Write (or show) a function that counts the number of non-overlapping
occurrences of a substring inside a string, returning an integer
count (e.g. countSubstring("ababababab", "abab") -> 2).
Task
Create a function, or show a built-in function, to count the number of non-overlapping occurrences of a substring inside a string.
The function should take two arguments: :::* the first argument being the string to search, and :::* the second a substring to be searched for.
It should return an integer count.
print countSubstring("the three truths","th")
3
// do not count substrings that overlap with previously-counted substrings:
print countSubstring("ababababab","abab")
2
The matching should yield the highest number of non-overlapping matches.
In general, this essentially means matching from left-to-right or right-to-left (see proof on talk page).
11l
print(‘the three truths’.count(‘th’))
print(‘ababababab’.count(‘abab’))
3
2
360 Assembly
The program uses two ASSIST macros (XDECO,XPRNT) to keep the code as short as possible.
* Count occurrences of a substring 05/07/2016
COUNTSTR CSECT
USING COUNTSTR,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
MVC HAYSTACK,=CL32'the three truths'
MVC LENH,=F'17' lh=17
MVC NEEDLE,=CL8'th' needle='th'
MVC LENN,=F'2' ln=2
BAL R14,SHOW call show
MVC HAYSTACK,=CL32'ababababab'
MVC LENH,=F'11' lh=11
MVC NEEDLE,=CL8'abab' needle='abab'
MVC LENN,=F'4' ln=4
BAL R14,SHOW call show
L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
HAYSTACK DS CL32 haystack
NEEDLE DS CL8 needle
LENH DS F length(haystack)
LENN DS F length(needle)
*------- ---- show---------------------------------------------------
SHOW ST R14,SAVESHOW save return address
BAL R14,COUNT count(haystack,needle)
LR R11,R0 ic=count(haystack,needle)
MVC PG(20),HAYSTACK output haystack
MVC PG+20(5),NEEDLE output needle
XDECO R11,PG+25 output ic
XPRNT PG,80 print buffer
L R14,SAVESHOW restore return address
BR R14 return to caller
SAVESHOW DS A return address of caller
PG DC CL80' ' buffer
*------- ---- count--------------------------------------------------
COUNT ST R14,SAVECOUN save return address
SR R7,R7 n=0
LA R6,1 istart=1
L R10,LENH lh
S R10,LENN ln
LA R10,1(R10) lh-ln+1
LOOPI CR R6,R10 do istart=1 to lh-ln+1
BH ELOOPI
LA R8,NEEDLE @needle
L R9,LENN ln
LA R4,HAYSTACK-1 @haystack[0]
AR R4,R6 +istart
LR R5,R9 ln
CLCL R4,R8 if substr(haystack,istart,ln)=needle
BNE NOTEQ
LA R7,1(R7) n=n+1
A R6,LENN istart=istart+ln
NOTEQ LA R6,1(R6) istart=istart+1
B LOOPI
ELOOPI LR R0,R7 return(n)
L R14,SAVECOUN restore return address
BR R14 return to caller
SAVECOUN DS A return address of caller
* ---- -------------------------------------------------------
YREGS
END COUNTSTR
the three truths th 3
ababababab abab 2
Ada
with Ada.Strings.Fixed, Ada.Integer_Text_IO;
procedure Substrings is
begin
Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source => "the three truths",
Pattern => "th"));
Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source => "ababababab",
Pattern => "abab"));
end Substrings;
3 2
ALGOL 68
Algol68 has no build in function to do this task, hence the next to create a ''count string in string'' routine.
#!/usr/local/bin/a68g --script #
PROC count string in string = (STRING needle, haystack)INT: (
INT start:=LWB haystack, next, out:=0;
FOR count WHILE string in string(needle, next, haystack[start:]) DO
start+:=next+UPB needle-LWB needle;
out:=count
OD;
out
);
printf(($d" "$,
count string in string("th", "the three truths"), # expect 3 #
count string in string("abab", "ababababab"), # expect 2 #
count string in string("a*b", "abaabba*bbaba*bbab"), # expect 2 #
$l$
))
3 2 2
Apex
Apex example for 'Count occurrences of a substring'.
String substr = 'ABC';
String str = 'ABCZZZABCYABCABCXXABC';
Integer substrLen = substr.length();
Integer count = 0;
Integer index = str.indexOf(substr);
while (index >= 0) {
count++;
str = str.substring(index+substrLen);
index = str.indexOf(substr);
}
System.debug('Count String : '+count);
Count String : 5
AppleScript
This is a good example of the kind of problem to which standard libraries (a regex library in this case) would offer most languages a simple and immediate solution. AppleScript, however, for want of various basics like regex and math library functions, can require scripters to draw on the supplementary resources of Bash, using the built-in ''do shell script'' function.
A slightly faster approach to seeking outside library help has, however, become possible since OS X 10.10 added JavaScript as an osalang sibling for AppleScript. Rather than using the resources of Bash, we can more quickly use AppleScript's built in ObjC interface to pass a subproblem over to the more richly endowed JavaScript core of JavaScript for Automation.
Here we use a generic ''evalOSA(language, code)'' function to apply a JavaScript for Automation regex to a pair of AppleScript strings, using OSAKit.
use framework "OSAKit"
on run
{countSubstring("the three truths", "th"), ¬
countSubstring("ababababab", "abab")}
end run
on countSubstring(str, subStr)
return evalOSA("JavaScript", "var matches = '" & str & "'" & ¬
".match(new RegExp('" & subStr & "', 'g'));" & ¬
"matches ? matches.length : 0") as integer
end countSubstring
-- evalOSA :: ("JavaScript" | "AppleScript") -> String -> String
on evalOSA(strLang, strCode)
set ca to current application
set oScript to ca's OSAScript's alloc's initWithSource:strCode ¬
|language|:(ca's OSALanguage's languageForName:(strLang))
set {blnCompiled, oError} to oScript's compileAndReturnError:(reference)
if blnCompiled then
set {oDesc, oError} to oScript's executeAndReturnError:(reference)
if (oError is missing value) then return oDesc's stringValue as text
end if
return oError's NSLocalizedDescription as text
end evalOSA
{3, 2}
AutoHotkey
While it is simple enough to parse the string, AutoHotkey has a rather unconventional method which outperforms this. StringReplace sets the number of replaced strings to ErrorLevel.
MsgBox % countSubstring("the three truths","th") ; 3
MsgBox % countSubstring("ababababab","abab") ; 2
CountSubstring(fullstring, substring){
StringReplace, junk, fullstring, %substring%, , UseErrorLevel
return errorlevel
}
AWK
#
# countsubstring(string, pattern)
# Returns number of occurrences of pattern in string
# Pattern treated as a literal string (regex characters not expanded)
#
function countsubstring(str, pat, len, i, c) {
c = 0
if( ! (len = length(pat) ) )
return 0
while(i = index(str, pat)) {
str = substr(str, i + len)
c++
}
return c
}
#
# countsubstring_regex(string, regex_pattern)
# Returns number of occurrences of pattern in string
# Pattern treated as regex
#
function countsubstring_regex(str, pat, c) {
c = 0
c += gsub(pat, "", str)
return c
}
BEGIN {
print countsubstring("[do&d~run?d!run&>run&]", "run&")
print countsubstring_regex("[do&d~run?d!run&>run&]", "run[&]")
print countsubstring("the three truths","th")
}
$ awk -f countsubstring.awk
2
2
3
BaCon
FUNCTION Uniq_Tally(text$, part$)
LOCAL x
WHILE TALLY(text$, part$)
INCR x
text$ = MID$(text$, INSTR(text$, part$)+LEN(part$))
WEND
RETURN x
END FUNCTION
PRINT "the three truths - th: ", Uniq_Tally("the three truths", "th")
PRINT "ababababab - abab: ", Uniq_Tally("ababababab", "abab")
the three truths - th: 3
ababababab - abab: 2
BASIC
In FreeBASIC, this needs to be compiled with -lang qb or -lang fblite.
DECLARE FUNCTION countSubstring& (where AS STRING, what AS STRING)
PRINT "the three truths, th:", countSubstring&("the three truths", "th")
PRINT "ababababab, abab:", countSubstring&("ababababab", "abab")
FUNCTION countSubstring& (where AS STRING, what AS STRING)
DIM c AS LONG, s AS LONG
s = 1 - LEN(what)
DO
s = INSTR(s + LEN(what), where, what)
IF 0 = s THEN EXIT DO
c = c + 1
LOOP
countSubstring = c
END FUNCTION
the three truths, th: 3 ababababab, abab: 2
See also: [[#Liberty BASIC|Liberty BASIC]], [[#PowerBASIC|PowerBASIC]], [[#PureBasic|PureBasic]].
Applesoft BASIC
=
10 F$ = "TH"
20 S$ = "THE THREE TRUTHS"
30 GOSUB 100"COUNT SUBSTRING
40 PRINT R
50 F$ = "ABAB"
60 S$ = "ABABABABAB"
70 GOSUB 100"COUNT SUBSTRING
80 PRINT R
90 END
100 R = 0
110 F = LEN(F$)
120 S = LEN(S$)
130 IF F > S THEN RETURN
140 IF F = 0 THEN RETURN
150 IF F = S AND F$ = S$ THEN R = 1 : RETURN
160 FOR I = 1 TO S - F
170 IF F$ = MID$(S$, I, F) THEN R = R + 1 : I = I + F - 1
180 NEXT I
190 RETURN
==={{header|IS-BASIC}}===
=
## Sinclair ZX81 BASIC
=
Works with 1k of RAM.
```basic
10 LET S$="THE THREE TRUTHS"
20 LET U$="TH"
30 GOSUB 100
40 PRINT N
50 LET S$="ABABABABAB"
60 LET U$="ABAB"
70 GOSUB 100
80 PRINT N
90 STOP
100 LET N=0
110 LET I=0
120 LET I=I+1
130 IF I+LEN U$>LEN S$ THEN RETURN
140 IF S$(I TO I+LEN U$-1)<>U$ THEN GOTO 120
150 LET N=N+1
160 LET I=I+LEN U$
170 GOTO 130
Batch File
@echo off
setlocal enabledelayedexpansion
::Main
call :countString "the three truths","th"
call :countString "ababababab","abab"
pause>nul
exit /b
::/Main
::Procedure
:countString
set input=%~1
set cnt=0
:count_loop
set trimmed=!input:*%~2=!
if "!trimmed!"=="!input!" (echo.!cnt!&goto :EOF)
set input=!trimmed!
set /a cnt+=1
goto count_loop
3
2
BBC BASIC
tst$ = "the three truths"
sub$ = "th"
PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"
tst$ = "ababababab"
sub$ = "abab"
PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"
END
DEF FNcountSubstring(A$, B$)
LOCAL I%, N%
I% = 1 : N% = 0
REPEAT
I% = INSTR(A$, B$, I%)
IF I% THEN N% += 1 : I% += LEN(B$)
UNTIL I% = 0
= N%
3 "th" in "the three truths"
2 "abab" in "ababababab"
Bracmat
( count-substring
= n S s p
. 0:?n:?p
& !arg:(?S.?s)
& @( !S
: ?
( [!p ? !s [?p ?
& !n+1:?n
& ~
)
)
| !n
)
& out$(count-substring$("the three truths".th))
& out$(count-substring$(ababababab.abab))
& ;
3
2
C
#include <stdio.h>
#include <string.h>
int match(const char *s, const char *p, int overlap)
{
int c = 0, l = strlen(p);
while (*s != '\0') {
if (strncmp(s++, p, l)) continue;
if (!overlap) s += l - 1;
c++;
}
return c;
}
int main()
{
printf("%d\n", match("the three truths", "th", 0));
printf("overlap:%d\n", match("abababababa", "aba", 1));
printf("not: %d\n", match("abababababa", "aba", 0));
return 0;
}
Alternate version:
#include <stdio.h>
#include <string.h>
// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const char *str, const char *sub)
{
int length = strlen(sub);
if (length == 0) return 0;
int count = 0;
for (str = strstr(str, sub); str; str = strstr(str + length, sub))
++count;
return count;
}
int main()
{
printf("%d\n", countSubstring("the three truths", "th"));
printf("%d\n", countSubstring("ababababab", "abab"));
printf("%d\n", countSubstring("abaabba*bbaba*bbab", "a*b"));
return 0;
}
3
2
2
C++
#include <iostream>
#include <string>
// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const std::string& str, const std::string& sub)
{
if (sub.length() == 0) return 0;
int count = 0;
for (size_t offset = str.find(sub); offset != std::string::npos;
offset = str.find(sub, offset + sub.length()))
{
++count;
}
return count;
}
int main()
{
std::cout << countSubstring("the three truths", "th") << '\n';
std::cout << countSubstring("ababababab", "abab") << '\n';
std::cout << countSubstring("abaabba*bbaba*bbab", "a*b") << '\n';
return 0;
}
3
2
2
C#
using System;
class SubStringTestClass
{
public static int CountSubStrings(this string testString, string testSubstring)
{
int count = 0;
if (testString.Contains(testSubstring))
{
for (int i = 0; i < testString.Length; i++)
{
if (testString.Substring(i).Length >= testSubstring.Length)
{
bool equals = testString.Substring(i, testSubstring.Length).Equals(testSubstring);
if (equals)
{
count++;
i += testSubstring.Length - 1; // Fix: Don't count overlapping matches
}
}
}
}
return count;
}
}
Using C# 6.0's expression-bodied member, null-conditional operator, and coalesce operator features:
using System;
class SubStringTestClass
{
public static int CountSubStrings(this string testString, string testSubstring) =>
testString?.Split(new [] { testSubstring }, StringSplitOptions.None)?.Length - 1 ?? 0;
}
Clojure
Use a sequence of regexp matches to count occurrences.
(defn count-substring [txt sub]
(count (re-seq (re-pattern 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)))
COBOL
INSPECT can be used for this task without having to create a function.
IDENTIFICATION DIVISION.
PROGRAM-ID. testing.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 occurrences PIC 99.
PROCEDURE DIVISION.
INSPECT "the three truths" TALLYING occurrences FOR ALL "th"
DISPLAY occurrences
MOVE 0 TO occurrences
INSPECT "ababababab" TALLYING occurrences FOR ALL "abab"
DISPLAY occurrences
MOVE 0 TO occurrences
INSPECT "abaabba*bbaba*bbab" TALLYING occurrences
FOR ALL "a*b"
DISPLAY occurrences
GOBACK
.
03
02
02
CoffeeScript
countSubstring = (str, substr) ->
n = 0
i = 0
while (pos = str.indexOf(substr, i)) != -1
n += 1
i = pos + substr.length
n
console.log countSubstring "the three truths", "th"
console.log countSubstring "ababababab", "abab"
Common Lisp
(defun count-sub (str pat)
(loop with z = 0 with s = 0 while s do
(when (setf s (search pat str :start2 s))
(incf z) (incf s (length pat)))
finally (return z)))
(count-sub "ababa" "ab") ; 2
(count-sub "ababa" "aba") ; 1
D
void main() {
import std.stdio, std.algorithm;
"the three truths".count("th").writeln;
"ababababab".count("abab").writeln;
}
3
2
Delphi
program OccurrencesOfASubstring;
{$APPTYPE CONSOLE}
uses StrUtils;
function CountSubstring(const aString, aSubstring: string): Integer;
var
lPosition: Integer;
begin
Result := 0;
lPosition := PosEx(aSubstring, aString);
while lPosition <> 0 do
begin
Inc(Result);
lPosition := PosEx(aSubstring, aString, lPosition + Length(aSubstring));
end;
end;
begin
Writeln(CountSubstring('the three truths', 'th'));
Writeln(CountSubstring('ababababab', 'abab'));
end.
=={{header|Déjà Vu}}==
!. count "the three truths" "th"
!. count "ababababab" "abab"
3
2
Dyalect
func countSubstring(str, val) {
var idx = 0
var count = 0
while true {
idx = str.indexOf(val, idx)
if idx == -1 {
break
}
idx += val.len()
count += 1
}
return count
}
print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3
2
EchoLisp
;; from Racket
(define count-substring
(compose length regexp-match*))
(count-substring "aab" "graabaabdfaabgh") ;; substring
→ 3
(count-substring "/ .e/" "Longtemps je me suis couché de bonne heure") ;; regexp
→ 4
Eiffel
class
APPLICATION
inherit
ARGUMENTS
create
make
feature {NONE} -- Initialization
make
-- Run application.
do
occurance := 0
from
index := 1
until
index > text.count
loop
temp := text.fuzzy_index(search_for, index, 0)
if
temp /= 0
then
index := temp + search_for.count
occurance := occurance + 1
else
index := text.count + 1
end
end
print(occurance)
end
index:INTEGER
temp:INTEGER
occurance:INTEGER
text:STRING = "ababababab"
search_for:STRING = "abab"
end
Elixir
countSubstring = fn(_, "") -> 0
(str, sub) -> length(String.split(str, sub)) - 1 end
data = [ {"the three truths", "th"},
{"ababababab", "abab"},
{"abaabba*bbaba*bbab", "a*b"},
{"abaabba*bbaba*bbab", "a"},
{"abaabba*bbaba*bbab", " "},
{"abaabba*bbaba*bbab", ""},
{"", "a"},
{"", ""} ]
Enum.each(data, fn{str, sub} ->
IO.puts countSubstring.(str, sub)
end)
3
2
2
7
0
0
0
0
Erlang
%% Count non-overlapping substrings in Erlang for the rosetta code wiki.
%% Implemented by J.W. Luiten
-module(substrings).
-export([main/2]).
%% String and Sub exhausted, count a match and present result
match([], [], _OrigSub, Acc) ->
Acc+1;
%% String exhausted, present result
match([], _Sub, _OrigSub, Acc) ->
Acc;
%% Sub exhausted, count a match
match(String, [], Sub, Acc) ->
match(String, Sub, Sub, Acc+1);
%% First character matches, advance
match([X|MainTail], [X|SubTail], Sub, Acc) ->
match(MainTail, SubTail, Sub, Acc);
%% First characters do not match. Keep scanning for sub in remainder of string
match([_X|MainTail], [_Y|_SubTail], Sub, Acc)->
match(MainTail, Sub, Sub, Acc).
main(String, Sub) ->
match(String, Sub, Sub, 0).
Command:
substrings:main("ababababab","abab").
2
Alternative using built in functions:
main( String, Sub ) -> erlang:length( binary:split(binary:list_to_bin(String), binary:list_to_bin(Sub), [global]) ) - 1.
Euphoria
function countSubstring(sequence s, sequence sub)
integer from,count
count = 0
from = 1
while 1 do
from = match_from(sub,s,from)
if not from then
exit
end if
from += length(sub)
count += 1
end while
return count
end function
? countSubstring("the three truths","th")
? countSubstring("ababababab","abab")
3
2
EGL
The "remove and count the difference" and "manual loop" methods. Implementation includes protection from empty source and search strings.
program CountStrings
function main()
SysLib.writeStdout("Remove and Count:");
SysLib.writeStdout(countSubstring("th", "the three truths"));
SysLib.writeStdout(countSubstring("abab", "ababababab"));
SysLib.writeStdout(countSubstring("a*b", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring("a", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring(" ", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring("", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstring("a", ""));
SysLib.writeStdout(countSubstring("", ""));
SysLib.writeStdout("Manual Loop:");
SysLib.writeStdout(countSubstringWithLoop("th", "the three truths"));
SysLib.writeStdout(countSubstringWithLoop("abab", "ababababab"));
SysLib.writeStdout(countSubstringWithLoop("a*b", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop("a", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop(" ", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop("", "abaabba*bbaba*bbab"));
SysLib.writeStdout(countSubstringWithLoop("a", ""));
SysLib.writeStdout(countSubstringWithLoop("", ""));
end
function countSubstring(substr string in, str string in) returns(int)
if(str.length() > 0 and substr.length() > 0)
return (str.length() - str.replaceStr(subStr, "").length()) / subStr.length();
else
return 0;
end
end
function countSubstringWithLoop(substr string in, str string in) returns(int)
count int = 0;
loc, index int = 1;
strlen int = str.length();
substrlen int = substr.length();
if(strlen > 0 and substrlen > 0)
while(loc != 0 and index <= strlen)
loc = str.indexOf(substr, index);
if(loc > 0)
count += 1;
index = loc + substrlen;
end
end
end
return count;
end
end
Remove and Count:
3
2
2
7
0
0
0
0
Manual Loop:
3
2
2
7
0
0
0
0
Factor
USING: math sequences splitting ;
: occurences ( seq subseq -- n ) split-subseq length 1 - ;
=={{header|Fōrmulæ}}==
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). 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 transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Forth
: str-count ( s1 len s2 len -- n )
2swap 0 >r
begin 2over search
while 2over nip /string
r> 1+ >r
repeat 2drop 2drop r> ;
s" the three truths" s" th" str-count . \ 3
s" ababababab" s" abab" str-count . \ 2
Fortran
program Example
implicit none
integer :: n
n = countsubstring("the three truths", "th")
write(*,*) n
n = countsubstring("ababababab", "abab")
write(*,*) n
n = countsubstring("abaabba*bbaba*bbab", "a*b")
write(*,*) n
contains
function countsubstring(s1, s2) result(c)
character(*), intent(in) :: s1, s2
integer :: c, p, posn
c = 0
if(len(s2) == 0) return
p = 1
do
posn = index(s1(p:), s2)
if(posn == 0) return
c = c + 1
p = p + posn + len(s2)
end do
end function
end program
3
2
2
FreeBASIC
' FB 1.05.0 Win64
Function countSubstring(s As String, search As String) As Integer
If s = "" OrElse search = "" Then Return 0
Dim As Integer count = 0, length = Len(search)
For i As Integer = 1 To Len(s)
If Mid(s, i, length) = Search Then
count += 1
i += length - 1
End If
Next
Return count
End Function
Print countSubstring("the three truths","th")
Print countSubstring("ababababab","abab")
Print countSubString("zzzzzzzzzzzzzzz", "z")
Print
Print "Press any key to quit"
Sleep
3
2
15
=={{header|F_Sharp|F#}}== "Remove and count the difference" method, as shown by J, Java, ...
open System
let countSubstring (where :string) (what : string) =
match what with
| "" -> 0 // just a definition; infinity is not an int
| _ -> (where.Length - where.Replace(what, @"").Length) / what.Length
[<EntryPoint>]
let main argv =
let show where what =
printfn @"countSubstring(""%s"", ""%s"") = %d" where what (countSubstring where what)
show "the three truths" "th"
show "ababababab" "abab"
show "abc" ""
0
countSubstring("the three truths", "th") = 3
countSubstring("ababababab", "abab") = 2
countSubstring("abc", "") = 0
FunL
import util.Regex
def countSubstring( str, substr ) = Regex( substr ).findAllMatchIn( str ).length()
println( countSubstring("the three truths", "th") )
println( countSubstring("ababababab", "abab") )
3
2
Go
Using strings.Count() method:
package main
import (
"fmt"
"strings"
)
func main() {
fmt.Println(strings.Count("the three truths", "th")) // says: 3
fmt.Println(strings.Count("ababababab", "abab")) // says: 2
}
Groovy
Solution, uses the Groovy "find" operator (=~), and the Groovy-extended Matcher property "count":
println (('the three truths' =~ /th/).count)
println (('ababababab' =~ /abab/).count)
println (('abaabba*bbaba*bbab' =~ /a*b/).count)
println (('abaabba*bbaba*bbab' =~ /a\*b/).count)
3
2
9
2
Haskell
=== Text-based solution ===
import Data.Text hiding (length)
-- Return the number of non-overlapping occurrences of sub in str.
countSubStrs str sub = length $ breakOnAll (pack sub) (pack str)
main = do
print $ countSubStrs "the three truths" "th"
print $ countSubStrs "ababababab" "abab"
3
2
Alternatively, in a language built around currying, it might make more sense to reverse the suggested order of arguments.
import Data.Text hiding (length)
countAll :: String -> String -> Int
countAll needle haystack = length (breakOnAll n h)
where
[n, h] = pack <$> [needle, haystack]
main :: IO ()
main =
print $ countAll "ab" <$> ["ababababab", "abelian absurdity", "babel kebab"]
[5,2,2]
=== List-based solution === Even though list-based strings are not "the right" way of representing texts, the problem of counting subsequences in a list is generally useful.
count :: Eq a =
[a] -> [a] -> Int
count [] = error "empty substring"
count sub = go
where
go = scan sub . dropWhile (/= head sub)
scan _ [] = 0
scan [] xs = 1 + go xs
scan (x:xs) (y:ys) | x == y = scan xs ys
| otherwise = go ys
λ> count "th" "the three truths"
3
λ> count "abab" "ababababab"
2
λ> count [2,3] [1,2,1,2,3,4,3,2,3,4,3,2]
2
λ> count "123456" $ foldMap show [1..1000000]
7
=== List-based solution using Data.List === The following solution is almost two times faster than the previous one.
import Data.List (tails, stripPrefix)
import Data.Maybe (catMaybes)
count :: Eq a => [a] -> [a] -> Int
count sub = length . catMaybes . map (stripPrefix sub) . tails
=={{header|Icon}} and {{header|Unicon}}==
procedure main()
every A := ![ ["the three truths","th"], ["ababababab","abab"] ] do
write("The string ",image(A[2])," occurs as a non-overlapping substring ",
countSubstring!A , " times in ",image(A[1]))
end
procedure countSubstring(s1,s2) #: return count of non-overlapping substrings
c := 0
s1 ? while tab(find(s2)) do {
move(*s2)
c +:= 1
}
return c
end
The string "th" occurs as a non-overlapping substring 3 times in "the three truths"
The string "abab" occurs as a non-overlapping substring 2 times in "ababababab"
J
require'strings'
countss=: #@] %~ #@[ - [ #@rplc '';~]
In other words: find length of original string, replace the string to be counted with the empty string, find the difference in lengths and divide by the length of the string to be counted.
Example use:
'the three truths' countss 'th'
3
'ababababab' countss 'abab'
2
Java
The "remove and count the difference" method:
public class CountSubstring {
public static int countSubstring(String subStr, String str){
return (str.length() - str.replace(subStr, "").length()) / subStr.length();
}
public static void main(String[] args){
System.out.println(countSubstring("th", "the three truths"));
System.out.println(countSubstring("abab", "ababababab"));
System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
}
}
3
2
2
The "split and count" method:
import java.util.regex.Pattern;
public class CountSubstring {
public static int countSubstring(String subStr, String str){
// the result of split() will contain one more element than the delimiter
// the "-1" second argument makes it not discard trailing empty strings
return str.split(Pattern.quote(subStr), -1).length - 1;
}
public static void main(String[] args){
System.out.println(countSubstring("th", "the three truths"));
System.out.println(countSubstring("abab", "ababababab"));
System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
}
}
3
2
2
Manual looping
public class CountSubstring {
public static int countSubstring(String subStr, String str){
int count = 0;
for (int loc = str.indexOf(subStr); loc != -1;
loc = str.indexOf(subStr, loc + subStr.length()))
count++;
return count;
}
public static void main(String[] args){
System.out.println(countSubstring("th", "the three truths"));
System.out.println(countSubstring("abab", "ababababab"));
System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
}
}
3
2
2
JavaScript
Using regexes:
function countSubstring(str, subStr) {
var matches = str.match(new RegExp(subStr, "g"));
return matches ? matches.length : 0;
}
Using 'split' and ES6 notation:
const countSubString = (str, subStr) => str.split(subStr).length - 1;
jq
Using regexes (available in jq versions after June 19, 2014):
def countSubstring(sub):
[match(sub; "g")] | length;
Example:
"the three truths" | countSubstring("th")
Julia
'''Built-in Function'''
matchall(r::Regex, s::String[, overlap::Bool=false]) -> Vector{String}
Return a vector of the matching substrings from eachmatch.
'''Main'''
ts = ["the three truths", "ababababab"]
tsub = ["th", "abab"]
println("Test of non-overlapping substring counts.")
for i in 1:length(ts)
print(ts[i], " (", tsub[i], ") => ")
println(length(matchall(Regex(tsub[i]), ts[i])))
end
println()
println("Test of overlapping substring counts.")
for i in 1:length(ts)
print(ts[i], " (", tsub[i], ") => ")
println(length(matchall(Regex(tsub[i]), ts[i], true)))
end
Test of non-overlapping substring counts.
the three truths (th) => 3
ababababab (abab) => 2
Test of overlapping substring counts.
the three truths (th) => 3
ababababab (abab) => 4
K
The dyadic verb _ss gives the positions of substring y in string x.
"the three truths" _ss "th"
0 4 13
#"the three truths" _ss "th"
3
"ababababab" _ss "abab"
0 4
#"ababababab" _ss "abab"
2
Kotlin
// version 1.0.6
fun countSubstring(s: String, sub: String): Int = s.split(sub).size - 1
fun main(args: Array<String>) {
println(countSubstring("the three truths","th"))
println(countSubstring("ababababab","abab"))
println(countSubstring("",""))
}
3
2
1
Lasso
define countSubstring(str::string, substr::string)::integer => {
local(i = 1, foundpos = -1, found = 0)
while(#i < #str->size && #foundpos != 0) => {
protect => {
handle_error => { #foundpos = 0 }
#foundpos = #str->find(#substr, -offset=#i)
}
if(#foundpos > 0) => {
#found += 1
#i = #foundpos + #substr->size
else
#i++
}
}
return #found
}
define countSubstring_bothways(str::string, substr::string)::integer => {
local(found = countSubstring(#str,#substr))
#str->reverse
local(found2 = countSubstring(#str,#substr))
#found > #found2 ? return #found | return #found2
}
countSubstring_bothways('the three truths','th')
//3
countSubstring_bothways('ababababab','abab')
//2
Liberty BASIC
print countSubstring( "the three truths", "th")
print countSubstring( "ababababab", "abab")
end
function countSubstring( a$, s$)
c =0
la =len( a$)
ls =len( s$)
for i =1 to la -ls
if mid$( a$, i, ls) =s$ then c =c +1: i =i +ls -1
next i
countSubstring =c
end function
Logtalk
Using atoms for string representation:
:- object(counting).
:- public(count/3).
count(String, SubString, Count) :-
count(String, SubString, 0, Count).
count(String, SubString, Count0, Count) :-
( sub_atom(String, Before, Length, After, SubString) ->
Count1 is Count0 + 1,
Start is Before + Length,
sub_atom(String, Start, After, 0, Rest),
count(Rest, SubString, Count1, Count)
; Count is Count0
).
:- end_object.
| ?- counting::count('the three truths', th, N).
N = 3
yes
| ?- counting::count(ababababab, abab, N).
N = 2
yes
Lua
Solution 1:
function countSubstring(s1, s2)
return select(2, s1:gsub(s2, ""))
end
print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3
2
Solution 2:
function countSubstring(s1, s2)
local count = 0
for eachMatch in s1:gmatch(s2) do
count = count + 1
end
return count
end
print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3
2
Maple
f:=proc(s::string,c::string,count::nonnegint) local n;
n:=StringTools:-Search(c,s);
if n>0 then 1+procname(s[n+length(c)..],c,count);
else 0; end if;
end proc:
f("the three truths","th",0);
f("ababababab","abab",0);
3
2
=={{header|Mathematica}} / {{header|Wolfram Language}}==
StringPosition["the three truths","th",Overlaps->False]//Length
3
StringPosition["ababababab","abab",Overlaps->False]//Length
2
=={{header|MATLAB}} / {{header|Octave}}==
% Count occurrences of a substring without overlap
length(findstr("ababababab","abab",0))
length(findstr("the three truths","th",0))
% Count occurrences of a substring with overlap
length(findstr("ababababab","abab",1))
>> % Count occurrences of a substring without overlap
>> length(findstr("ababababab","abab",0))
ans = 2
>> length(findstr("the three truths","th",0))
ans = 3
>> % Count occurrences of a substring with overlap
>> length(findstr("ababababab","abab",1))
ans = 4
>>
Maxima
scount(e, s) := block(
[n: 0, k: 1],
while integerp(k: ssearch(e, s, k)) do (n: n + 1, k: k + 1),
n
)$
scount("na", "banana");
2
MiniScript
string.count = function(s)
return self.split(s).len - 1
end function
print "the three truths".count("th")
print "ababababab".count("abab")
3
2
Mirah
import java.util.regex.Pattern
import java.util.regex.Matcher
#The "remove and count the difference" method
def count_substring(pattern:string, source:string)
(source.length() - source.replace(pattern, "").length()) / pattern.length()
end
puts count_substring("th", "the three truths") # ==> 3
puts count_substring("abab", "ababababab") # ==> 2
puts count_substring("a*b", "abaabba*bbaba*bbab") # ==> 2
# The "split and count" method
def count_substring2(pattern:string, source:string)
# the result of split() will contain one more element than the delimiter
# the "-1" second argument makes it not discard trailing empty strings
source.split(Pattern.quote(pattern), -1).length - 1
end
puts count_substring2("th", "the three truths") # ==> 3
puts count_substring2("abab", "ababababab") # ==> 2
puts count_substring2("a*b", "abaabba*bbaba*bbab") # ==> 2
# This method does a match and counts how many times it matches
def count_substring3(pattern:string, source:string)
result = 0
Matcher m = Pattern.compile(Pattern.quote(pattern)).matcher(source);
while (m.find())
result = result + 1
end
result
end
puts count_substring3("th", "the three truths") # ==> 3
puts count_substring3("abab", "ababababab") # ==> 2
puts count_substring3("a*b", "abaabba*bbaba*bbab") # ==> 2
Nemerle
using System.Console;
module CountSubStrings
{
CountSubStrings(this text : string, target : string) : int
{
match (target) {
|"" => 0
|_ => (text.Length - text.Replace(target, "").Length) / target.Length
}
}
Main() : void
{
def text1 = "the three truths";
def target1 = "th";
def text2 = "ababababab";
def target2 = "abab";
WriteLine($"$target1 occurs $(text1.CountSubStrings(target1)) times in $text1");
WriteLine($"$target2 occurs $(text2.CountSubStrings(target2)) times in $text2");
}
}
th occurs 3 times in the three truths
abab occurs 2 times in ababababab
NetRexx
NetRexx provides the ''string''.countstr(''needle'') built-in function:
/* NetRexx */
options replace format comments java crossref symbols nobinary
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method countSubstring(inStr, findStr) public static
return inStr.countstr(findStr)
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method main(args = String[]) public static
strings = ''
find = 'FIND'
ix = 0
ix = ix + 1; strings[0] = ix; find[0] = ix; strings[ix] = 'the three truths'; strings[ix, find] = 'th'
ix = ix + 1; strings[0] = ix; find[0] = ix; strings[ix] = 'ababababab'; strings[ix, find] = 'abab'
loop ix = 1 to strings[0]
str = strings[ix]
fnd = strings[ix, find]
say 'there are' countSubstring(str, fnd) 'occurences of "'fnd'" in "'str'"'
end ix
return
there are 3 occurences of "th" in "the three truths"
there are 2 occurences of "abab" in "ababababab"
NewLISP
; file: stringcount.lsp
; url: https://rosettacode.org/wiki/Count_occurrences_of_a_substring
; author: oofoe 2012-01-29
; Obvious (and non-destructive...)
; Note that NewLISP performs an /implicit/ slice on a string or list
; with this form "(start# end# stringorlist)". If the end# is omitted,
; the slice will go to the end of the string. This is handy here to
; keep removing the front part of the string as it gets matched.
(define (scount needle haystack)
(let ((h (copy haystack)) ; Copy of haystack string.
(i 0) ; Cursor.
(c 0)) ; Count of occurences.
(while (setq i (find needle h))
(inc c)
(setq h ((+ i (length needle)) h)))
c)) ; Return count.
; Tricky -- Uses functionality from replace function to find all
; non-overlapping occurrences, replace them, and return the count of
; items replaced in system variable $0.
(define (rcount needle haystack)
(replace needle haystack "X") $0)
; Test
(define (test f needle haystack)
(println "Found " (f needle haystack)
" occurences of '" needle "' in '" haystack "'."))
(dolist (f (list scount rcount))
(test f "glart" "hinkerpop")
(test f "abab" "ababababab")
(test f "th" "the three truths")
(println)
)
(exit)
Found 0 occurences of 'glart' in 'hinkerpop'.
Found 2 occurences of 'abab' in 'ababababab'.
Found 3 occurences of 'th' in 'the three truths'.
Found 0 occurences of 'glart' in 'hinkerpop'.
Found 2 occurences of 'abab' in 'ababababab'.
Found 3 occurences of 'th' in 'the three truths'.
Nim
import strutils
proc count(s, sub: string): int =
var i = 0
while true:
i = s.find(sub, i)
if i < 0:
break
i += sub.len # i += 1 for overlapping substrings
inc result
echo count("the three truths","th")
echo count("ababababab","abab")
3
2
=={{header|Objective-C}}== The "split and count" method:
@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end
@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
return [[self componentsSeparatedByString:subStr] count] - 1;
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
}
return 0;
}
3
2
2
The "remove and count the difference" method:
@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end
@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
return ([self length] - [[self stringByReplacingOccurrencesOfString:subStr withString:@""] length]) / [subStr length];
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
}
return 0;
}
3
2
2
Manual looping:
@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end
@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
NSUInteger count = 0;
for (NSRange range = [self rangeOfString:subStr]; range.location != NSNotFound;
range.location += range.length,
range = [self rangeOfString:subStr options:0
range:NSMakeRange(range.location, [self length] - range.location)])
count++;
return count;
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
}
return 0;
}
3
2
2
OCaml
let count_substring str sub =
let sub_len = String.length sub in
let len_diff = (String.length str) - sub_len
and reg = Str.regexp_string sub in
let rec aux i n =
if i > len_diff then n else
try
let pos = Str.search_forward reg str i in
aux (pos + sub_len) (succ n)
with Not_found -> n
in
aux 0 0
let () =
Printf.printf "count 1: %d\n" (count_substring "the three truth" "th");
Printf.printf "count 2: %d\n" (count_substring "ababababab" "abab");
;;
Oforth
: countSubString(s, sub)
0 1 while(sub swap s indexOfAllFrom dup notNull) [ sub size + 1 under+ ]
drop ;
countSubString("the three truths", "th") println
3
countSubString("ababababab", "abab") println
2
ooRexx
bag="the three truths"
x="th"
say left(bag,30) left(x,15) 'found' bag~countstr(x)
bag="ababababab"
x="abab"
say left(bag,30) left(x,15) 'found' bag~countstr(x)
-- can be done caselessly too
bag="abABAbaBab"
x="abab"
say left(bag,30) left(x,15) 'found' bag~caselesscountstr(x)
the three truths th found 3
ababababab abab found 2
abABAbaBab abab found 2
PARI/GP
subvec(v,u)={
my(i=1,s);
while(i+#u<=#v,
for(j=1,#u,
if(v[i+j-1]!=u[j], i++; next(2))
);
s++;
i+=#u
);
s
};
substr(s1,s2)=subvec(Vec(s1),Vec(s2));
substr("the three truths","th")
substr("ababababab","abab")
%1 = 3
%2 = 2
Pascal
See [[Count_occurrences_of_a_substring#Delphi | Delphi]]
Perl
sub countSubstring {
my $str = shift;
my $sub = quotemeta(shift);
my $count = () = $str =~ /$sub/g;
return $count;
# or return scalar( () = $str =~ /$sub/g );
}
print countSubstring("the three truths","th"), "\n"; # prints "3"
print countSubstring("ababababab","abab"), "\n"; # prints "2"
Perl 6
sub count-substring($big,$little) { +$big.comb: ~$little }
say count-substring("the three truths","th"); # 3
say count-substring("ababababab","abab"); # 4
say count-substring(123123123,12); # 3
The ~ prefix operator converts $little to a Str if it isn't already, and .comb when given a Str as an argument returns instances of that substring. You can think of it as if the argument was a regex that matched the string literally /$little/. Also, prefix + forces numeric context in Perl 6 (it's a no-op in Perl 5). For the built in listy types that is the same as calling .elems method. One other style point: we now tend to prefer hyphenated names over camelCase.
Phix
sequence tests = {{"the three truths","th"},
{"ababababab","abab"},
{"ababababab","aba"},
{"ababababab","ab"},
{"ababababab","a"},
{"ababababab",""}}
integer start, count
string test, substring
for i=1 to length(tests) do
start = 1
count = 0
{test, substring} = tests[i]
while 1 do
start = match(substring,test,start)
if start=0 then exit end if
start += length(substring)
count += 1
end while
printf(1,"The string \"%s\" occurs as a non-overlapping substring %d times in \"%s\"\n",{substring,count,test})
end for
The string "th" occurs as a non-overlapping substring 3 times in "the three truths"
The string "abab" occurs as a non-overlapping substring 2 times in "ababababab"
The string "aba" occurs as a non-overlapping substring 2 times in "ababababab"
The string "ab" occurs as a non-overlapping substring 5 times in "ababababab"
The string "a" occurs as a non-overlapping substring 5 times in "ababababab"
The string "" occurs as a non-overlapping substring 0 times in "ababababab"
PHP
<?php
echo substr_count("the three truths", "th"), "\n"; // prints "3"
echo substr_count("ababababab", "abab"), "\n"; // prints "2"
?>
PicoLisp
(de countSubstring (Str Sub)
(let (Cnt 0 H (chop Sub))
(for (S (chop Str) S (cdr S))
(when (head H S)
(inc 'Cnt)
(setq S (map prog2 H S)) ) )
Cnt ) )
Test:
: (countSubstring "the three truths" "th")
-> 3
: (countSubstring "ababababab" "abab")
-> 2
PL/I
cnt: procedure options (main);
declare (i, tally) fixed binary;
declare (text, key) character (100) varying;
get edit (text) (L); put skip data (text);
get edit (key) (L); put skip data (key);
tally = 0; i = 1;
do until (i = 0);
i = index(text, key, i);
if i > 0 then do; tally = tally + 1; i = i + length(key); end;
end;
put skip list (tally);
end cnt;
Output for the two specified strings is as expected. {{out}} for the following data:
TEXT='AAAAAAAAAAAAAAA';
KEY='AA';
7
PowerBASIC
Windows versions of PowerBASIC (at least since PB/Win 7, and possibly earlier) provide the TALLY function, which does exactly what this task requires (count non-overlapping substrings).
[[PB/DOS]] can use the example under [[#BASIC|BASIC]], above.
Note that while this example is marked as working with PB/Win, the PRINT statement would need to be replaced with MSGBOX, or output to a file. (PB/Win does not support console output.)
FUNCTION PBMAIN () AS LONG
PRINT "the three truths, th:", TALLY("the three truths", "th")
PRINT "ababababab, abab:", TALLY("ababababab", "abab")
END FUNCTION
the three truths, th: 3 ababababab, abab: 2
PureBasic
a = CountString("the three truths","th")
b = CountString("ababababab","abab")
; a = 3
; b = 2
PowerShell
[regex]::Matches("the three truths", "th").count
Output:
3
[regex]::Matches("ababababab","abab").count
Output:
2
Prolog
Using SWI-Prolog's string facilities (this solution is very similar to the Logtalk solution that uses sub_atom/5):
count_substring(String, Sub, Total) :-
count_substring(String, Sub, 0, Total).
count_substring(String, Sub, Count, Total) :-
( substring_rest(String, Sub, Rest)
->
succ(Count, NextCount),
count_substring(Rest, Sub, NextCount, Total)
;
Total = Count
).
substring_rest(String, Sub, Rest) :-
sub_string(String, Before, Length, Remain, Sub),
DropN is Before + Length,
sub_string(String, DropN, Remain, 0, Rest).
Usage:
?- count_substring("the three truths","th",X).
X = 3.
?- count_substring("ababababab","abab",X).
X = 2.
Python
>>
"the three truths".count("th")
3
>>> "ababababab".count("abab")
2
R
The fixed parameter (and, in stringr, the function of the same name) is used to specify a search for a fixed string. Otherwise, the search pattern is interpreted as a POSIX regular expression. PCRE is also an option: use the perl parameter or function.
count = function(haystack, needle)
{v = attr(gregexpr(needle, haystack, fixed = T)[[1]], "match.length")
if (identical(v, -1L)) 0 else length(v)}
print(count("hello", "l"))
library(stringr)
print(str_count("hello", fixed("l")))
Racket
(define count-substring
(compose length regexp-match*))
> (count-substring "th" "the three truths")
3
> (count-substring "abab" "ababababab")
2
Red
Red []
;;-----------------------------------
count-sub1: func [hay needle][
;;-----------------------------------
prin rejoin ["hay: " pad copy hay 20 ",needle: " pad copy needle 6 ",count: " ]
i: 0
parse hay [ some [thru needle (i: i + 1)] ]
print i
]
;;-----------------------------------
count-sub2: func [hay needle][
;;-----------------------------------
prin rejoin ["hay: " pad copy hay 20 ",needle: " pad copy needle 6 ",count: " ]
i: 0
while [hay: find hay needle][
i: i + 1
hay: skip hay length? needle
]
print i
]
count-sub1 "the three truths" "th"
count-sub1 "ababababab" "abab"
print "^/version 2"
count-sub2 "the three truths" "th"
count-sub2 "ababababab" "abab"
hay: the three truths ,needle: th ,count: 3
hay: ababababab ,needle: abab ,count: 2
version 2
hay: the three truths ,needle: th ,count: 3
hay: ababababab ,needle: abab ,count: 2
>>
REXX
Some older REXXes don't have the built-in function '''countstr''', so one is included within the REXX program as a function.
The '''countstr''' subroutine (below) mimics the BIF in newer REXXes (except for error checking).
Either of the first two strings may be null.
The third argument is optional and is the ''start position'' to start counting (the default is '''1''', meaning the first character).
If specified, it must be a positive integer (and it may exceed the length of the 1st string).
The third argument was added here to be compatible with the newer REXXes BIF.
No checks are made (in the '''countstr''' subroutine) for: ::::* missing arguments ::::* too many arguments ::::* if '''start''' is a positive integer (when specified)
/*REXX program counts the occurrences of a (non─overlapping) substring in a string. */
w=. /*max. width so far.*/
bag= 'the three truths' ; x= "th" ; call showResult
bag= 'ababababab' ; x= "abab" ; call showResult
bag= 'aaaabacad' ; x= "aa" ; call showResult
bag= 'abaabba*bbaba*bbab' ; x= "a*b" ; call showResult
bag= 'abaabba*bbaba*bbab' ; x= " " ; call showResult
bag= ; x= "a" ; call showResult
bag= ; x= ; call showResult
bag= 'catapultcatalog' ; x= "cat" ; call showResult
bag= 'aaaaaaaaaaaaaa' ; x= "aa" ; call showResult
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
countstr: procedure; parse arg haystack,needle,start; if start=='' then start=1
width=length(needle)
do $=0 until p==0; p=pos(needle,haystack,start)
start=width + p /*prevent overlaps.*/
end /*$*/
return $ /*return the count.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
showResult: if w==. then do; w=30 /*W: largest haystack width.*/
say center('haystack',w) center('needle',w%2) center('count',5)
say left('', w, "═") left('', w%2, "═") left('', 5, "═")
end
if bag=='' then bag= " (null)" /*handle displaying of nulls.*/
if x=='' then x= " (null)" /* " " " " */
say left(bag, w) left(x, w%2) center(countstr(bag, x), 5)
return
'''output''' when using the default (internal) inputs:
haystack needle count
══════════════════════════════ ═══════════════ ═════
the three truths th 3
ababababab abab 2
aaaabacad aa 2
abaabba*bbaba*bbab a*b 2
abaabba*bbaba*bbab 0
(null) a 0
(null) (null) 1
catapultcatalog cat 2
aaaaaaaaaaaaaa aa 7
Ring
aString = "Ring Welcome Ring to the Ring Ring Programming Ring Language Ring"
bString = "Ring"
see count(aString,bString)
func count cString,dString
sum = 0
while substr(cString,dString) > 0
sum++
cString = substr(cString,substr(cString,dString)+len(string(sum)))
end
return sum
Output:
6
Ruby
def countSubstrings str, subStr
str.scan(subStr).length
end
p countSubstrings "the three truths", "th" #=> 3
p countSubstrings "ababababab", "abab" #=> 2
String#scan returns an array of substrings, and Array#length (or Array#size) counts them.
Run BASIC
print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")
FUNCTION countSubstring(s$,find$)
WHILE instr(s$,find$,i) <> 0
countSubstring = countSubstring + 1
i = instr(s$,find$,i) + len(find$)
WEND
END FUNCTION
3
2
Rust
fn main() {
println!("{}","the three truths".matches("th").count());
println!("{}","ababababab".matches("abab").count());
}
3
2
Scala
Using Recursion
import scala.annotation.tailrec
def countSubstring(str1:String, str2:String):Int={
@tailrec def count(pos:Int, c:Int):Int={
val idx=str1 indexOf(str2, pos)
if(idx == -1) c else count(idx+str2.size, c+1)
}
count(0,0)
}
Using Sliding
def countSubstring(str: String, sub: String): Int =
str.sliding(sub.length).count(_ == sub)
Using Regular Expressions
def countSubstring( str:String, substr:String ) = substr.r.findAllMatchIn(str).length
println(countSubstring("ababababab", "abab"))
println(countSubstring("the three truths", "th"))
2
3
Scheme
gosh
(use gauche.lazy)
#<undef>
gosh> (length (lrxmatch "th" "the three truths"))
3
gosh> (length (lrxmatch "abab" "ababababab"))
2
Seed7
$ include "seed7_05.s7i";
const func integer: countSubstring (in string: stri, in string: searched) is func
result
var integer: count is 0;
local
var integer: offset is 0;
begin
offset := pos(stri, searched);
while offset <> 0 do
incr(count);
offset := pos(stri, searched, offset + length(searched));
end while;
end func;
const proc: main is func
begin
writeln(countSubstring("the three truths", "th"));
writeln(countSubstring("ababababab", "abab"));
end func;
3
2
Sidef
'''Built-in:'''
say "the three truths".count("th");
say "ababababab".count("abab");
'''User-created function:'''
func countSubstring(s, ss) {
var re = Regex.new(ss.escape, 'g'); # 'g' for global
var counter = 0;
while (s =~ re) { ++counter };
return counter;
}
say countSubstring("the three truths","th");
say countSubstring("ababababab","abab");
3
2
SNOBOL4
DEFINE("countSubstring(t,s)")
OUTPUT = countSubstring("the three truths","th")
OUTPUT = countSubstring("ababababab","abab")
:(END)
countSubstring t ARB s = :F(RETURN)
countSubstring = countSubstring + 1 :(countSubstring)
END
3
2
Standard ML
fun count_substrings (str, sub) =
let
fun aux (str', count) =
let
val suff = #2 (Substring.position sub str')
in
if Substring.isEmpty suff then
count
else
aux (Substring.triml (size sub) suff, count + 1)
end
in
aux (Substring.full str, 0)
end;
print (Int.toString (count_substrings ("the three truths", "th")) ^ "\n");
print (Int.toString (count_substrings ("ababababab", "abab")) ^ "\n");
print (Int.toString (count_substrings ("abaabba*bbaba*bbab", "a*b")) ^ "\n");
Stata
function strcount(s, x) {
n = 0
k = 1-(i=strlen(x))
do {
if (k = ustrpos(s, x, k+i)) n++
} while(k)
return(n)
}
strcount("peter piper picked a peck of pickled peppers", "pe")
5
strcount("ababababab","abab")
2
TUSCRIPT
$$ MODE TUSCRIPT, {}
occurences=COUNT ("the three truths", ":th:")
occurences=COUNT ("ababababab", ":abab:")
occurences=COUNT ("abaabba*bbaba*bbab",":a\*b:")
3
2
2
Tcl
The regular expression engine is ideal for this task, especially as the ***= prefix makes it interpret the rest of the argument as a literal string to match:
proc countSubstrings {haystack needle} {
regexp -all ***=$needle $haystack
}
puts [countSubstrings "the three truths" "th"]
puts [countSubstrings "ababababab" "abab"]
puts [countSubstrings "abaabba*bbaba*bbab" "a*b"]
3
2
2
TXR
@(next :args)
@(do (defun count-occurrences (haystack needle)
(for* ((occurrences 0)
(old-pos 0)
(new-pos (search-str haystack needle old-pos nil)))
(new-pos occurrences)
((inc occurrences)
(set old-pos (+ new-pos (length needle)))
(set new-pos (search-str haystack needle old-pos nil))))))
@ndl
@hay
@(output)
@(count-occurrences hay ndl) occurrences(s) of @ndl inside @hay
@(end)
$ ./txr count-occurrences.txr "baba" "babababa"
2 occurence(s) of baba inside babababa
$ ./txr count-occurrences.txr "cat" "catapultcatalog"
2 occurence(s) of cat inside catapultcatalog
UNIX Shell
#!/bin/bash
function countString(){
input=$1
cnt=0
until [ "${input/$2/}" == "$input" ]; do
input=${input/$2/}
let cnt+=1
done
echo $cnt
}
countString "the three truths" "th"
countString "ababababab" "abab"
3
2
VBA
Function CountStringInString(stLookIn As String, stLookFor As String)
CountStringInString = UBound(Split(stLookIn, stLookFor))
End Function
VBScript
Function CountSubstring(str,substr)
CountSubstring = 0
For i = 1 To Len(str)
If Len(str) >= Len(substr) Then
If InStr(i,str,substr) Then
CountSubstring = CountSubstring + 1
i = InStr(i,str,substr) + Len(substr) - 1
End If
Else
Exit For
End If
Next
End Function
WScript.StdOut.Write CountSubstring("the three truths","th") & vbCrLf
WScript.StdOut.Write CountSubstring("ababababab","abab") & vbCrLf
3
2
Visual Basic .NET
Module Count_Occurrences_of_a_Substring
Sub Main()
Console.WriteLine(CountSubstring("the three truths", "th"))
Console.WriteLine(CountSubstring("ababababab", "abab"))
Console.WriteLine(CountSubstring("abaabba*bbaba*bbab", "a*b"))
Console.WriteLine(CountSubstring("abc", ""))
End Sub
Function CountSubstring(str As String, substr As String) As Integer
Dim count As Integer = 0
If (Len(str) > 0) And (Len(substr) > 0) Then
Dim p As Integer = InStr(str, substr)
Do While p <> 0
p = InStr(p + Len(substr), str, substr)
count += 1
Loop
End If
Return count
End Function
End Module
3
2
2
0
Wortel
@let {
c &[s t] #!s.match &(t)g
[[
!!c "the three truths" "th"
!!c "ababababab" "abab"
]]
}
Returns:
[3 2]
XPL0
include c:\cxpl\codes; \intrinsic 'code' declarations
string 0; \use zero-terminated strings, instead of MSb terminated
func StrNCmp(A, B, N); \Compare string A to string B up to N bytes long
\This returns:
\ >0 if A > B
\ =0 if A = B
\ <0 if A < B
char A, B; \strings to be compared
int N; \number of bytes to compare
int I;
[for I:= 0 to N-1 do
if A(I) # B(I) then
return A(I) - B(I);
return 0; \they're equal
]; \StrNCmp
func StrLen(Str); \Return the number of characters in an ASCIIZ string
char Str;
int I;
for I:= 0 to -1>>1-1 do
if Str(I) = 0 then return I;
func SubStr(A, B); \Count number of times string B occurs in A
char A, B;
int LA, LB, C, I;
[LA:= StrLen(A); LB:= StrLen(B);
C:= 0; I:= 0;
while I < LA do
if StrNCmp(B, A+I, LB) = 0 then [C:= C+1; I:= I+LB]
else I:= I+1;
return C;
];
[IntOut(0, SubStr("the three truths", "th")); CrLf(0);
IntOut(0, SubStr("ababababab", "abab")); CrLf(0);
]
3
2
>> % Count occurrences of a substring without overlap
>> length(findstr("ababababab","abab",0))
ans = 2
>> length(findstr("the three truths","th",0))
ans = 3
>> % Count occurrences of a substring with overlap
>> length(findstr("ababababab","abab",1))
ans = 4
>>
zkl
Two solutions:
fcn countSubstring(s,p){ pn:=p.len(); cnt:=n:=0;
while(Void!=(n:=s.find(p,n))){cnt+=1; n+=pn}
cnt
}
fcn countSubstring(s,p){ (pl:=p.len()) and (s.len()-(s-p).len())/pl }
zkl: println(countSubstring("the three truths","th"))
3
zkl: println(countSubstring("ababababab","abab"))
2
zkl: println(countSubstring("ababababab","v"))
0
ZX Spectrum Basic
10 LET t$="ABABABABAB": LET p$="ABAB": GO SUB 1000
20 LET t$="THE THREE TRUTHS": LET p$="TH": GO SUB 1000
30 STOP
1000 PRINT t$: LET c=0
1010 LET lp=LEN p$
1020 FOR i=1 TO LEN t$-lp+1
1030 IF (t$(i TO i+lp-1)=p$) THEN LET c=c+1: LET i=i+lp-1
1040 NEXT i
1050 PRINT p$;"=";c''
1060 RETURN