Find the occurrences of a character in an input string

Question

Challenge

Write a program that, given a string x which is 10 characters long and a character y, outputs the number of times character y occurs in string x.

The shortest program in bytes to do so wins.

Example

Input: tttggloyoi, t
Output: 3

---

Input: onomatopoe, o
Output: 4

Answer 1

Pyth, 3 bytes

/ww

Example run:

$ pyth -c '/ww'
sdhkfhjkkj
k
3

Of course, the user could input more or less than 10 letters on the first input, but we don't need to worry about what happens when the user violates the spec.

Answer 2

Pyth - 3 bytes

A different, less obvious, Pyth answer of the same size. It folds counting over the input.

/FQ

Test Suite.

Answer 3

JavaScript, 32

(p=prompt)().split(p()).length-1

Answer 4

Bash, 24 characters

x=${1//[^$2]}
echo ${#x}

Sample run:

bash-4.3$ bash letter-count.sh tttggloyoi t
3

bash-4.3$ bash letter-count.sh onomatopoe o
4

Answer 5

Retina, 12 bytes

(.)(?=.*\1$)

Simply a regex which matches a character which is equal to the last character in the input (except itself). When given a single regex, Retina simply returns the number of matches.

Answer 6

Python 3, 29 bytes

print(input().count(input()))

Meh, this was easy. Assumes that input is a ten letter string.

Answer 7

Labyrinth, 32 29 27 24 bytes

),}{)-
@ ,  +);__
!-`{:}

This reads the single character first, followed by the string in which to count, and assumes that there are no null-bytes in the string.

Explanation

The code starts with ),}, which sets the bottom of the stack to 1, reads the first character and moves it to the auxiliary stack for future use. The 1 will be our counter (the offset of 1 will be cancelled later and is necessary for the IP to take the required turns).

The IP will now move down to read the first character of the search string with ,. The value is negated with `, again to get the correct turning behaviour. While we're reading characters from STDIN, the IP will now follow this loop:

  }{)-
  ,  +);__
  `{:}

{:} makes a copy of the stored character code and + adds it to the current value. If the result is 0 (i.e. the current character is the one we're looking for), the IP moves straight ahead: - simply gets rid of the 0, ) increments the counter, {} is a no-op.

However, if the result after + is non-zero, we don't want to count the current character. So the IP takes a right-turn instead. That's a dead-end, so that code there is executed twice, once forwards and once backwards. That is, the actual code in this case becomes );___;)+-){}. ); just gets rid of that non-zero difference, ___ pushes 3 zeroes, but ; discards one of them. ) increments one of the two remaining two zeroes, + adds them into a single 1, - subtracts it from the counter and ) increments the counter. In other words, we've created a very elaborate no-op.

When we hit EOF, , pushes -1, which ` turns into 1 and the IP takes a right-turn. - subtracts the 1 from the counter (cancelling the initial offset). ! prints the counter and @ terminates the program.

Answer 8

C++ Template-Metaprogramming, 160 154 116 bytes

Just for the giggles.

Thanks to ex-bart for golfing it down!

template<int w,int x,int y,int...s>class A:A<w+(x==y),x,s...>{};A<0,'t','t','t','t','g','g','l','o','y','o','i',0>a;

Usage: The first char in the template instanciation is the character to search.

Complile with clang -std=c++11 -c -> the result is at the beginning of the error message.

Occurences.cpp:1:66: error: too few template arguments for class template 'A'
template<int w,char x,char y,char...s>class A{static const int a=A<w+((x==y)?1:0),x,s...>::a;};
                                                             ^
Occurences.cpp:1:66: note: in instantiation of template class 'A<3, 't', '\x00'>' requested here
template<int w,char x,char y,char...s>class A{static const int a=A<w+((x==y)?1:0),x,s...>::a;};

Complile with gcc -std=c++11 -c -> the result is at the bottom of the error message.

Occurences.cpp: In instantiation of ‘const int A<3, 't', '\000'>::a’:
Occurences.cpp:1:64:   recursively required from ‘const int A<1, 't', 't', 't', 'g', 'g', 'l', 'o', 'y', 'o', 'i', '\000'>::a’
Occurences.cpp:1:64:   required from ‘const int A<0, 't', 't', 't', 't', 'g', 'g', 'l', 'o', 'y', 'o', 'i', '\000'>::a’
Occurences.cpp:2:62:   required from here
Occurences.cpp:1:64: error: wrong number of template arguments (2, should be at least 3)

Search for the A<3, 't', '\000'> and A<3, 't', '\x00'>

154 byte version

template<int w,char x,char y,char...s>class A{static const int a=A<w+(x==y),x,s...>::a;};                          
int a=A<0,'t','t','t','t','g','g','l','o','y','o','i','\0'>::a;

160 byte version:

template<int w,char x,char y,char...s>class A{static const int a=A<w+((x==y)?1:0),x,s...>::a;};                          
int a=A<0,'t','t','t','t','g','g','l','o','y','o','i','\0'>::a;

Answer 9

Snowman 1.0.2, 16 characters

~vgvgaSaLNdEtSsP

Surprisingly short. Explanation:

~      make all vars active (even though we only need two, we don't really care)
vgvg   get two lines of input
aS     split first line on second line
aL     length of the new array
NdE    decrement (because ex. "axbxc""x"aS -> ["a" "b" "c"] which is length 3)
tSsP   to-string and print

Answer 10

PHP, 36 35 bytes

<?=substr_count($argv[1],$argv[2]);

Usage:
Call the script with two arguments.
php script.php qwertzqwertz q

PHP, 23 bytes

If you register global Variables (only possible in PHP 5.3 and below) you can save 12 bytes (thanks to Martijn)

<?=substr_count($a,$b);

Usage:
Call the script and declare global variables php script.php?a=qwertzqwertz&b=q

Answer 11

Perl, 21 16 characters

(13 characters code + 3 character command line option.)

$_=0+s/$^I//g

Sample run:

bash-4.3$ perl -it -pe '$_=0+s/$^I//g' <<< tttggloyoi
3

bash-4.3$ perl -io -pe '$_=0+s/$^I//g' <<< onomatopoe
4

bash-4.3$ perl -i5 -pe '$_=0+s/$^I//g' <<< 1234
0

Answer 12

Bash + grep, 26 bytes

grep -o "$1"<<<"$2"|wc -l

Answer 13

Javascript (ES6), 26 bytes

(a,b)=>a.split(b).length-1

This quick'n'easy solution defines an anonymous function. To use it, add a variable declaration to the beginning. Try it out:

z=(a,b)=>(a.split(b)||[0]).length-1;

input1=document.getElementById("input1");
input2=document.getElementById("input2");
p=document.getElementById("a");
q=function(){
  setTimeout(function(){
    p.innerHTML=z(input1.value,input2.value)||0;
  },10);
};
input1.addEventListener("keydown",q);
input2.addEventListener("keydown",q);

<form>Text to search: <input type="text" id="input1" value="hello world!"/><br>Char to find: <input type="text" id="input2" value="l"/></form>

<h3>Output:</h3>
<p id="a">3</p>

EDIT: Oh, I see there's a very similar solution already. I hope that's OK.

Answer 14

Haskell, 21 bytes

a!b=sum[1|x<-a,x==b]

Answer 15

Julia, 26 25 bytes

f(s,c)=endof(findin(s,c))

The findin function returns the indices in the first argument at which the second argument is found as a vector. The length of the vector is the number of occurrences.

Saved one byte thanks to Glen O.

Answer 16

APL, 7 3 bytes

+/⍷

This creates a function train. It works by creating a vector of zeros and ones corresponding to the indices at which the character appears in the string (⍷). The vector is then summed (+/).

Saved 4 bytes thanks to kirbyfan64sos and NBZ!

Answer 17

T-SQL, 99 40 Bytes

SELECT 11-LEN(REPLACE(s,c,'')+'x')FROM t

Simply does a difference between the input string and the string with the character removed. Takes input from table t

Edit changed to remove an issue with counting spaces and to take into account current acceptable inputs for SQL. Thanks @BradC for all the changes and savings

Answer 18

C++, 78 bytes

int main(int,char**v){int c=0,i=0;while(i<10)v[1][i++]==*v[2]&&++c;return c;}

Call like this:

$ g++ -std=c++14 -O2 -Wall -pedantic -pthread main.cpp && ./a.out tttggloyoi t; echo $?
3

Answer 19

Element, 23 bytes

__);11'[)\
~="0 1@][+]`

The newline is part of the program. I am actually using it as a variable name.

This program basically works by storing the target character in a variable, keeping the current string on the top of the stack, and then looping the "chop, compare, and move result underneath" process, adding up the results in the end.

The newline as a variable name comes from using the newline at the end of input by chopping it off and storing into it. The newline in the code is where I read from it.

Input is like this:

qqqqwwweee
q

Output is like this

4

Answer 20

Dyalog APL, 3 bytes

      +/=

I.e. "The sum of the equal bytes". E.g.:

      f ← +/=
      'onomatopoe' f 'o'
4

or just

      'onomatopoe'(+/=)'o'
4

K doesn't beat APL this time.

Try it online.

Answer 21

Japt, 2 bytes

èV

Try it online!

Answer 22

Octave / Matlab, 33 bytes

sum(input('','s')==input('','s'))

Answer 23

Ruby, 22 20 bytes

p gets.count(gets)-1

Demo: http://ideone.com/MEeTd2

The -1 is due to the fact that gets retrieves the input, plus a newline character. Ruby's String#count counts the number of times any character from the argument occurs in the string.

For example, for the input [test\n, t\n], the t occurs twice and the \n occurs once, and needs to be subtracted.

Answer 24

PowerShell, 32 Bytes

A four-for-one! And they're all the same length! :)

($args[0]-split$args[1]).Count-1

or

param($a,$b)($a-split$b).Count-1

Alternatively,

$args[0].Split($args[1]).Count-1

or

param($a,$b)$a.Split($b).Count-1

The first two styles use the inline operator -split, while the second two implicitly casts the first argument as a String and uses the .Split() string-based operator. In all instances an array is returned, where we must decrement Count by one, since we're getting back one more array item than occurrences of the second argument.

_{This one was kinda fun...}

Answer 25

J, 5 bytes

+/@:=

I feel like J would have a built-in for this, but I haven't been able to find one - maybe one of the active J users can enlighten me. So instead this first applies = to the inputs, turning each character into 1 if it's equal to the requested one or 0 otherwise. Then +/ computes the sum of that list.

Answer 26

Batch File, 121 Bytes

Because I'm a masochist ...

SET c=0
SET e=_
SET t=%1%%e%
:l
SET a=%t:~0,1%
IF "%a%"=="%2" SET /A c+=1
SET t=%t:~1%
IF NOT "%t%"=="%e%" GOTO l
ECHO %c%

Warning: Assumes that _ doesn't occur in the input string. If it does, then the variable e needs to be adjusted appropriately.

This sets up our counter variable, c, and our end-of-string demarcation as _, before appending that to our input string %1 and setting the concatenated string to t. Then, we're entering loop :l, we set a temporary character variable a to be the first character of t, check if it matches our second input string %2 and increment c if true, then trim the first character off of t. Our end-of-loop condition checks t against our end-of-string demarcation, and loops back if not. We then echo out the value of our counter.

It would probably be possible to use a FOR loop instead, but that would necessitate enabling DelayedExpansion, which I think will actually be longer byte-wise than this. Verification of that is left as an exercise to the reader.

Answer 27

CJam, 5 bytes

ll/,(

Explanation

l      e# read x
 l     e# read y
  /    e# split x by y
   ,   e# count
    (  e# subtract one

Answer 28

><> (Fish), 30 bytes

0&v
=?\ilb
=?\:@=&+&l1
n&/;

Takes the string, then character to count. Input isn't separated (at least in the online interpreter). Try it on the online interpreter: http://fishlanguage.com I counted the bytes by hand, so let me know if I'm wrong.

Explanation

First off, ><> is 2 dimensional and and loops through a line or column until it hits a ; or error. This means that if it's proceeding left to right (like it does at the beginning of a program), it will wrap around the line if it reaches the end and is not moved or told to stop the program. Some characters per line will be repeated because they have different functions depending on the direction of the pointer, and the fourth line will have characters in reverse order because the pointer moves right to left.

A summary of the program is provided below. Look at the instructions listed for ><> on esolangs to see what each individual character does.

Line 1: 0&v

0&v -put 0 into the register and change direction to down-up

Line 2: =?\ilb

(starting where line 1 moves the pointer to, i.e. the third character)

\ -reflect the pointer and make it move left-right
i -read input
lb=?\ -reflect downwards if there are 11 values in the stack

line 3: =?\:@=&+&l1

(starting at the third character)

:@ -duplicate y and shift the stack e.g. ['x','y','y'] -> ['y','x','y']
=&+& -increment the register if the character popped from x = y
l1=?\ -reflect downwards if there is 1 value in the stack

Line 4: n&/;

(starting at the third character)

/ -reflect right-left
&n; -print value of the register

Answer 29

Julia, 21 bytes

f(s,c)=sum(i->c==i,s)

Note that it requires that c be a char, not a single-character string. So you use it as f("test me",'e') (which returns 2) and not f("test me","e") (which returns 0, because 'e'!="e").

Answer 30

Ruby, 18 bytes

->s,c{p s.count c}

Usage:

->s,c{p s.count c}.call 'tttggloyoi', 't'

->s,c{p s.count c}.call 'onomatopoe', 'o'

Answer 31

APL (Dyalog Extended), 1 byte^SBCS

⍧

Try it online!

⍧ is Count In

Answer 32

05AB1E, 1 byte

¢

First input is the character, second the string.

Try it online or verify both test cases at once.

Explanation:

¢  # Count the amount of occurrences of the first (implicit) input-character
   # in the second (implicit) input-string
   # (after which the result is output implicitly)

Answer 33

Zsh, 15 bytes

<<<${#1//[!$2]}

Try it online!

Similar to the bash answer, but more compact thanks to Zsh allowing both ${ //} and ${# } in the same expansion.

Answer 34

Retina, 11 bytes

w`^(.)¶.*\1

Try it online!

Takes input as two lines - the first line is the character, the second is the string.

Could be 10 bytes if taking input as the character preprended to the string, but that felt a bit too cheaty.

Explanation

w`^(.)¶.*\1     # (implicit count stage as there is just one line)
w`              # Include overlapping matches for the pattern
   (.)          # Match a single character...
  ^             # ...at the start of the input...
      ¶         # ... and followed by a newline...
       .*       # ...then zero or more arbitrary characters...
         \1     # ...and then the first character again
                # (implicitly output the number of matches)

Answer 35

Burlesque, 4 bytes

peCN

Try it online!

Takes input as "string" 'i to search string for i.

If special inputs are not allowed:

Burlesque, 6 bytes

pe-]CN

Try it online!

Takes input as two quoted strings.

pe  # Parse and evaluate
CN  # Count

Answer 36

rs, 33 bytes

+(.)(.* (?!\1).)/\2
(.*)../(^^\1)

Live demo and test cases.

This is pretty simple. The first line removes all characters except the ones we're counting. The next line counts the number of characters that are left, excluding the one we're searching for.

Takes input separated by spaces, e.g.:

onomatopoe o

Answer 37

C#, 36 51 48

Debug.Write(args[0].split(args[1][0]).Length-1);

Answer 38

K, 3 bytes

+/=

K actually beat APL!!! :D Well, now it's a tie... :/

You can call it like this:

+/=["hello";"l"]

Or assign it to a variable first:

f:+/=
f["hello";"l"]

Also note that this does not work in oK. It does work in the official K2 and K5 interpreters, as well as Kona.

Answer 39

O, 23 10 bytes

ie\i-e@-_p

Original:

0K;i""/iJ;l{J=K+:K;}dKp

I don't think I can get this any shorter... :/

Ah, I'm stupid. I forgot the traditional method of counting occurrences: subtract the string lengths.

Answer 40

STATA, 52 bytes

di _r(a)_r(b)
di 10-length(subinstr("$a","$b","",.))

Gets the two strings as inputs. Replaces all occurrences of the second string (the character) with empty string. Subtract that from 10, since the length of the first string is 10.

Unfortunately functions in STATA can't be shortened the same way that variable names and commands can. Also, this doesn't work in the online interpreter, since it doesn't include these functions.

Answer 41

scala, 31 bytes

(x:String,y:Char)=>x.count(y==)

Answer 42

Pip, 3 bytes

Joining the 3-byte cavalcade...

bNa

Read as "b in a". a and b are command-line arguments. The N operator, unlike its Python counterpart, doesn't just return a truth value; it returns the count.

Answer 43

R, 27 bytes

sum(strsplit(x,"")[[1]]==y)

Usage

sum(strsplit("tttggloyoi","")[[1]]=="t")
[1] 3
sum(strsplit("onomatopoe","")[[1]]=="o")
[1] 4

Or, Another one with 42 bytes

f<-function(x,y)length(gregexpr(y,x)[[1]])

Usage

f("tttggloyoi", "t")
[1] 3
f("onomatopoe", "o")
[1] 4

Answer 44

Javascript

Program, 48 chars

alert(prompt().match(/(.)(?=.*\1$)|$/g).length-1)

Function with 1 argument (ES6), 39 chars

f=s=>s.match(/(.)(?=.*\1$)|$/g).length-1

Function with 2 arguments (ES6), 28 chars

f=(s,c)=>s.split(c).length-1

Answer 45

Jelly, 1 byte (non-competing)

Non-competing since the question predates the language.

ċ

TryItOnline

The obvious implementation would be 3 bytes: s1ċ, which splits a string or list (left argument) into slices of length 1 with s1 and counts occurrences of the character (right argument) with ċ.

However, since the character will only ever be of length 1, we could, instead, inspect all non-empty contiguous slices of a string with Ẇ; the character will only ever match those slices that are of length 1.

...and since a string is an array of characters, the function can just be ċ.

Answer 46

Java 7, 69 68 54 52 bytes

int c(String...a){return a[0].split(a[1]).length-1;}

Surprisingly enough there wasn't a Java answer yet..
14 bytes saved thanks to @Geobits!

If a completely flexible input is allowed, we can save an additional byte by having a String-array parameter (51 bytes):

int c(String[]a){return a[0].split(a[1]).length-1;}

Ungolfed & test code:

Try it here.

class M{
  static int c(String... a){
    return a[0].split(a[1]).length-1;
  }

  public static void main(String[] a){
    System.out.println(c("tttggloyoi", "t"));
    System.out.println(c("onomatopoe", "o"));
  }
}

Output:

3
4

Answer 47

C, 55 bytes

t;c(char* i,char z){t=0;for(;*i;)t+=*i++==z;return t;}

Answer 48

Jelly, 2 bytes, non-competing

There is a one-byte Jelly solution, but that takes in the argument in a rather particular way. With 1 byte extra, we can just input test instead of ["t", ...]

fL

Try it online!

f   filter all elements from the first implicit input that are not in the second implicit input
    "onomatopoe" f "o" filters out all non-"o" chars --> oooo
 L  Get the length of the remainder.

Answer 49

Wren, 29 bytes

Very hard to golf.

Fn.new{|a,b|a.count{|i|i==b}}

Try it online!

Explanation

Fn.new{|a,b|                  // Anonymous function with parameters a & b
            a.count           // Count every item
                   {|i|       // that...
                       i==b}} // is equal to b

Wren, 36 bytes

Fn.new{|a,b|a.trim(a.trim(b)).count}

Try it online!

Explanation

Fn.new{|a,b|                       } // Uninteresting anonymous function
                   a.trim(b)         // Remove all occurences of b in a
            a.trim(         )        // Trim this result with a, keeping the removed occurences
                             .count  // Find the length of this result

Answer 50

W, 4 bytes

Basically the same algorithm as my second Wren answer.

Sttk

Answer 51

naz, 92 bytes

2x1v2a2x2v8a2x3v1r2x4v1x1f3r3x2v3e3x4v2e1f0x1x2f1v1a2x1v1f0x1x3f1v3x3v4e1o0x1x4f9s1o1s1o0x1f

Works for any valid input string terminated with the control character STX (U+0002), with the character to search for appearing first (e.g. t, tttggloyoi).

Explanation (with 0x commands removed)

2x1v                 # Set variable 1 equal to 0
2a2x2v               # Set variable 2 equal to 2
8a2x3v               # Set variable 3 equal to 10
1r2x4v               # Store the first byte of the input string in variable 4
1x1f                 # Function 1
    3r               # Read the 3rd byte of input, then remove it from the input
                     # This has the effect of skipping over the comma and space
      3x2v3e         # Jump to function 3 if it equals variable 2
            3x4v2e   # Jump to function 2 if it equals variable 4
                  1f # Otherwise, jump back to the start of the function
1x2f                 # Function 2
    1v1a2x1v1f       # Load variable 1 into the register, add 1,
                     # and store the new value in variable 1
                     # Then, jump to function 1
1x3f                 # Function 3
    1v3x3v4e1o       # Load variable 1 into the register
                     # Jump to function 4 if the register equals variable 3
                     # Otherwise, output once
1x4f                 # Function 4
    9s1o1s1o         # Subtract 9 and output, then subtract 1 and output
                     # (outputs "10")
1f                   # Call function 1

Answer 52

Ahead, 16 bytes

The first character on stdin is the counted character, and the remainder of stdin is the string.

i&>jlir
 @Or+=t<

Try it online!

Answer 53

Fortran (GFortran), 60 bytes

character a(10),b
read('(10a,a)'),a,b
print*,count(a==b)
end

Try it online!

Takes input as an 11 length string, the last char being b.

Answer 54

I have been outgolfed.

PowerShell, 75 Bytes (Single Run)

nv a (read-host).split(", ");foreach($b in $a[0]){if($b=$a[2]){$c++}};$c|oh

The above will only work once, will error if variables still have values from last run(run twice in same environment).

PowerShell, 82 Bytes (Infinite Use)

$c=0;nv a (read-host).split(", ")-f;foreach($b in $a[0]){if($b=$a[2]){$c++}};$c|oh

The above can be used any number of times in the same environment.