Character Counting System [duplicate]

Question

GOAL : To Implement a character counting program

This is how this counting system works :

input of the program is a set of printable ASCII characters . the program counts each type of character , and outputs the number of each character , starting from first mentioned character to last mentioned character .

Example :

input : aaaabbbbbcccc    123445b
output : 4 6 4 4 1 1 1 2 1

Which 4 is number of as , 6 is number of bs , 4 is number of cs , 4 is number of spaces () and so on . Also you can split answers with space (like example) , EOL , or anything other than numerals .

Rules

Your program must log to STDOUT or an acceptable alternative, if STDOUT is not available.

Your program must be a full, runnable program, and not a function or snippet

It's optional to mention the ungolfed version of program and/or a short explanation .

Test input : ThIs Is ThE iNPUT , WITH 72/2 CHARS!

Test output : 4 2 3 2 7 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1

Answer 1

MATL, 3 bytes

8#u

Input is a string enclosed with quotation marks (which is allowed by default)

Try it online!

Explanation

u is the unique function, which essentially removes duplicates. It can produce up to four outputs. The fourth output is the count of unique characters, identified in order of appearance.

8# specifies that the fourth output of u should be produced. In general, 2# means "produce two outputs", 3# means "produce three outputs" etc. When the maximum number of function outputs is reached, larger numbers mean "take the first output only", or "the second only" etc. Thus, since u has four possible outputs, 4# would produce all four outputs; 5# produces the first, 6# the second, etc.

Answer 2

05AB1E, 6 bytes

Ùv¹y¢,

Explained

Ùv      # for each unique char in original order
  ¹y¢   # count number of occurances in input string
     ,  # print on newline

Try it online

Answer 3

Pyth, 4 bytes

/LQ{

Test suite.

Explanation

/LQ{    Input: Q
/LQ{Q   Implicitly fill arguments

   {Q   Yield the unique elements of Q.
 L      For each unique element:
/       Yield its number of occurrences
  Q     in Q.

Answer 4

J, 18 15 bytes

echo#/.~stdin''

Following the challenge spec to write a program, the above is a one-line script in J that reads a line from stdin, computes the result, and prints it to stdout.

If using an online interpreter, it would require 4 bytes since without stdio, only function arguments could be used.

#/.~

Straight-forward application of adverbs /. and ~ to the tally # verb.

Usage

As a script,

$ echo -n 'aaaabbbbbcccc    123445b' | jconsole golf.j 
4 6 4 4 1 1 1 2 1

As a function,

   (#/.~) 'aaaabbbbbcccc    123445b'
4 6 4 4 1 1 1 2 1

Explanation

#/.~  Input: s
   ~  Reflects the input
 /.   An adverb that executes a verb on each set of identical items in s
      Operates in the order of first-seen distinct items in s
#     Get the size of each set of identical items
      Returns a list of the sizes of identical items

Answer 5

CJam, 9 bytes

l__&\fe=p

Test it here.

Explanation

l    e# Read input.
__   e# Make two copies.
&    e# Set intersection - removes duplicates.
\    e# Swap with input.
fe=  e# For each (unique) character, count its occurrences.
p    e# Print as array literal.

Answer 6

Dyalog APL, 10 bytes

Shortest by byte count

+⌿t∘.=∪t←⍞

t←⍞ get character input and store as t

`∪' unique elements of

t∘.= table of which elements of t are equal to (the unique elements of t)

+⌿ sum the columns

5 character solution which unfortunately is 15 bytes:

⊢∘≢⌸⍞

⍞ get character input

⌸ make a table of contents containing...

⊢∘≢ the tally (≢) of the indices while the entries are ignored (⊢)

Answer 7

Jelly, 4 bytes

Qċ@€

Try it online

Explanation

Qċ@€   Main link. Argument: S

Q      Remove duplicate characters from S.
       Current value v contains each character of S once sorted by first appearance.
  @€   For each character in v.
 ċ       count its occurrence in S.

Answer 8

Mathematica, 56 bytes

Print@StringRiffle[Last/@Tally@Characters@InputString[]]

Essentially, Tally does exactly what the challenge asks for (although it gives pairs of {element, count}). The rest is just for working with strings and proper I/O for a full program.

Answer 9

Retina, 30 28 bytes

2 bytes thanks to Martin Ender.

+`((.).+)\2
$2$1
(.)\1*
$.&¶

Try it online!

Answer 10

Brachylog, 32 bytes

:efd:?:{tL,?he:L:{t:.m~h?}fl.}f.

Awfully long but there's no simple way of counting elements of something yet…

Answer 11

Java 7, 246 235 231 213 bytes

Loads of bytes removed thanks to cliffroot.

import java.util.*;class M{public static void main(String[]a){Map<Long,Long>m=new LinkedHashMap();for(long c:a[0].toCharArray())m.put(c,m.get(c)!=null?m.get(c)+1:1);for(Long e:m.values())System.out.print(e+" ");}}

Ungolfed & test code:

Try it here.

import java.util.*;

class Main{
  public static void main(String[] a){
    // Test code:
    a = new String[]{ "ThIs Is ThE iNPUT , WITH 72/2 CHARS!" };

    Map<Long, Long> m = new LinkedHashMap();
    for(long c : a[0].toCharArray()){
      m.put(c, m.get(c) != null ? m.get(c) + 1 : 1);
    }
    for(Long e : m.values()){
      System.out.print(e + " ");
    }
  }
}

Output:

4 2 3 2 7 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 

Answer 12

Python 3, 134 81 bytes

s,r=input(),[]
for c in s:
 k,s=s.count(c),s.replace(c,'')
 if k:r+=[k]
print(*r)

Output:

4 2 3 2 7 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1

Not that short :(

---

Another approach: 109 bytes

Not shorter than the others but the counting happens in a one-liner.

s=input()
k=[i[1]for i in sorted({i:str(s.count(i))for i in s}.items(),key=lambda v:s.index(v[0]))]
print(*k)

---

Edit: Golfed 53 bytes thanks to @Dr Green Eggs and Iron Man Edit 2: Added one-liner method

Answer 13

Java 7, 172 168 bytes

class M{public static void main(String[]a){int[]m=new int[256];byte[]b=a[0].getBytes();for(int z:b)m[z]++;for(int z:b){System.out.print(m[z]>0?(m[z]+" "):"");m[z]=0;}}}

Stores number of occurrences in an int array, then for each char print number of occurrences and replace this value with zero so it won't be printed again.

Whitespaces and comments added:

class M{
    public static void main(String[]a){
        int[]m=new int[256]; // map, ascii value to number of occurrences
        byte[b]=a[0].getBytes();
        for(int z:b)m[z]++; // count number of occurrences
        for(int z:b){
            System.out.print(m[z]>0?(m[z]+" "):""); // print result 
            m[z]=0; // make it not be printed again
        }
    }
}

See it online: https://ideone.com/7X58F9

Answer 14

C#

public static void Main(string[] args)
    {
        string str = "aaaabbbbbcccc    123445b";
        Dictionary<char, int> counts = new Dictionary<char, int>();

        for (int i = 0; i < str.Length; i++)
           if (counts.ContainsKey(str[i]))
               counts[str[i]]++;
           else
              counts.Add(str[i], 1);

        foreach (var count in counts)
         Console.WriteLine("{0} = {1}", count.Key, count.Value.ToString());
    }

Answer 15

ES6, 90 bytes

a=prompt();b=new Map();for(c of a)b.set(c,(b.get(c)|0)+1);alert([...b.values()].join(' '))

Answer 16

Python 3, 76 73 bytes

3 bytes shorter:

s,a=input(),[]
for y in s:a+=(y,'')[y in a]
print([s.count(x)for x in a])

Old version:

s=input();print([s.count(x)for x in sorted(set(s),key=lambda i:s.index(i))])

Iterates over a sorted set of unique characters from the input, counts the occurrences, and outputs a list of the counts.

Answer 17

Python 3, 93 bytes

s=input();c,o=[],""
for i in s:
 if i not in c:c+=i
for i in c:o+=str(s.count(i))+" "
print o

Answer 18

Matlab, 47 chars

@(y)arrayfun(@(x)sum(y==x),unique(y,'stable'))

Reasonably straightforward, though the default behavior of Matlab's unique function to sort the values makes the function rather longer.

Answer 19

Pyke, 3 bytes

}F/

Try it here!

for i in uniquify(input): print input.count(i)

Answer 20

Python 2.7 - 103 Bytes (Ungolfed)

i=list(raw_input())
print ''.join([str(e) for e in map(lambda x,y:x!=y and i.count(x)or' ',i,['']+i) ])

edit: Thanks to Dr Green Eggs and Iron Man for helping to reduce 47 Bytes!