/input.Length;
infoC+=freq*logtwo(freq);
}
infoC*=-1;
Console.WriteLine("The Entropy of {0} is {1}",input,infoC);
goto label1;

}
}
}

Output:
The Entropy of 1223334444 is 1.84643934467102

Without using Hashtables or Dictionaries:

using System;
namespace Entropy
{
class Program
{
public static double logtwo(double num)
{
return Math.Log(num)/Math.Log(2);
}
static double Contain(string x,char k)
{
double count=0;
foreach (char Y in x)
{
if(Y.Equals(k))
count++;
}
return count;
}
public static void Main(string[] args)
{
label1:
string input = Console.ReadLine();
double infoC=0;
double freq;
string k="";
foreach (char c1 in input)
{
if (!(k.Contains(c1.ToString())))
k+=c1;
}
foreach (char c in k)
{
freq=Contain(input,c)/(double)input.Length;
infoC+=freq*logtwo(freq);
}
infoC/=-1;
Console.WriteLine("The Entropy of {0} is {1}",input,infoC);
goto label1;

}
}
}


## C++

#include <string>
#include <map>
#include <iostream>
#include <algorithm>
#include <cmath>

double log2( double number ) {
return log( number ) / log( 2 ) ;
}

int main( int argc , char *argv[ ] ) {
std::string teststring( argv[ 1 ] ) ;
std::map<char , int> frequencies ;
for ( char c : teststring )
frequencies[ c ] ++ ;
int numlen = teststring.length( ) ;
double infocontent = 0 ;
for ( std::pair<char , int> p : frequencies ) {
double freq = static_cast<double>( p.second ) / numlen ;
infocontent -= freq * log2( freq ) ;
}

std::cout << "The information content of " << teststring
<< " is " << infocontent << std::endl ;
return 0 ;
}

Output:
(entropy "1223334444")
The information content of 1223334444 is 1.84644

## Clojure

(defn entropy [s]
(let [len (count s), log-2 (Math/log 2)]
(->> (frequencies s)
(map (fn [[_ v]]
(let [rf (/ v len)]
(-> (Math/log rf) (/ log-2) (* rf) Math/abs))))
(reduce +))))

Output:
(entropy "1223334444")
1.8464393446710154


## CLU

% NOTE: when compiling with Portable CLU,
% this program needs to be merged with 'useful.lib' to get log()
%
% pclu -merge $CLUHOME/lib/useful.lib -compile entropy.clu shannon = proc (s: string) returns (real) % find the frequency of each character freq: array[int] := array[int]$fill(0, 256, 0)
for c: char in string$chars(s) do i: int := char$c2i(c)
freq[i] := freq[i] + 1
end

% calculate the component for each character
h: real := 0.0
rlen: real := real$i2r(string$size(s))
for i: int in array[int]$indexes(freq) do if freq[i] ~= 0 then f: real := real$i2r(freq[i]) / rlen
h := h - f * log(f) / log(2.0)
end
end
return (h)
end shannon

start_up = proc ()
po: stream := stream$primary_output() stream$putl(po, f_form(shannon("1223334444"), 1, 6))
end start_up
Output:
1.846439

## CoffeeScript

entropy = (s) ->
freq = (s) ->
result = {}
for ch in s.split ""
result[ch] ?= 0
result[ch]++
return result

frq = freq s
n = s.length
((frq[f]/n for f of frq).reduce ((e, p) -> e - p * Math.log(p)), 0) * Math.LOG2E

console.log "The entropy of the string '1223334444' is #{entropy '1223334444'}"

Output:
The entropy of the string '1223334444' is 1.8464393446710157

## Common Lisp

Not very Common Lisp-y version:

(defun entropy (string)
(let ((table (make-hash-table :test 'equal))
(entropy 0))
(mapc (lambda (c) (setf (gethash c table) (+ (gethash c table 0) 1)))
(coerce string 'list))
(maphash (lambda (k v)
(decf entropy (* (/ v (length input-string))
(log (/ v (length input-string)) 2))))
table)
entropy))


More like Common Lisp version:

(defun entropy (string &aux (length (length string)))
(declare (type string string))
(let ((table (make-hash-table)))
(loop for char across string
do (incf (gethash char table 0)))
(- (loop for freq being each hash-value in table
for freq/length = (/ freq length)
sum (* freq/length (log freq/length 2))))))


## Crystal

# Method to calculate sum of Float64 array
def sum(array : Array(Float64))
res = 0
array.each do |n|
res += n
end
res
end

# Method to calculate which char appears how often
def histogram(source : String)
hist = {} of Char => Int32
l = 0
source.each_char do |e|
if !hist.has_key? e
hist[e] = 0
end
hist[e] += 1
end
return Tuple.new(source.size, hist)
end

# Method to calculate entropy from histogram
def entropy(hist : Hash(Char, Int32), l : Int32)
elist = [] of Float64
hist.each do |el|
v = el[1]
c = v / l
elist << (-c * Math.log(c, 2))
end
return sum elist
end

source = "1223334444"
hist_res = histogram source
l = hist_res[0]
h = hist_res[1]
puts ".[Results]."
puts "Length: " + l.to_s
puts "Entropy: " + (entropy h, l).to_s


## D

import std.stdio, std.algorithm, std.math;

double entropy(T)(T[] s)
pure nothrow if (__traits(compiles, s.sort())) {
immutable sLen = s.length;
return s
.sort()
.group
.map!(g => g[1] / double(sLen))
.map!(p => -p * p.log2)
.sum;
}

void main() {
"1223334444"d.dup.entropy.writeln;
}

Output:
1.84644

## Delphi

Library: StrUtils
Library: Math
Translation of: Pascal

Just fix Pascal code to run in Delphi.

program Entropytest;

uses
StrUtils,
Math;

type
FArray = array of CARDINAL;

var
strng: string = '1223334444';

// list unique characters in a string
function uniquechars(str: string): string;
var
n: CARDINAL;
begin
Result := '';
for n := 1 to length(str) do
if (PosEx(str[n], str, n) > 0) and (PosEx(str[n], Result, 1) = 0) then
Result := Result + str[n];
end;

// obtain a list of character-frequencies for a string
//  given a string containing its unique characters
function frequencies(str, ustr: string): FArray;
var
u, s, p, o: CARDINAL;
begin
SetLength(Result, Length(ustr) + 1);
p := 0;
for u := 1 to length(ustr) do
for s := 1 to length(str) do
begin
o := p;
p := PosEx(ustr[u], str, s);
if (p > o) then
INC(Result[u]);
end;
end;

// Obtain the Shannon entropy of a string
function entropy(s: string): EXTENDED;
var
pf: FArray;
us: string;
i, l: CARDINAL;
begin
us := uniquechars(s);
pf := frequencies(s, us);
l := length(s);
Result := 0.0;
for i := 1 to length(us) do
Result := Result - pf[i] / l * log2(pf[i] / l);
end;

begin
Writeln('Entropy of "', strng, '" is ', entropy(strng): 2: 5, ' bits.');
readln;
end.


## EasyLang

func entropy s$. len d[] 255 for c$ in strchars s$d[strcode c$] += 1
.
for cnt in d[]
if cnt > 0
prop = cnt / len s$entr -= (prop * log10 prop / log10 2) . . return entr . print entropy "1223334444" ## EchoLisp (lib 'hash) ;; counter: hash-table[key]++ (define (count++ ht k ) (hash-set ht k (1+ (hash-ref! ht k 0)))) (define (hi count n ) (define pi (// count n)) (- (* pi (log2 pi)))) ;; (H [string|list]) → entropy (bits) (define (H info) (define S (if(string? info) (string->list info) info)) (define ht (make-hash)) (define n (length S)) (for ((s S)) (count++ ht s)) (for/sum ((s (make-set S))) (hi (hash-ref ht s) n)))  Output: ;; by increasing entropy (H "🔴") → 0 (H "🔵🔴") → 1 (H "1223334444") → 1.8464393446710154 (H "♖♘♗♕♔♗♘♖♙♙♙♙♙♙♙♙♙") → 2.05632607578088 (H "EchoLisp") → 3 (H "Longtemps je me suis couché de bonne heure") → 3.860828877124944 (H "azertyuiopmlkjhgfdsqwxcvbn") → 4.700439718141092 (H (for/list ((i 1000)) (random 1000))) → 9.13772704467521 (H (for/list ((i 100_000)) (random 100_000))) → 15.777516877140766 (H (for/list ((i 1000_000)) (random 1000_000))) → 19.104028424596976  ## Elena Translation of: C# ELENA 6.x : import system'math; import system'collections; import system'routines; import extensions; extension op { logTwo() = self.ln() / 2.ln(); } public program() { var input := console.readLine(); var infoC := 0.0r; var table := Dictionary.new(); input.forEach::(ch) { var n := table[ch]; if (nil == n) { table[ch] := 1 } else { table[ch] := n + 1 } }; var freq := 0; table.forEach::(letter) { freq := letter.toInt().realDiv(input.Length); infoC += (freq * freq.logTwo()) }; infoC *= -1; console.printLine("The Entropy of ", input, " is ", infoC) } Output: The Entropy of 1223334444 is 1.846439344671  ## Elixir Works with: Erlang/OTP version 18 :math.log2 was added in OTP 18. defmodule RC do def entropy(str) do leng = String.length(str) String.graphemes(str) |> Enum.group_by(&(&1)) |> Enum.map(fn{_,value} -> length(value) end) |> Enum.reduce(0, fn count, entropy -> freq = count / leng entropy - freq * :math.log2(freq) end) end end IO.inspect RC.entropy("1223334444")  Output: 1.8464393446710154  ## Emacs Lisp (defun shannon-entropy (input) (let ((freq-table (make-hash-table)) (entropy 0) (length (+ (length input) 0.0))) (mapcar (lambda (x) (puthash x (+ 1 (gethash x freq-table 0)) freq-table)) input) (maphash (lambda (k v) (set 'entropy (+ entropy (* (/ v length) (log (/ v length) 2))))) freq-table) (- entropy)))  Output: After adding the above to the emacs runtime, you can run the function interactively in the scratch buffer as shown below (type ctrl-j at the end of the first line and the output will be placed by emacs on the second line). (shannon-entropy "1223334444") 1.8464393446710154  ## Erlang -module( entropy ). -export( [shannon/1, task/0] ). shannon( String ) -> shannon_information_content( lists:foldl(fun count/2, dict:new(), String), erlang:length(String) ). task() -> shannon( "1223334444" ). count( Character, Dict ) -> dict:update_counter( Character, 1, Dict ). shannon_information_content( Dict, String_length ) -> {_String_length, Acc} = dict:fold( fun shannon_information_content/3, {String_length, 0.0}, Dict ), Acc / math:log( 2 ). shannon_information_content( _Character, How_many, {String_length, Acc} ) -> Frequency = How_many / String_length, {String_length, Acc - (Frequency * math:log(Frequency))}.  Output: 24> entropy:task(). 1.8464393446710157  ## Euler Math Toolbox >function entropy (s) ...$  v=strtochar(s);
$m=getmultiplicities(unique(v),v);$  m=m/sum(m);
$return sum(-m*logbase(m,2))$endfunction
>entropy("1223334444")
1.84643934467

## Excel

This solution uses the LAMBDA, LET, and MAP functions introduced into the Microsoft 365 version of Excel in 2021. The LET function is able to use functions as first class citizens. Taking advantage of this makes the solution much simpler. The solution below looks for the string in cell A1.

=LET(
_MainS,A1,
_N,LEN(_MainS),
_Chars,UNIQUE(MID(_MainS,SEQUENCE(LEN(_MainS),1,1,1),1)),
calcH,LAMBDA(_c,(_c/_N)*LOG(_c/_N,2)),
getCount,LAMBDA(_i,LEN(_MainS)-LEN(SUBSTITUTE(_MainS,_i,""))),
_CharMap,MAP(_Chars,LAMBDA(a, calcH(getCount(a)))),
-SUM(_CharMap)
)

_Chars uses the SEQUENCE function to split the text into an array. The UNIQUE function then returns a list of unique characters in the string.

calcH applies the calculation described at the top of the page that will then be summed for each character

getCount uses the SUBSTITUTE method to count the occurrences of a character within the string.

If you needed to re-use this calculation then you could wrap it in a LAMBDA function within the name manager, changing A1 to a variable name (e.g. String):

ShannonEntropyH2=LAMBDA(String,LET(_MainS,String,_N,LEN(_MainS),_Chars,UNIQUE(MID(_MainS,SEQUENCE(LEN(_MainS),1,1,1),1)),calcH,LAMBDA(_c,(_c/_N)*LOG(_c/_N,2)),getCount,LAMBDA(_i,LEN(_MainS)-LEN(SUBSTITUTE(_MainS,_i,""))),_CharMap,MAP(_Chars,LAMBDA(a, calcH(getCount(a)))),-SUM(_CharMap)))

Then you can just use the named lambda. E.g. If A1 = 1223334444 then:

=ShannonEntropyH2(A1)

Returns 1.846439345

## F#

open System

let ld x = Math.Log x / Math.Log 2.

let entropy (s : string) =
let n = float s.Length
Seq.groupBy id s
|> Seq.map (fun (_, vals) -> float (Seq.length vals) / n)
|> Seq.fold (fun e p -> e - p * ld p) 0.

printfn "%f" (entropy "1223334444")

Output:
1.846439

## Factor

USING: assocs kernel math math.functions math.statistics
prettyprint sequences ;
IN: rosetta-code.entropy

: shannon-entropy ( str -- entropy )
[ length ] [ histogram >alist [ second ] map ] bi
[ swap / ] with map
[ dup log 2 log / * ] map-sum neg ;

"1223334444" shannon-entropy .
"Factor is my favorite programming language." shannon-entropy .

Output:
1.846439344671015
4.04291723248433


## Forth

: flog2 ( f -- f ) fln 2e fln f/ ;

create freq 256 cells allot

: entropy ( str len -- f )
freq 256 cells erase
tuck
bounds do
i c@ cells freq +
1 swap +!
loop
0e
256 0 do
i cells freq + @ ?dup if
s>f dup s>f f/
fdup flog2 f* f-
then
loop
drop ;

s" 1223334444" entropy f.     \ 1.84643934467102  ok


## Fortran

Please find the GNU/linux compilation instructions along with sample run among the comments at the start of the FORTRAN 2008 source. This program acquires input from the command line argument, thereby demonstrating the fairly new get_command_argument intrinsic subroutine. The expression of the algorithm is a rough translated of the j solution. Thank you.

!-*- mode: compilation; default-directory: "/tmp/" -*-
!Compilation started at Tue May 21 21:43:12
!
!a=./f && make $a && OMP_NUM_THREADS=2$a 1223334444
!gfortran -std=f2008 -Wall -ffree-form -fall-intrinsics f.f08 -o f
! Shannon entropy of 1223334444 is    1.84643936
!
!Compilation finished at Tue May 21 21:43:12

program shannonEntropy
implicit none
integer :: num, L, status
character(len=2048) :: s
num = 1
call get_command_argument(num, s, L, status)
if ((0 /= status) .or. (L .eq. 0)) then
write(0,*)'Expected a command line argument with some length.'
else
write(6,*)'Shannon entropy of '//(s(1:L))//' is ', se(s(1:L))
endif

contains
!     algebra
!
! 2**x = y
! x*log(2) = log(y)
! x = log(y)/log(2)

!   NB. The j solution
!   entropy=:  +/@:-@(* 2&^.)@(#/.~ % #)
!   entropy '1223334444'
!1.84644

real function se(s)
implicit none
character(len=*), intent(in) :: s
integer, dimension(256) :: tallies
real, dimension(256) :: norm
tallies = 0
call TallyKey(s, tallies)
! J's #/. works with the set of items in the input.
! TallyKey is sufficiently close that, with the merge, gets the correct result.
norm = tallies / real(len(s))
se = sum(-(norm*log(merge(1.0, norm, norm .eq. 0))/log(2.0)))
end function se

subroutine TallyKey(s, counts)
character(len=*), intent(in) :: s
integer, dimension(256), intent(out) :: counts
integer :: i, j
counts = 0
do i=1,len(s)
j = iachar(s(i:i))
counts(j) = counts(j) + 1
end do
end subroutine TallyKey

end program shannonEntropy


## FreeBASIC

' version 25-06-2015
' compile with: fbc -s console

Sub calc_entropy(source As String, base_ As Integer)

Dim As Integer i, sourcelen = Len(source), totalchar(255)
Dim As Double prop, entropy

For i = 0 To sourcelen -1
totalchar(source[i]) += 1
Next

Print "Char    count"
For i = 0 To 255
If totalchar(i) = 0 Then Continue For
Print "   "; Chr(i); Using "   ######"; totalchar(i)
prop = totalchar(i) / sourcelen
entropy = entropy - (prop * Log (prop) / Log(base_))
Next

Print : Print "The Entropy of "; Chr(34); source; Chr(34); " is"; entropy

End Sub

' ------=< MAIN >=------

calc_entropy("1223334444", 2)
Print

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
Char    count
1        1
2        2
3        3
4        4

The Entropy of "1223334444" is 1.846439344671015

## friendly interactive shell

Sort of hacky, but friendly interactive shell isn't really optimized for mathematic tasks (in fact, it doesn't even have associative arrays).

function entropy
for arg in $argv set name count_$arg
if not count $$name > /dev/null set name 0 set values values arg end set name (math$$name + 1)
end
set entropy 0
for value in $values set name count_$value
set entropy (echo "
scale = 50
p = "$$name" / "(count argv)" entropy - p * l(p) " | bc -l) end echo "entropy / l(2)" | bc -l end entropy (echo 1223334444 | fold -w1)  Output: 1.84643934467101549345 ## 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. Solution Test case ## Go ### Go: Slice version package main import ( "fmt" "math" "strings" ) func main(){ fmt.Println(H("1223334444")) } // for ASCII strings func H(data string) (entropy float64) { if data == "" { return 0 } for i := 0; i < 256; i++ { px := float64(strings.Count(data, string(byte(i)))) / float64(len(data)) if px > 0 { entropy += -px * math.Log2(px) } } return entropy }  Output: 1.8464393446710154  ### Go: Map version package main import ( "fmt" "math" ) func main() { const s = "1223334444" l := float64(0) m := map[rune]float64{} for _, r := range s { m[r]++ l++ } var hm float64 for _, c := range m { hm += c * math.Log2(c) } fmt.Println(math.Log2(l) - hm/l) }  Output: 1.8464393446710152  ## Groovy String.metaClass.getShannonEntrophy = { -delegate.inject([:]) { map, v -> map[v] = (map[v] ?: 0) + 1; map }.values().inject(0.0) { sum, v -> def p = (BigDecimal)v / delegate.size() sum + p * Math.log(p) / Math.log(2) } }  Testing [ '1223334444': '1.846439344671', '1223334444555555555': '1.969811065121', '122333': '1.459147917061', '1227774444': '1.846439344671', aaBBcccDDDD: '1.936260027482', '1234567890abcdefghijklmnopqrstuvwxyz': '5.169925004424', 'Rosetta Code': '3.084962500407' ].each { s, expected -> println "Checking s has a shannon entrophy of expected" assert sprintf('%.12f', s.shannonEntrophy) == expected }  Output: Checking 1223334444 has a shannon entrophy of 1.846439344671 Checking 1223334444555555555 has a shannon entrophy of 1.969811065121 Checking 122333 has a shannon entrophy of 1.459147917061 Checking 1227774444 has a shannon entrophy of 1.846439344671 Checking aaBBcccDDDD has a shannon entrophy of 1.936260027482 Checking 1234567890abcdefghijklmnopqrstuvwxyz has a shannon entrophy of 5.169925004424 Checking Rosetta Code has a shannon entrophy of 3.084962500407 ## Haskell import Data.List main = print  entropy "1223334444" entropy :: (Ord a, Floating c) => [a] -> c entropy = sum . map lg . fq . map genericLength . group . sort where lg c = -c * logBase 2 c fq c = let sc = sum c in map (/ sc) c  Or, inlining with an applicative expression (turns out to be fractionally faster): import Data.List (genericLength, group, sort) entropy :: (Ord a, Floating c) => [a] -> c entropy = sum . map (negate . ((*) <*> logBase 2)) . (map =<< flip (/) . sum) . map genericLength . group . sort main :: IO () main = print  entropy "1223334444"  Output: 1.8464393446710154 ## Icon and Unicon Hmmm, the 2nd equation sums across the length of the string (for the example, that would be the sum of 10 terms). However, the answer cited is for summing across the different characters in the string (sum of 4 terms). The code shown here assumes the latter and works in Icon and Unicon. This assumption is consistent with the Wikipedia description. procedure main(a) s := !a | "1223334444" write(H(s)) end procedure H(s) P := table(0.0) every P[!s] +:= 1.0/*s every (h := 0.0) -:= P[c := key(P)] * log(P[c],2) return h end  Output: ->en 1.846439344671015 ->  ## J Solution:  entropy=: +/@(-@* 2&^.)@(#/.~ % #)  Example:  entropy '1223334444' 1.84644 entropy i.256 8 entropy 2569 0 entropy 2560 1 1 entropy 2560 1 2 3 2  So it looks like entropy is roughly the number of bits which would be needed to distinguish between each item in the argument (for example, with perfect compression). Note that in some contexts this might not be the same thing as information because the choice of the items themselves might matter. But it's good enough in contexts with a fixed set of symbols. ## Java Translation of: NetRexx Translation of: REXX Works with: Java version 7+ import java.lang.Math; import java.util.Map; import java.util.HashMap; public class REntropy { @SuppressWarnings("boxing") public static double getShannonEntropy(String s) { int n = 0; Map<Character, Integer> occ = new HashMap<>(); for (int c_ = 0; c_ < s.length(); ++c_) { char cx = s.charAt(c_); if (occ.containsKey(cx)) { occ.put(cx, occ.get(cx) + 1); } else { occ.put(cx, 1); } ++n; } double e = 0.0; for (Map.Entry<Character, Integer> entry : occ.entrySet()) { char cx = entry.getKey(); double p = (double) entry.getValue() / n; e += p * log2(p); } return -e; } private static double log2(double a) { return Math.log(a) / Math.log(2); } public static void main(String[] args) { String[] sstr = { "1223334444", "1223334444555555555", "122333", "1227774444", "aaBBcccDDDD", "1234567890abcdefghijklmnopqrstuvwxyz", "Rosetta Code", }; for (String ss : sstr) { double entropy = REntropy.getShannonEntropy(ss); System.out.printf("Shannon entropy of %40s: %.12f%n", "\"" + ss + "\"", entropy); } return; } } Output: Shannon entropy of "1223334444": 1.846439344671 Shannon entropy of "1223334444555555555": 1.969811065278 Shannon entropy of "122333": 1.459147917027 Shannon entropy of "1227774444": 1.846439344671 Shannon entropy of "aaBBcccDDDD": 1.936260027532 Shannon entropy of "1234567890abcdefghijklmnopqrstuvwxyz": 5.169925001442 Shannon entropy of "Rosetta Code": 3.084962500721  ## JavaScript Works with: ECMAScript 2015 Calculate the entropy of a string by determining the frequency of each character, then summing each character's probability multiplied by the log base 2 of that same probability, taking the negative of the sum. // Shannon entropy in bits per symbol. function entropy(str) { const len = str.length // Build a frequency map from the string. const frequencies = Array.from(str) .reduce((freq, c) => (freq[c] = (freq[c] || 0) + 1) && freq, {}) // Sum the frequency of each character. return Object.values(frequencies) .reduce((sum, f) => sum - f/len * Math.log2(f/len), 0) } console.log(entropy('1223334444')) // 1.8464393446710154 console.log(entropy('0')) // 0 console.log(entropy('01')) // 1 console.log(entropy('0123')) // 2 console.log(entropy('01234567')) // 3 console.log(entropy('0123456789abcdef')) // 4  Output: 1.8464393446710154 0 1 2 3 4 Another variant const entropy = (s) => { const split = s.split(''); const counter = {}; split.forEach(ch => { if (!counter[ch]) counter[ch] = 1; else counter[ch]++; }); const lengthf = s.length * 1.0; const counts = Object.values(counter); return -1 * counts .map(count => count / lengthf * Math.log2(count / lengthf)) .reduce((a, b) => a + b); };  Output: console.log(entropy("1223334444")); // 1.8464393446710154 ## jq For efficiency with long strings, we use a hash (a JSON object) to compute the frequencies. The helper function, counter, could be defined as an inner function of entropy, but for the sake of clarity and because it is independently useful, it is defined separately. # Input: an array of strings. # Output: an object with the strings as keys, the values of which are the corresponding frequencies. def counter: reduce .[] as item ( {}; .[item] += 1 ) ; # entropy in bits of the input string def entropy: (explode | map( [.] | implode ) | counter | [ .[] | . * log ] | add) as sum | ((length|log) - (sum / length)) / (2|log) ; Example: "1223334444" | entropy # => 1.8464393446710154 ## Jsish From Javascript entry. /* Shannon entropy, in Jsish */ function values(obj:object):array { var vals = []; for (var key in obj) vals.push(obj[key]); return vals; } function entropy(s) { var split = s.split(''); var counter = {}; split.forEach(function(ch) { if (!counter[ch]) counter[ch] = 1; else counter[ch]++; }); var lengthf = s.length * 1.0; var counts = values(counter); return -1 * counts.map(function(count) { return count / lengthf * (Math.log(count / lengthf) / Math.log(2)); }) .reduce(function(a, b) { return a + b; } ); }; if (Interp.conf('unitTest')) { ; entropy('1223334444'); ; entropy('Rosetta Code'); ; entropy('password'); }  Output: prompt jsish --U entropy.jsi entropy('1223334444') ==> 1.84643934467102 entropy('Rosetta Code') ==> 3.08496250072116 entropy('password') ==> 2.75 ## Julia Works with: Julia version 0.6 entropy(s) = -sum(x -> x * log(2, x), count(x -> x == c, s) / length(s) for c in unique(s)) @show entropy("1223334444") @show entropy([1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5])  Output: entropy("1223334444") = 1.8464393446710154 entropy([1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5]) = 2.103909910282364 ## K Works with: ngn/k entropy: {(ln[#x]-(+/{x*ln@x}@+/{x=\:?x}x)%#x)%ln@2} entropy "1223334444"  Output: 1.8464393446710161 ## Kotlin // version 1.0.6 fun log2(d: Double) = Math.log(d) / Math.log(2.0) fun shannon(s: String): Double { val counters = mutableMapOf<Char, Int>() for (c in s) { if (counters.containsKey(c)) counters[c] = counters[c]!! + 1 else counters.put(c, 1) } val nn = s.length.toDouble() var sum = 0.0 for (key in counters.keys) { val term = counters[key]!! / nn sum += term * log2(term) } return -sum } fun main(args: Array<String>) { val samples = arrayOf( "1223334444", "1223334444555555555", "122333", "1227774444", "aaBBcccDDDD", "1234567890abcdefghijklmnopqrstuvwxyz", "Rosetta Code" ) println(" String Entropy") println("------------------------------------ ------------------") for (sample in samples) println("{sample.padEnd(36)} -> {"%18.16f".format(shannon(sample))}") }  Output:  String Entropy ------------------------------------ ------------------ 1223334444 -> 1.8464393446710154 1223334444555555555 -> 1.9698110652780971 122333 -> 1.4591479170272448 1227774444 -> 1.8464393446710154 aaBBcccDDDD -> 1.9362600275315274 1234567890abcdefghijklmnopqrstuvwxyz -> 5.1699250014423095 Rosetta Code -> 3.0849625007211556  ## Ksh Works with: ksh93 function entropy { typeset -i i len={#1} typeset -X13 r=0 typeset -Ai counts for ((i = 0; i < len; ++i)) do counts[{1:i:1}]+=1 done for i in "{counts[@]}" do r+='i * log2(i)' done r='log2(len) - r / len' print -r -- "r" } printf '%g\n' "(entropy '1223334444')"  Output: 1.84644 ## Lambdatalk {def entropy {def entropy.count {lambda {:s :c :i} {let { {:c {/ {A.get :i :c} {A.length :s}}} } {* :c {log2 :c}}}}} {def entropy.sum {lambda {:s :c} {- {+ {S.map {entropy.count :s :c} {S.serie 0 {- {A.length :c} 1}}}}}}} {lambda {:s} {entropy.sum {A.split :s} {cdr {W.frequency :s}}}}} -> entropy The W.frequency function is explained in rosettacode.org/wiki/Letter_frequency#Lambdatalk {def txt 1223334444} -> txt {def F {W.frequency {txt}}} -> F characters: {car {F}} -> [1,2,3,4] frequencies: {cdr {F}} -> [1,2,3,4] {entropy {txt}} -> 1.8464393446710154 {entropy 0} -> 0 {entropy 00000000000000} -> 0 {entropy 11111111111111} -> 0 {entropy 01} -> 1 {entropy Lambdatalk} -> 2.8464393446710154 {entropy entropy} -> 2.807354922057604 {entropy abcdefgh} -> 3 {entropy Rosetta Code} -> 3.084962500721156 {entropy Longtemps je me suis couché de bonne heure} -> 3.8608288771249444 {entropy abcdefghijklmnopqrstuvwxyz} -> 4.70043971814109 {entropy abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz} -> 4.70043971814109  ## Lang5 : -rot rot rot ; [] '__A set : dip swap __A swap 1 compress append '__A set execute __A -1 extract nip ; : nip swap drop ; : sum '+ reduce ; : 2array 2 compress ; : comb "" split ; : lensize length nip ; : <group> #( a -- 'a ) grade subscript dup 's dress distinct strip length 1 2array reshape swap 'A set : filter(*) A in A swap select ; 'filter apply ; : elements(*) lensize ; : entropy #( s -- n ) length "<group> 'elements apply" dip / dup neg swap log * 2 log / sum ; "1223334444" comb entropy . # 1.84643934467102 ## Liberty BASIC dim countOfChar( 255) ' all possible one-byte ASCII chars source ="1223334444" charCount =len( source) usedChar ="" for i =1 to len( source) ' count which chars are used in source ch =mid( source, i, 1) if not( instr( usedChar, ch)) then usedChar =usedChar +ch 'currentCh =mid( j =instr( usedChar, ch) countOfChar( j) =countOfChar( j) +1 next i l =len( usedChar) for i =1 to l probability =countOfChar( i) /charCount entropy =entropy -( probability *logBase( probability, 2)) next i print " Characters used and the number of occurrences of each " for i =1 to l print " '"; mid( usedChar, i, 1); "'", countOfChar( i) next i print " Entropy of '"; source; "' is "; entropy; " bits." print " The result should be around 1.84644 bits." end function logBase( x, b) ' in LB log() is base 'e'. logBase =log( x) /log( 2) end function Output:  Characters used and the number of occurrences of each '1' 1 '2' 2 '3' 3 '4' 4 Entropy of '1223334444' is 1.84643934 bits. The result should be around 1.84644 bits. ## Lua function log2 (x) return math.log(x) / math.log(2) end function entropy (X) local N, count, sum, i = X:len(), {}, 0 for char = 1, N do i = X:sub(char, char) if count[i] then count[i] = count[i] + 1 else count[i] = 1 end end for n_i, count_i in pairs(count) do sum = sum + count_i / N * log2(count_i / N) end return -sum end print(entropy("1223334444"))  ## Mathematica / Wolfram Language shE[s_String] := -Plus @@ ((# Log[2., #]) & /@ ((Length /@ Gather[#])/ Length[#]) &[Characters[s]])  Example:  shE["1223334444"] 1.84644 shE["Rosetta Code"] 3.08496  ## MATLAB / Octave This version allows for any input vectors, including strings, floats, negative integers, etc. function E = entropy(d) if ischar(d), d=abs(d); end; [Y,I,J] = unique(d); H = sparse(J,1,1); p = full(H(H>0))/length(d); E = -sum(p.*log2(p)); end;  Usage: > entropy('1223334444') ans = 1.8464  ## MiniScript entropy = function(s) count = {} for c in s if count.hasIndex(c) then count[c] = count[c]+1 else count[c] = 1 end for sum = 0 for x in count.values countOverN = x / s.len sum = sum + countOverN * log(countOverN, 2) end for return -sum end function print entropy("1223334444")  Output: 1.846439 ## Modula-2 MODULE Entropy; FROM InOut IMPORT WriteString, WriteLn; FROM RealInOut IMPORT WriteReal; FROM Strings IMPORT Length; FROM MathLib IMPORT ln; PROCEDURE entropy(s: ARRAY OF CHAR): REAL; VAR freq: ARRAY [0..255] OF CARDINAL; i, length: CARDINAL; h, f: REAL; BEGIN (* the entropy of the empty string is zero *) length := Length(s); IF length = 0 THEN RETURN 0.0; END; (* find the frequency of each character *) FOR i := 0 TO 255 DO freq[i] := 0; END; FOR i := 0 TO length-1 DO INC(freq[ORD(s[i])]); END; (* calculate the component for each character *) h := 0.0; FOR i := 0 TO 255 DO IF freq[i] # 0 THEN f := FLOAT(freq[i]) / FLOAT(length); h := h - f * (ln(f) / ln(2.0)); END; END; RETURN h; END entropy; BEGIN WriteReal(entropy("1223334444"), 14); WriteLn; END Entropy.  Output:  1.8464394E+00 ## NetRexx Translation of: REXX /* NetRexx */ options replace format comments java crossref savelog symbols runSample(Arg) return -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ /* REXX *************************************************************** * 28.02.2013 Walter Pachl **********************************************************************/ method getShannonEntropy(s = "1223334444") public static --trace var occ c chars n cn i e p pl Numeric Digits 30 occ = 0 chars = '' n = 0 cn = 0 Loop i = 1 To s.length() c = s.substr(i, 1) If chars.pos(c) = 0 Then Do cn = cn + 1 chars = chars || c End occ[c] = occ[c] + 1 n = n + 1 End i p = '' Loop ci = 1 To cn c = chars.substr(ci, 1) p[c] = occ[c] / n End ci e = 0 Loop ci = 1 To cn c = chars.substr(ci, 1) pl = log2(p[c]) e = e + p[c] * pl End ci Return -e -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method log2(a = double) public static binary returns double return Math.log(a) / Math.log(2) -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(Arg) public static parse Arg sstr if sstr = '' then sstr = '1223334444' - '1223334444555555555' - '122333' - '1227774444' - 'aaBBcccDDDD' - '1234567890abcdefghijklmnopqrstuvwxyz' - 'Rosetta_Code' say 'Calculating Shannon''s entropy for the following list:' say '['(sstr.space(1, ',')).changestr(',', ', ')']' say entropies = 0 ssMax = 0 -- This crude sample substitutes a '_' character for a space in the input strings loop w_ = 1 to sstr.words() ss = sstr.word(w_) ssMax = ssMax.max(ss.length()) ss_ = ss.changestr('_', ' ') entropy = getShannonEntropy(ss_) entropies[ss] = entropy end w_ loop report = 1 to sstr.words() ss = sstr.word(report) ss_ = ss.changestr('_', ' ') Say 'Shannon entropy of' ('"'ss_'"').right(ssMax + 2)':' entropies[ss].format(null, 12) end report return Output: Calculating Shannon's entropy for the following list: [1223334444, 1223334444555555555, 122333, 1227774444, aaBBcccDDDD, 1234567890abcdefghijklmnopqrstuvwxyz, Rosetta_Code] Shannon entropy of "1223334444": 1.846439344671 Shannon entropy of "1223334444555555555": 1.969811065278 Shannon entropy of "122333": 1.459147917027 Shannon entropy of "1227774444": 1.846439344671 Shannon entropy of "aaBBcccDDDD": 1.936260027532 Shannon entropy of "1234567890abcdefghijklmnopqrstuvwxyz": 5.169925001442 Shannon entropy of "Rosetta Code": 3.084962500721  ## Nim import tables, math proc entropy(s: string): float = var t = initCountTable[char]() for c in s: t.inc(c) for x in t.values: result -= x/s.len * log2(x/s.len) echo entropy("1223334444")  ## Objeck use Collection; class Entropy { function : native : GetShannonEntropy(result : String) ~ Float { frequencies := IntMap->New(); each(i : result) { c := result->Get(i); if(frequencies->Has(c)) { count := frequencies->Find(c)->As(IntHolder); count->Set(count->Get() + 1); } else { frequencies->Insert(c, IntHolder->New(1)); }; }; length := result->Size(); entropy := 0.0; counts := frequencies->GetValues(); each(i : counts) { count := counts->Get(i)->As(IntHolder)->Get(); freq := count->As(Float) / length; entropy += freq * (freq->Log() / 2.0->Log()); }; return -1 * entropy; } function : Main(args : String[]) ~ Nil { inputs := [ "1223334444", "1223334444555555555", "122333", "1227774444", "aaBBcccDDDD", "1234567890abcdefghijklmnopqrstuvwxyz", "Rosetta Code"]; each(i : inputs) { input := inputs[i]; "Shannon entropy of '{input}': "->Print(); GetShannonEntropy(inputs[i])->PrintLine(); }; } } Output: Shannon entropy of '1223334444': 1.84644 Shannon entropy of '1223334444555555555': 1.96981 Shannon entropy of '122333': 1.45915 Shannon entropy of '1227774444': 1.84644 Shannon entropy of 'aaBBcccDDDD': 1.93626 Shannon entropy of '1234567890abcdefghijklmnopqrstuvwxyz': 5.16993 Shannon entropy of 'Rosetta Code': 3.08496  ## OCaml By using a map, purely functional module CharMap = Map.Make(Char) let entropy s = let count map c = CharMap.update c (function Some n -> Some (n +. 1.) | None -> Some 1.) map and calc _ n sum = sum +. n *. Float.log2 n in let sum = CharMap.fold calc (String.fold_left count CharMap.empty s) 0. and len = float (String.length s) in Float.log2 len -. sum /. len let () = entropy "1223334444" |> string_of_float |> print_endline  By using a mutable Hashtbl (* pre-bake & return an inner-loop function to bin & assemble a character frequency map *) let get_fproc (m: (char, int) Hashtbl.t) :(char -> unit) = (fun (c:char) -> try Hashtbl.replace m c ( (Hashtbl.find m c) + 1) with Not_found -> Hashtbl.add m c 1) (* pre-bake and return an inner-loop function to do the actual entropy calculation *) let get_calc (slen:int) :(float -> float) = let slen_float = float_of_int slen in let log_2 = log 2.0 in (fun v -> let pt = v /. slen_float in pt *. ((log pt) /. log_2) ) (* main function, given a string argument it: builds a (mutable) frequency map (initial alphabet size of 255, but it's auto-expanding), extracts the relative probability values into a list, folds-in the basic entropy calculation and returns the result. *) let shannon (s:string) :float = let freq_hash = Hashtbl.create 255 in String.iter (get_fproc freq_hash) s; let relative_probs = Hashtbl.fold (fun k v b -> (float v)::b) freq_hash [] in let calc = get_calc (String.length s) in -1.0 *. List.fold_left (fun b x -> b +. calc x) 0.0 relative_probs  Output: 1.84643934467 ## Oforth : entropy(s) -- f | freq sz | s size dup ifZero: [ return ] asFloat ->sz ListBuffer initValue(255, 0) ->freq s apply( #[ dup freq at 1+ freq put ] ) 0.0 freq applyIf( #[ 0 <> ], #[ sz / dup ln * - ] ) Ln2 / ; entropy("1223334444") . Output: 1.84643934467102 ## ooRexx Translation of: REXX /* REXX */ Numeric Digits 16 Parse Arg s If s='' Then s="1223334444" occ.=0 chars='' n=0 cn=0 Do i=1 To length(s) c=substr(s,i,1) If pos(c,chars)=0 Then Do cn=cn+1 chars=chars||c End occ.c=occ.c+1 n=n+1 End do ci=1 To cn c=substr(chars,ci,1) p.c=occ.c/n /* say c p.c */ End e=0 Do ci=1 To cn c=substr(chars,ci,1) e=e+p.c*rxcalclog(p.c)/rxcalclog(2) End Say s 'Entropy' format(-e,,12) Exit ::requires 'rxmath' LIBRARY Output: 1223334444 Entropy 1.846439344671 ## PARI/GP entropy(s)=s=Vec(s);my(v=vecsort(s,,8));-sum(i=1,#v,(x->x*log(x))(sum(j=1,#s,v[i]==s[j])/#s))/log(2) >entropy("1223334444") %1 = 1.8464393446710154934341977463050452232 ## Pascal Free Pascal (http://freepascal.org). PROGRAM entropytest; USES StrUtils, Math; TYPE FArray = ARRAY of CARDINAL; VAR strng: STRING = '1223334444'; // list unique characters in a string FUNCTION uniquechars(str: STRING): STRING; VAR n: CARDINAL; BEGIN uniquechars := ''; FOR n := 1 TO length(str) DO IF (PosEx(str[n],str,n)>0) AND (PosEx(str[n],uniquechars,1)=0) THEN uniquechars += str[n]; END; // obtain a list of character-frequencies for a string // given a string containing its unique characters FUNCTION frequencies(str,ustr: STRING): FArray; VAR u,s,p,o: CARDINAL; BEGIN SetLength(frequencies, Length(ustr)+1); p := 0; FOR u := 1 TO length(ustr) DO FOR s := 1 TO length(str) DO BEGIN o := p; p := PosEx(ustr[u],str,s); IF (p>o) THEN INC(frequencies[u]); END; END; // Obtain the Shannon entropy of a string FUNCTION entropy(s: STRING): EXTENDED; VAR pf : FArray; us : STRING; i,l: CARDINAL; BEGIN us := uniquechars(s); pf := frequencies(s,us); l := length(s); entropy := 0.0; FOR i := 1 TO length(us) DO entropy -= pf[i]/l * log2(pf[i]/l); END; BEGIN Writeln('Entropy of "',strng,'" is ',entropy(strng):2:5, ' bits.'); END.  Output: Entropy of "1223334444" is 1.84644 bits.  ## Perl sub entropy { my %count; count{_}++ for @_; my entropy = 0; for (values %count) { my p = _/@_; entropy -= p * log p; } entropy / log 2 } print entropy split //, "1223334444";  ## Phix with javascript_semantics function entropy(sequence s) sequence symbols = {}, counts = {} integer N = length(s) for i=1 to N do object si = s[i] integer k = find(si,symbols) if k=0 then symbols = append(symbols,si) counts = append(counts,1) else counts[k] += 1 end if end for atom H = 0 integer n = length(counts) for i=1 to n do atom ci = counts[i]/N H -= ci*log2(ci) end for return H end function ?entropy("1223334444")  Output: 1.846439345  ## PHP <?php function shannonEntropy(string) { h = 0.0; len = strlen(string); foreach (count_chars(string, 1) as count) { h -= (double) (count / len) * log((double) (count / len), 2); } return h; } strings = array( '1223334444', '1225554444', 'aaBBcccDDDD', '122333444455555', 'Rosetta Code', '1234567890abcdefghijklmnopqrstuvwxyz', ); foreach (strings AS string) { printf( '%36s : %s' . PHP_EOL, string, number_format(shannonEntropy(string), 6) ); }  Output:  1223334444 : 1.846439 1225554444 : 1.846439 aaBBcccDDDD : 1.936260 122333444455555 : 2.149255 Rosetta Code : 3.084963 1234567890abcdefghijklmnopqrstuvwxyz : 5.169925 ## Picat go => ["1223334444", "Rosetta Code is the best site in the world!", "1234567890abcdefghijklmnopqrstuvwxyz", "Picat is fun"].map(entropy).println(), nl. % probabilities of each element/character in L entropy(L) = Entropy => Len = L.length, Occ = new_map(), % # of occurrences foreach(E in L) Occ.put(E, Occ.get(E,0) + 1) end, Entropy = -sum([P2*log2(P2) : _C=P in Occ, P2 = P/Len]). Output: [1.846439344671016,3.646513010214172,5.169925001442309,3.251629167387823] ## PicoLisp PicoLisp only supports fixed point arithmetic, but it does have the ability to call libc transcendental functions (for log) (scl 8) (load "@lib/math.l") (setq LN2 0.693147180559945309417) (de tabulate-chars (Str) (let Map NIL (for Ch (chop Str) (if (assoc Ch Map) (con @ (inc (cdr @))) (setq Map (cons (cons Ch 1) Map)))) Map)) (de entropy (Str) (let ( Sz (length Str) Hist (tabulate-chars Str) ) (*/ (sum '((Pair) (let R (*/ (cdr Pair) 1. Sz) (- (*/ R (log R) 1.)))) Hist) 1. LN2))) Output: : (format (entropy "1223334444") *Scl) -> "1.84643934"  ## PL/I *process source xref attributes or(!); /*-------------------------------------------------------------------- * 08.08.2014 Walter Pachl translated from REXX version 1 *-------------------------------------------------------------------*/ ent: Proc Options(main); Dcl (index,length,log2,substr) Builtin; Dcl sysprint Print; Dcl occ(100) Bin fixed(31) Init((100)0); Dcl (n,cn,ci,i,pos) Bin fixed(31) Init(0); Dcl chars Char(100) Var Init(''); Dcl s Char(100) Var Init('1223334444'); Dcl c Char(1); Dcl (occf,p(100)) Dec Float(18); Dcl e Dec Float(18) Init(0); Do i=1 To length(s); c=substr(s,i,1); pos=index(chars,c); If pos=0 Then Do; pos=length(chars)+1; cn+=1; chars=chars!!c; End; occ(pos)+=1; n+=1; End; do ci=1 To cn; occf=occ(ci); p(ci)=occf/n; End; Do ci=1 To cn; e=e+p(ci)*log2(p(ci)); End; Put Edit('s='''!!s!!''' Entropy=',-e)(Skip,a,f(15,12)); End; Output: s='1223334444' Entropy= 1.846439344671 ## PowerShell function entropy (string) { n = string.Length string.ToCharArray() | group | foreach{ p = _.Count/n i = [Math]::Log(p,2) -p*i } | measure -Sum | foreach Sum } entropy "1223334444"  Output: 1.84643934467102  ## Prolog Works with: Swi-Prolog version 7.3.3 This solution calculates the run-length encoding of the input string to get the relative frequencies of its characters. :-module(shannon_entropy, [shannon_entropy/2]). %! shannon_entropy(+String, -Entropy) is det. % % Calculate the Shannon Entropy of String. % % Example query: % == % ?- shannon_entropy(1223334444, H). % H = 1.8464393446710154. % == % shannon_entropy(String, Entropy):- atom_chars(String, Cs) ,relative_frequencies(Cs, Frequencies) ,findall(CI ,(member(_C-F, Frequencies) ,log2(F, L) ,CI is F * L ) ,CIs) ,foldl(sum, CIs, 0, E) ,Entropy is -E. %! frequencies(+Characters,-Frequencies) is det. % % Calculates the relative frequencies of elements in the list of % Characters. % % Frequencies is a key-value list with elements of the form: % C-F, where C a character in the list and F its relative % frequency in the list. % % Example query: % == % ?- relative_frequencies([a,a,a,b,b,b,b,b,b,c,c,c,a,a,f], Fs). % Fs = [a-0.3333333333333333, b-0.4, c-0.2,f-0.06666666666666667]. % == % relative_frequencies(List, Frequencies):- run_length_encoding(List, Rle) % Sort Run-length encoded list and aggregate lengths by element ,keysort(Rle, Sorted_Rle) ,group_pairs_by_key(Sorted_Rle, Elements_Run_lengths) ,length(List, Elements_in_list) ,findall(E-Frequency_of_E ,(member(E-RLs, Elements_Run_lengths) % Sum the list of lengths of runs of E ,foldl(plus, RLs, 0, Occurences_of_E) ,Frequency_of_E is Occurences_of_E / Elements_in_list ) ,Frequencies). %! run_length_encoding(+List, -Run_length_encoding) is det. % % Converts a list to its run-length encoded form where each "run" % of contiguous repeats of the same element is replaced by that % element and the length of the run. % % Run_length_encoding is a key-value list, where each element is a % term: % % Element:term-Repetitions:number. % % Example query: % == % ?- run_length_encoding([a,a,a,b,b,b,b,b,b,c,c,c,a,a,f], RLE). % RLE = [a-3, b-6, c-3, a-2, f-1]. % == % run_length_encoding([], []-0):- !. % No more results needed. run_length_encoding([Head|List], Run_length_encoded_list):- run_length_encoding(List, [Head-1], Reversed_list) % The resulting list is in reverse order due to the head-to-tail processing ,reverse(Reversed_list, Run_length_encoded_list). %! run_length_encoding(+List,+Initialiser,-Accumulator) is det. % % Business end of run_length_encoding/3. Calculates the run-length % encoded form of a list and binds the result to the Accumulator. % Initialiser is a list [H-1] where H is the first element of the % input list. % run_length_encoding([], Fs, Fs). % Run of F consecutive occurrences of C run_length_encoding([C|Cs],[C-F|Fs], Acc):- % Backtracking would produce successive counts % of runs of C at different indices in the list. ! ,F_ is F + 1 ,run_length_encoding(Cs, [C-F_| Fs], Acc). % End of a run of consecutive identical elements. run_length_encoding([C|Cs], Fs, Acc):- run_length_encoding(Cs,[C-1|Fs], Acc). /* Arithmetic helper predicates */ %! log2(N, L2_N) is det. % % L2_N is the logarithm with base 2 of N. % log2(N, L2_N):- L_10 is log10(N) ,L_2 is log10(2) ,L2_N is L_10 / L_2. %! sum(+A,+B,?Sum) is det. % % True when Sum is the sum of numbers A and B. % % Helper predicate to allow foldl/4 to do addition. The following % call will raise an error (because there is no predicate +/3): % == % foldl(+, [1,2,3], 0, Result). % == % % This will not raise an error: % == % foldl(sum, [1,2,3], 0, Result). % == % sum(A, B, Sum):- must_be(number, A) ,must_be(number, B) ,Sum is A + B.  Example query: ?- shannon_entropy(1223334444, H). H = 1.8464393446710154.  ## PureBasic #TESTSTR="1223334444" NewMap uchar.i() : Define.d e Procedure.d nlog2(x.d) : ProcedureReturn Log(x)/Log(2) : EndProcedure Procedure countchar(s, Map uchar()) If Len(s) uchar(Left(s,1))=CountString(s,Left(s,1)) s=RemoveString(s,Left(s,1)) ProcedureReturn countchar(s, uchar()) EndIf EndProcedure countchar(#TESTSTR,uchar()) ForEach uchar() e-uchar()/Len(#TESTSTR)*nlog2(uchar()/Len(#TESTSTR)) Next OpenConsole() Print("Entropy of ["+#TESTSTR+"] = "+StrD(e,15)) Input() Output: Entropy of [1223334444] = 1.846439344671015 ## Python ### Python: Longer version from __future__ import division import math def hist(source): hist = {}; l = 0; for e in source: l += 1 if e not in hist: hist[e] = 0 hist[e] += 1 return (l,hist) def entropy(hist,l): elist = [] for v in hist.values(): c = v / l elist.append(-c * math.log(c ,2)) return sum(elist) def printHist(h): flip = lambda (k,v) : (v,k) h = sorted(h.iteritems(), key = flip) print 'Sym\thi\tfi\tInf' for (k,v) in h: print '%s\t%f\t%f\t%f'%(k,v,v/l,-math.log(v/l, 2)) source = "1223334444" (l,h) = hist(source); print '.[Results].' print 'Length',l print 'Entropy:', entropy(h, l) printHist(h)  Output: .[Results]. Length 10 Entropy: 1.84643934467 Sym hi fi Inf 1 1.000000 0.100000 3.321928 2 2.000000 0.200000 2.321928 3 3.000000 0.300000 1.736966 4 4.000000 0.400000 1.321928  ### Python: More succinct version The Counter module is only available in Python >= 2.7. from math import log2 from collections import Counter def entropy(s): p, lns = Counter(s), float(len(s)) return log2(lns) - sum(count * log2(count) for count in p.values()) / lns print(entropy("1223334444"))  Output: 1.8464393446710154 ### Uses Python 2 def Entropy(text): import math log2=lambda x:math.log(x)/math.log(2) exr={} infoc=0 for each in text: try: exr[each]+=1 except: exr[each]=1 textlen=len(text) for k,v in exr.items(): freq = 1.0*v/textlen infoc+=freq*log2(freq) infoc*=-1 return infoc while True: print Entropy(raw_input('>>>'))  ## R entropy <- function(str) { vec <- strsplit(str, "")[[1]] N <- length(vec) p_xi <- table(vec) / N -sum(p_xi * log(p_xi, 2)) } Output: > entropy("1223334444") [1] 1.846439  ## Racket #lang racket (require math) (provide entropy hash-entropy list-entropy digital-entropy) (define (hash-entropy h) (define (log2 x) (/ (log x) (log 2))) (define n (for/sum [(c (in-hash-values h))] c)) (- (for/sum ([c (in-hash-values h)] #:unless (zero? c)) (* (/ c n) (log2 (/ c n)))))) (define (list-entropy x) (hash-entropy (samples->hash x))) (define entropy (compose list-entropy string->list)) (define digital-entropy (compose entropy number->string)) (module+ test (require rackunit) (check-= (entropy "1223334444") 1.8464393446710154 1E-8) (check-= (digital-entropy 1223334444) (entropy "1223334444") 1E-8) (check-= (digital-entropy 1223334444) 1.8464393446710154 1E-8) (check-= (entropy "xggooopppp") 1.8464393446710154 1E-8)) (module+ main (entropy "1223334444"))  Output:  1.8464393446710154 ## Raku (formerly Perl 6) Works with: rakudo version 2015-09-09 sub entropy(@a) { [+] map -> \p { p * -log p }, bag(@a).values »/» +@a; } say log(2) R/ entropy '1223334444'.comb;  Output: 1.84643934467102 In case we would like to add this function to Raku's core, here is one way it could be done: use MONKEY-TYPING; augment class Bag { method entropy { [+] map -> \p { - p * log p }, self.values »/» +self; } } say '1223334444'.comb.Bag.entropy / log 2;  ## REXX ### version 1 /* REXX *************************************************************** * 28.02.2013 Walter Pachl * 12.03.2013 Walter Pachl typo in log corrected. thanx for testing * 22.05.2013 -"- extended the logic to accept other strings * 25.05.2013 -"- 'my' log routine is apparently incorrect * 25.05.2013 -"- problem identified & corrected **********************************************************************/ Numeric Digits 30 Parse Arg s If s='' Then s="1223334444" occ.=0 chars='' n=0 cn=0 Do i=1 To length(s) c=substr(s,i,1) If pos(c,chars)=0 Then Do cn=cn+1 chars=chars||c End occ.c=occ.c+1 n=n+1 End do ci=1 To cn c=substr(chars,ci,1) p.c=occ.c/n /* say c p.c */ End e=0 Do ci=1 To cn c=substr(chars,ci,1) e=e+p.c*log(p.c,30,2) End Say 'Version 1:' s 'Entropy' format(-e,,12) Exit log: Procedure /*********************************************************************** * Return log(x) -- with specified precision and a specified base * Three different series are used for the ranges 0 to 0.5 * 0.5 to 1.5 * 1.5 to infinity * 03.09.1992 Walter Pachl * 25.05.2013 -"- 'my' log routine is apparently incorrect * 25.05.2013 -"- problem identified & corrected ***********************************************************************/ Parse Arg x,prec,b If prec='' Then prec=9 Numeric Digits (2*prec) Numeric Fuzz 3 Select When x<=0 Then r='*** invalid argument ***' When x<0.5 Then Do z=(x-1)/(x+1) o=z r=z k=1 Do i=3 By 2 ra=r k=k+1 o=o*z*z r=r+o/i If r=ra Then Leave End r=2*r End When x<1.5 Then Do z=(x-1) o=z r=z k=1 Do i=2 By 1 ra=r k=k+1 o=-o*z r=r+o/i If r=ra Then Leave End End Otherwise /* 1.5<=x */ Do z=(x+1)/(x-1) o=1/z r=o k=1 Do i=3 By 2 ra=r k=k+1 o=o/(z*z) r=r+o/i If r=ra Then Leave End r=2*r End End If b<>'' Then r=r/log(b,prec) Numeric Digits (prec) r=r+0 Return r  ### version 2 REXX doesn't have a BIF for LOG or LN, so the subroutine (function) LOG2 is included herein. The LOG2 subroutine is only included here for functionality, not to document how to calculate LOG2 using REXX. /*REXX program calculates the information entropy for a specified character string. */ numeric digits length( e() ) % 2 - length(.) /*use 1/2 of the decimal digits of E. */ parse arg ; if ='' then = 1223334444 /*obtain the optional input from the CL*/ #=0; @.= 0; L= length() /*define handy-dandy REXX variables. */$$=                                              /*initialize the   $$list. */ do j=1 for L; _= substr(, j, 1) /*process each character in  string.*/ if @._==0 then do; #= # + 1 /*Unique? Yes, bump character counter.*/$$= $$|| _ /*add this character to the$$  list. */
end
@._= @._ + 1                              /*keep track of this character's count.*/
end   /*j*/
sum= 0                                           /*calculate info entropy for each char.*/
do i=1  for #;        _= substr(, i, 1) /*obtain a character from unique list. */
sum= sum  -   @._/L * log2(@._/L)         /*add (negatively) the char entropies. */
end   /*i*/
say ' input string: '   $say 'string length: ' L say ' unique chars: ' #; say say 'the information entropy of the string ──► ' format(sum,,12) " bits." exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ e: e= 2.718281828459045235360287471352662497757247093699959574966967627724076630; return e /*──────────────────────────────────────────────────────────────────────────────────────*/ log2: procedure; parse arg x 1 ox; ig= x>1.5; ii= 0; is= 1 - 2 * (ig\==1) numeric digits digits()+5; call e /*the precision of E must be≥digits(). */ do while ig & ox>1.5 | \ig&ox<.5; _= e; do j=-1; iz= ox * _ ** -is if j>=0 & (ig & iz<1 | \ig&iz>.5) then leave; _= _ * _; izz= iz; end /*j*/ ox=izz; ii=ii+is*2**j; end /*while*/; x= x * e** -ii -1; z= 0; _= -1; p= z do k=1; _= -_ * x; z= z+_/k; if z=p then leave; p= z; end /*k*/ r= z + ii; if arg()==2 then return r; return r / log2(2, .)  output when using the default input of: 1223334444  input string: 1223334444 string length: 10 unique chars: 4 the information entropy of the string ──► 1.846439344671 bits.  output when using the input of: Rosetta Code  input string: Rosetta Code string length: 12 unique chars: 9 the information entropy of the string ──► 3.084962500721 bits.  ## Ring decimals(8) entropy = 0 countOfChar = list(255) source="1223334444" charCount =len( source) usedChar ="" for i =1 to len( source) ch =substr(source, i, 1) if not(substr( usedChar, ch)) usedChar =usedChar +ch ok j =substr( usedChar, ch) countOfChar[j] =countOfChar[j] +1 next l =len(usedChar) for i =1 to l probability =countOfChar[i] /charCount entropy =entropy - (probability *logBase(probability, 2)) next see "Characters used and the number of occurrences of each " + nl for i =1 to l see "'" + substr(usedChar, i, 1) + "' " + countOfChar[i] + nl next see " Entropy of " + source + " is " + entropy + " bits." + nl see " The result should be around 1.84644 bits." + nl func logBase (x, b) logBase =log( x) /log( 2) return logBase Output: Characters used and the number of occurrences of each '1' 1 '2' 2 '3' 3 '4' 4 Entropy of 1223334444 is 1.84643934 bits. The result should be around 1.84644 bits.  ## RPL Works with: Halcyon Calc version 4.2.7 Code Comments ≪ DUP SIZE 2 LN → str len log2 ≪ { 255 } 0 CON 1 len FOR j str j DUP SUB NUM DUP2 GET 1 + PUT NEXT 0 1 255 FOR j IF OVER j GET THEN LAST len / DUP LN log2 / * + END NEXT NEG SWAP DROP ≫ ≫ 'NTROP' STO  NTROP ( "string" -- entropy ) Initialize local variables Initialize a vector with 255 counters For each character in the string... ... increase the counter according to ASCII code For each non-zero counter calculate term Change sign and forget the vector  The following line of code delivers what is required: "1223334444" NTROP  Output: 1: 1.84643934467  ## Ruby def entropy(s) counts = s.chars.tally leng = s.length.to_f counts.values.reduce(0) do |entropy, count| freq = count / leng entropy - freq * Math.log2(freq) end end p entropy("1223334444")  Output: 1.8464393446710154  ## Run BASIC dim chrCnt( 255) ' possible ASCII chars source$		= "1223334444"
numChar		= len(source$) for i = 1 to len(source$)   		' count which chars are used in source
ch$= mid$(source$,i,1) if not( instr(chrUsed$, ch$)) then chrUsed$ = chrUsed$+ ch$
j	= instr(chrUsed$, ch$)
chrCnt(j) =chrCnt(j) +1
next i

lc	= len(chrUsed$) for i = 1 to lc odds = chrCnt(i) /numChar entropy = entropy - (odds * (log(odds) / log(2))) next i print " Characters used and times used of each " for i = 1 to lc print " '"; mid$(chrUsed$,i,1); "'";chr$(9);chrCnt(i)
next i

print " Entropy of '"; source$; "' is "; entropy; " bits." end Characters used and times used of each '1' 1 '2' 2 '3' 3 '4' 4 Entropy of '1223334444' is 1.84643939 bits.  ## Rust fn entropy(s: &[u8]) -> f32 { let mut histogram = [0u64; 256]; for &b in s { histogram[b as usize] += 1; } histogram .iter() .cloned() .filter(|&h| h != 0) .map(|h| h as f32 / s.len() as f32) .map(|ratio| -ratio * ratio.log2()) .sum() } fn main() { let arg = std::env::args().nth(1).expect("Need a string."); println!("Entropy of {} is {}.", arg, entropy(arg.as_bytes())); }  Output: $ ./entropy 1223334444
Entropy of 1223334444 is 1.8464394.


## Scala

import scala.math._

def entropy( v:String ) = { v
.groupBy (a => a)
.values
.map( i => i.length.toDouble / v.length )
.map( p => -p * log10(p) / log10(2))
.sum
}

// Confirm that "1223334444" has an entropy of about 1.84644
assert( math.round( entropy("1223334444") * 100000 ) * 0.00001 == 1.84644 )


## scheme

A version capable of calculating multidimensional entropy.

(define (entropy input)
(define (close? a b)
(define (norm x y)
(define (infinite_norm m n)
(define (absminus p q)
(cond ((null? p) '())
(else (cons (abs (- (car p) (car q))) (absminus (cdr p) (cdr q))))))
(define (mm l)
(cond ((null? (cdr l)) (car l))
((> (car l) (cadr l)) (mm (cons (car l) (cddr l))))
(else (mm (cdr l)))))
(mm (absminus m n)))
(if (pair? x) (infinite_norm x y) (abs (- x y))))
(let ((epsilon 0.2))
(< (norm a b) epsilon)))
(define (freq-list x)
(define (f x)
(define (count a b)
(cond ((null? b) 1)
(else (+ (if (close? a (car b)) 1 0) (count a (cdr b))))))
(let ((t (car x)) (tt (cdr x)))
(count t tt)))
(define (g x)
(define (filter a b)
(cond ((null? b) '())
((close? a (car b)) (filter a (cdr b)))
(else (cons (car b) (filter a (cdr b))))))
(let ((t (car x)) (tt (cdr x)))
(filter t tt)))
(cond ((null? x) '())
(else (cons (f x) (freq-list (g x))))))
(define (scale x)
(define (sum x)
(if (null? x) 0.0 (+ (car x) (sum (cdr x)))))
(let ((z (sum x)))
(map (lambda(m) (/ m z)) x)))
(define (cal x)
(if (null? x) 0 (+ (* (car x) (/ (log (car x)) (log 2))) (cal (cdr x)))))
(- (cal (scale (freq-list input)))))

(entropy (list 1 2 2 3 3 3 4 4 4 4))
(entropy (list (list 1 1) (list 1.1 1.1) (list 1.2 1.2) (list 1.3 1.3) (list 1.5 1.5) (list 1.6 1.6)))

Output:
1.8464393446710154 bits

1.4591479170272448 bits


## Scilab

function E = entropy(d)
d=strsplit(d);
n=unique(string(d));
N=size(d,'r');

count=zeros(n);
n_size = size(n,'r');
for i = 1:n_size
count(i) = sum ( d == n(i) );
end

E=0;
for i=1:length(count)
E = E - count(i)/N * log(count(i)/N) / log(2);
end
endfunction

word ='1223334444';
E = entropy(word);
disp('The entropy of '+word+' is '+string(E)+'.');

Output:
 The entropy of 1223334444 is 1.8464393.

$include "seed7_05.s7i"; include "float.s7i"; include "math.s7i"; const func float: entropy (in string: stri) is func result var float: entropy is 0.0; local var hash [char] integer: count is (hash [char] integer).value; var char: ch is ' '; var float: p is 0.0; begin for ch range stri do if ch in count then incr(count[ch]); else count @:= [ch] 1; end if; end for; for key ch range count do p := flt(count[ch]) / flt(length(stri)); entropy -:= p * log(p) / log(2.0); end for; end func ; const proc: main is func begin writeln(entropy("1223334444") digits 5); end func; Output: 1.84644  ## SETL program shannon_entropy; print(entropy "1223334444"); op entropy(symbols); hist := {}; loop for symbol in symbols do hist(symbol) +:= 1; end loop; h := 0.0; loop for count = hist(symbol) do f := count / #symbols; h -:= f * log f / log 2; end loop; return h; end op; end program; Output: 1.84643934467102 ## Sidef func entropy(s) { var counts = Hash.new; s.each { |c| counts{c} := 0 ++ }; var len = s.len; [0, counts.values.map {|count| var freq = count/len; freq * freq.log2 }... ]«-»; } say entropy("1223334444");  Output: 1.846439344671015493434197746305045223237 ## Standard ML val Entropy = fn input => let val N = Real.fromInt (String.size input) ; val term = fn a => Math.ln (a/N) * a / ( Math.ln 2.0 * N ) ; val v0 = Vector.tabulate (255,fn i=>0) ; val freq = Vector.map Real.fromInt (* List.foldr: count occurrences *) (List.foldr (fn (i,v) => Vector.update( v, ord i, Vector.sub(v,ord i) + 1) ) v0 (explode input) ) in ~ (Vector.foldr (fn (a,s) => if a > 0.0 then term a + s else s) 0.0 freq ) end ; Entropy "1223334444" ; val it = 1.846439345: real  ## Swift import Foundation func entropy(of x: String) -> Double { return x .reduce(into: [String: Int](), {cur, char in cur[String(char), default: 0] += 1 }) .values .map({i in Double(i) / Double(x.count) } as (Int) -> Double) .map({p in -p * log2(p) } as (Double) -> Double) .reduce(0.0, +) } print(entropy(of: "1223334444"))  Output: 1.8464393446710154 ## Tcl proc entropy {str} { set log2 [expr log(2)] foreach char [split$str ""] {dict incr counts $char} set entropy 0.0 foreach count [dict values$counts] {
set freq [expr {$count / double([string length$str])}]
set entropy [expr {$entropy -$freq * log($freq)/$log2}]
}
return $entropy }  Demonstration: puts [format "entropy = %.5f" [entropy "1223334444"]] puts [format "entropy = %.5f" [entropy "Rosetta Code"]]  Output: entropy = 1.84644 entropy = 3.08496  ## V (Vlang) ### Vlang: Map version import math import arrays fn hist(source string) map[string]int { mut hist := map[string]int{} for e in source.split('') { if e !in hist { hist[e] = 0 } hist[e]+=1 } return hist } fn entropy(hist map[string]int, l int) f64 { mut elist := []f64{} for _,v in hist { c := f64(v) / f64(l) elist << -c * math.log2(c) } return arrays.sum<f64>(elist) or {-1} } fn main(){ input := "1223334444" h := hist(input) e := entropy(h, input.len) println(e) } Output: 1.8464393446710152  ## Wren Translation of: Go var s = "1223334444" var m = {} for (c in s) { var d = m[c] m[c] = (d) ? d + 1 : 1 } var hm = 0 for (k in m.keys) { var c = m[k] hm = hm + c * c.log2 } var l = s.count System.print(l.log2 - hm/l)  Output: 1.846439344671  ## XPL0 code real RlOut=48, Ln=54; \intrinsic routines string 0; \use zero-terminated strings func StrLen(A); \Return number of characters in an ASCIIZ string char A; int I; for I:= 0, -1>>1-1 do if A(I) = 0 then return I; func real Entropy(Str); \Return Shannon entropy of string char Str; int Len, I, Count(128); real Sum, Prob; [Len:= StrLen(Str); for I:= 0 to 127 do Count(I):= 0; for I:= 0 to Len-1 do \count number of each character in string Count(Str(I)):= Count(Str(I)) + 1; Sum:= 0.0; for I:= 0 to 127 do if Count(I) # 0 then \(avoid Ln(0.0) error) [Prob:= float(Count(I)) / float(Len); \probability of char in string Sum:= Sum + Prob*Ln(Prob); ]; return -Sum/Ln(2.0); ]; RlOut(0, Entropy("1223334444")) Output:  1.84644  ## Zig const std = @import("std"); const math = std.math; pub fn main() !void { const stdout = std.io.getStdOut().outStream(); try stdout.print("{d:.12}\n", .{H("1223334444")}); } fn H(s: []const u8) f64 { var counts = [_]u16{0} ** 256; for (s) |ch| counts[ch] += 1; var h: f64 = 0; for (counts) |c| if (c != 0) { const p = @intToFloat(f64, c) / @intToFloat(f64, s.len); h -= p * math.log2(p); }; return h; }  Output: 1.846439344671  ## zkl Translation of: D fcn entropy(text){ text.pump(Void,fcn(c,freq){ c=c.toAsc(); freq[c]+=1; freq } .fp1( (0).pump(256,List,0.0).copy() )) // array[256] of 0.0 .filter() // remove all zero entries from array .apply('/(text.len())) // (num of char)/len .apply(fcn(p){-p*p.log()}) // |p*ln(p)| .sum(0.0)/(2.0).log(); // sum * ln(e)/ln(2) to convert to log2 } entropy("1223334444").println(" bits"); Output: 1.84644 bits  ## ZX Spectrum Basic Translation of: FreeBASIC 10 LET s$="1223334444": LET base=2: LET entropy=0
20 LET sourcelen=LEN s$30 DIM t(255) 40 FOR i=1 TO sourcelen 50 LET number= CODE s$(i)
60 LET t(number)=t(number)+1
70 NEXT i
80 PRINT "Char";TAB (6);"Count"
90 FOR i=1 TO 255
100 IF t(i)<>0 THEN PRINT CHR$i;TAB (6);t(i): LET prop=t(i)/sourcelen: LET entropy=entropy-(prop*(LN prop)/(LN base)) 110 NEXT i 120 PRINT '"The Entropy of """;s$;""" is ";entropy`