Count the words in a text and display them

Question

The code should take input a text (not mandatory can be anything file, stdin, string for JavaScript, etc):

This is a text and a number: 31.

The output should contain the words with their number of occurrence, sorted by the number of occurrences in descending order:

a:2
and:1
is:1
number:1
This:1
text:1
31:1

Notice that 31 is a word, so a word is anything alpha-numeric, number are not acting as separators so for example 0xAF qualifies as a word. Separators will be anything that is not alpha-numeric including .(dot) and -(hyphen) thus i.e. or pick-me-up would result in 2 respectively 3 words. Should be case sensitive, This and this would be two different words, ' would also be separator so wouldnand t will be 2 different words from wouldn't.

Write the shortest code in your language of choice.

Shortest correct answer so far:

• grep and coreutils - 42 bytes

Answer 1

J, 35+?

(~.(([\:]);]\:]),.@(+/"1@=))@(>@;:)

Doesn't fully work though. Problem is that the splitting into words ';:' monad doesn't handle non-alplanumeric characters in quite the right way. Any suggestions?

Here's how you use it:

(~.(([\:]);]\:]),.@(+/"1@=))@(>@;:) 'This is a text and a number: 31.'
┌───────┬─┐
│a      │2│
│This   │1│
│is     │1│
│text   │1│
│and    │1│
│number:│1│
│31.    │1│
└───────┴─┘ 

Answer 2

Clojure

(defn wc [s]
  (let [mc #(assoc % %2 (inc (get % %2 0)))]
    (sort-by #(- (val %))
             (reduce mc {} (re-seq #"\w+" (.toLowerCase s))))))

example:

(wc "hi mom hi dad hello peter hello dad hi")
;; (["hi" 3] ["hello" 2] ["dad" 2] ["peter" 1] ["mom" 1])

Answer 3

Python (95):

a=sorted(raw_input().split(__import__("string").punctuation))
for i in set(a):print i,a.count(i)

Pretty straightforward, I'd say.

Answer 4

LiveScript - 74 (translation of ECMA one)

s.match(/[^_\W]+/g,a={})map (->-~=a[it]),keys(a)map(->[it,a[it]])sort (.1-&1.1)

Answer 5

Mathematica, 56 bytes

What, no Mathematica answer yet? This one is not quite a winner, but it's both concise and quite expressive:

f=SortBy[Tally@StringCases[#,WordCharacter..],-Last@#&]&

Calling f["This is a text and a number: 31."] yields

{
 {"a", 2},
 {"31", 1},
 {"and", 1},
 {"is", 1},
 {"number", 1},
 {"text", 1},
 {"This", 1}
}

Answer 6

shell utils, also 42 characters

tr -sc [:alnum:] \\n|sort|uniq -c|sort -rn

Another approach that equals Thor's answer.

Answer 7

Smalltalk, 91 78

input in s

(s allRegexMatches:'\w+')asBag valuesAndCountsDo:[:c :n|(c,$:)print.n printNL]

input:

pick-me-up This is a text and a number: 31.

output:

31:1
pick:1
text:1
me:1
number:1
up:1
is:1
a:2
and:1
This:1

Answer 8

Will never win but wanted to try anyway...

C# (118)

var i = "This is a text and a number: 31.";
//The above line is not counted
Regex.Split(i,@"[\W_]").Where(w=>w!="").GroupBy(g=>g).OrderBy(o=>-o.Count()).Select(s=>new{s.Key,V=s.Count()}).Dump();

Ungolfed:

Regex.Split(i, @"[\W_]")                         //split by special chars
     .Where(w => w != "")                        //remove empty
     .GroupBy(g => g)                            //group by word
     .OrderBy(o => -o.Count())                   //order by reversed count
     .Select( s => new { s.Key, V = s.Count() }) //select value and count
     .Dump();                                    //write to screen (LinQPad)

Answer 9

J 69

Should handle corner-cases like ' - etc, at the huge cost of including all alphanumeric characters (u:62$,65 97 48+/i.26):

(>@~.,.':',.":@#/.~)(#~*@#&>)(<;._1~-.@e.&(u:62$,65 97 48+/i.26))'.',

Usage: just append whatever string to be counted bewteen single quotes (mind that you need to double single quotes in the string).

Example:

   (>@~.,.':',.":@#/.~)(#~*@#&>)(<;._1~-.@e.&(u:62$,65 97 48+/i.26))'.','This is a text and a number: 31. More-tests wouldn''t be bad'
This  :1
is    :1
a     :2
text  :1
and   :1
number:1
31    :1
More  :1
tests :1
wouldn:1
t     :1
be    :1
bad   :1

Answer 10

Python 2 - 61

Assuming the input is in variable s, which is more realistic in real programming than user input anyway.

import re,collections as c
print c.Counter(re.split('\W+',s))

output

Counter({'a': 2, 'and': 1, '': 1, 'This': 1, 'text': 1, 'is': 1, 'number': 1, '31': 1})

This isn't really good output me thinks. It has an empty word and isn't readable. Here's a version with neat output (90):

Python 2 (neat) - 90

import re,collections as c
d=c.Counter(re.split('\W+',s))
for w in d:
  if w:print w+':'+`d[w]`

output

a:2
and:1
This:1
text:1
is:1
number:1
31:1

Answer 11

Clojure

(defn count-words [string]
  (as-> string s
        (clojure.string/replace s #"[^a-zA-Z0-9 ]" "")
        (clojure.string/split s #"\W")
        (frequencies s)
        (sort-by val s)
        (reverse s)))

(clojure.pprint/pprint (count-words "This is a text and a number: 31."))

;; => (["a" 2] ["31" 1] ["number" 1] ["and" 1] ["text" 1] ["is" 1] ["This" 1])

I thought I would beat markw's concision with this approach, but I didn't.

Answer 12

Perl 6, 53 39 bytes

{.comb(/<-[_\W]>+/).Bag.sort:{-.value}}

I use .comb to find every word matching the regexp <-[_\W]>+. In Perl 6, character classes are written <[]> instead of [], and negative character classes <-[]> instead of [^].

We then transform the list of words to a Bag (a set that keeps the number of occurences), and we sort said bag by their value.

Answer 13

C# (332)

This is my original program after excluding white spaces. Please pardon me if I did mistake in counting

using System;
using System.Collections.Generic;
using System.Linq;

class P
{        
    static void Main(String[] A)
    {           
        Dictionary<string,int> D=new Dictionary<string,int>();
        foreach(string s in A)
        {
            if(!D.ContainsKey(s))
                D.Add(s,1);
            else D[s]+= 1;
        }
        foreach(KeyValuePair<string,int> i in D.OrderByDescending(k=>k.Value))
            Console.WriteLine(i.Key+":" + i.Value);
    }
}

This is program with lots of character saving after suggestions from my dear friend jimbobmcgee

using System;
using System.Collections.Generic;
using System.Linq;

class P
{
    static void Main(string[] A)
    {        
       var D = new Dictionary<string, int>();
       foreach (var v in A)
       {
          if (!D.ContainsKey(v))
            D[v] = 1;
          else D[v] += 1;
       }
     foreach (var v in D.OrderBy(k=>-k.Value))
     {
        Console.WriteLine(v.Key+": "+v.Value);
     }
   }
}

Output

Answer 14

JavaScript (ES5), 146

 F=Function;a={};r="return ";b="b";x=prompt().match(/\w+/g);alert(""+x.filter(F(b,r+"!~(a[b]=~-a[b])")).map(F(b,r+"-a[b]+':'+b")).sort().reverse())

=>| This is a text and a number: 31.
<=| 2:a,1:text,1:number,1:is,1:and,1:This,1:31

Answer 15

Python3 (244 bytes)

import re, collections as c, sys
t=''
with open(sys.argv[1]) as f:
    t+=f.read()
d=c.Counter([s for s in (re.sub('[\s\W]','\n',t).split('\n')) if s])
for x in sorted(d.items(), key=lambda x: x[1])[::-1]:
    print('{}:{}'.format(x[0],x[1]))

Works also with newlines, tabs etc. in the sample text.

Output

> $ python3 wc.py test.txt
a:2
text:1
is:1
This:1
and:1
number:1
31:1

Answer 16

JAVA CODE

 String s="This is a text and a number: 31";
                String[] stringArray = s.split(" ");
                final Map<String, Integer> counter = new HashMap<String, Integer>();
                for (String str : stringArray)
                    counter.put(str, 1 + (counter.containsKey(str)counter.get(str): 0));
                List<String> list = new ArrayList<String>(counter.keySet());
                Collections.sort(list, new Comparator<String>() {
                    @Override
                    public int compare(String x, String y) {
                        return counter.get(y) - counter.get(x);
                    }
                });
                list.toArray(new String[list.size()]);
                for (String str : list) {
                    int frequency = counter.get(str);
                    System.out.println(str + ":" + frequency);
                }

OUTPUT

 a:2
 text:1
 is:1
 31:1
 number::1
 This:1
 and:1