Word count

From Rosetta Code
Word count is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Given a text file and an integer n, print the n most common words in the file (and the number of their occurrences) in decreasing frequency.

For the purposes of this task:

  • A word is a sequence of one or more contiguous letters
  • You are free to define what a letter is. Underscores, accented letters, apostrophes, and other special characters can be handled at the example writer's discretion. For example, you may treat a compound word like "well-dressed" as either one word or two. The word "it's" could also be one or two words as you see fit. You may also choose not to support non US-ASCII characters. Feel free to explicitly state the thoughts behind the program decisions.
  • Assume words will not span multiple lines.
  • Do not worry about normalization of word spelling differences. Treat "color" and "colour" as two distinct words.
  • Uppercase letters are considered equivalent to their lowercase counterparts
  • Words of equal frequency can be listed in any order


Show example output using Les Misérables from Project Gutenberg as the text file input and display the top 10 most used words.


History

This task was originally taken from programming pearls from Communications of the ACM June 1986 Volume 29 Number 6 where this problem is solved by Donald Knuth using literate programming and then critiqued by Doug McIlroy, demonstrating solving the problem in a 6 line Unix shell script (provided as an example below).


References



AutoHotkey[edit]

URLDownloadToFile, http://www.gutenberg.org/files/135/135-0.txt, % A_temp "\tempfile.txt"
FileRead, H, % A_temp "\tempfile.txt"
FileDelete,  % A_temp "\tempfile.txt"
words := []
while pos := RegExMatch(H, "\b[[:alpha:]]+\b", m, A_Index=1?1:pos+StrLen(m))
words[m] := words[m] ? words[m] + 1 : 1
for word, count in words
list .= count "`t" word "`r`n"
Sort, list, RN
loop, parse, list, `n, `r
{
result .= A_LoopField "`r`n"
if A_Index = 10
break
}
MsgBox % "Freq`tWord`n" result
return
Outputs:
Freq	Word
41036	The
19946	of
14940	and
14589	A
13939	TO
11204	in
9645	HE
8619	WAS
7922	THAT
6659	it


C#[edit]

Translation of: D
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text.RegularExpressions;
 
namespace WordCount {
class Program {
static void Main(string[] args) {
var text = File.ReadAllText("135-0.txt").ToLower();
 
var match = Regex.Match(text, "\\w+");
Dictionary<string, int> freq = new Dictionary<string, int>();
while (match.Success) {
string word = match.Value;
if (freq.ContainsKey(word)) {
freq[word]++;
} else {
freq.Add(word, 1);
}
 
match = match.NextMatch();
}
 
Console.WriteLine("Rank Word Frequency");
Console.WriteLine("==== ==== =========");
int rank = 1;
foreach (var elem in freq.OrderByDescending(a => a.Value).Take(10)) {
Console.WriteLine("{0,2} {1,-4} {2,5}", rank++, elem.Key, elem.Value);
}
}
}
}
Output:
Rank  Word  Frequency
====  ====  =========
 1    the     41035
 2    of      19946
 3    and     14940
 4    a       14577
 5    to      13939
 6    in      11204
 7    he       9645
 8    was      8619
 9    that     7922
10    it       6659

Clojure[edit]

(defn count-words [file n]
(->> file
slurp
clojure.string/lower-case
(re-seq #"\w+")
frequencies
(sort-by val >)
(take n)))
Output:
user=> (count-words "135-0.txt" 10)
(["the" 41036] ["of" 19946] ["and" 14940] ["a" 14589] ["to" 13939]
 ["in" 11204] ["he" 9645] ["was" 8619] ["that" 7922] ["it" 6659])

D[edit]

import std.algorithm : sort;
import std.array : appender, split;
import std.range : take;
import std.stdio : File, writefln, writeln;
import std.typecons : Tuple;
import std.uni : toLower;
 
//Container for a word and how many times it has been seen
alias Pair = Tuple!(string, "k", int, "v");
 
void main() {
int[string] wcnt;
 
//Read the file line by line
foreach (line; File("135-0.txt").byLine) {
//Split the words on whitespace
foreach (word; line.split) {
//Increment the times the word has been seen
wcnt[word.toLower.idup]++;
}
}
 
//Associative arrays cannot be sort, so put the key/value in an array
auto wb = appender!(Pair[]);
foreach(k,v; wcnt) {
wb.put(Pair(k,v));
}
Pair[] sw = wb.data.dup;
 
//Sort the array, and display the top ten values
writeln("Rank Word Frequency");
int rank=1;
foreach (word; sw.sort!"a.v>b.v".take(10)) {
writefln("%4s  %-10s  %9s", rank++, word.k, word.v);
}
}
Output:
Rank  Word        Frequency
   1  the             40368
   2  of              19863
   3  and             14470
   4  a               14277
   5  to              13587
   6  in              11019
   7  he               9212
   8  was              8346
   9  that             7251
  10  his              6414

F#[edit]

 
open System.IO
open System.Text.RegularExpressions
let g=Regex("[A-Za-zÀ-ÿ]+").Matches(File.ReadAllText "135-0.txt")
[for n in g do yield n.Value.ToLower()]|>List.countBy(id)|>List.sortBy(fun n->(-(snd n)))|>List.take 10|>List.iter(fun n->printfn "%A" n)
 
Output:
("the", 41088)
("of", 19949)
("and", 14942)
("a", 14596)
("to", 13951)
("in", 11214)
("he", 9648)
("was", 8621)
("that", 7924)
("it", 6661)

Factor[edit]

This program expects stdin to read from a file via the command line. ( e.g. invoking the program in Windows: >factor word-count.factor < input.txt ) The definition of a word here is simply any string surrounded by some combination of spaces, punctuation, or newlines.

 
USING: ascii io math.statistics prettyprint sequences
splitting ;
IN: rosetta-code.word-count
 
lines " " join " .,?!:;()\"-" split harvest [ >lower ] map
sorted-histogram <reversed> 10 head .
 
Output:
{
    { "the" 41021 }
    { "of" 19945 }
    { "and" 14938 }
    { "a" 14522 }
    { "to" 13938 }
    { "in" 11201 }
    { "he" 9600 }
    { "was" 8618 }
    { "that" 7822 }
    { "it" 6532 }
}

Go[edit]

Translation of: Kotlin
package main
 
import (
"fmt"
"io/ioutil"
"log"
"regexp"
"sort"
"strings"
)
 
type keyval struct {
key string
val int
}
 
func main() {
reg := regexp.MustCompile(`\p{Ll}+`)
bs, err := ioutil.ReadFile("135-0.txt")
if err != nil {
log.Fatal(err)
}
text := strings.ToLower(string(bs))
matches := reg.FindAllString(text, -1)
groups := make(map[string]int)
for _, match := range matches {
groups[match]++
}
var keyvals []keyval
for k, v := range groups {
keyvals = append(keyvals, keyval{k, v})
}
sort.Slice(keyvals, func(i, j int) bool {
return keyvals[i].val > keyvals[j].val
})
fmt.Println("Rank Word Frequency")
fmt.Println("==== ==== =========")
for rank := 1; rank <= 10; rank++ {
word := keyvals[rank-1].key
freq := keyvals[rank-1].val
fmt.Printf("%2d  %-4s  %5d\n", rank, word, freq)
}
}
Output:
Rank  Word  Frequency
====  ====  =========
 1    the     41088
 2    of      19949
 3    and     14942
 4    a       14596
 5    to      13951
 6    in      11214
 7    he       9648
 8    was      8621
 9    that     7924
10    it       6661

Groovy[edit]

Solution:

def topWordCounts = { String content, int n ->
def mapCounts = [:]
content.toLowerCase().split(/\W+/).each {
mapCounts[it] = (mapCounts[it] ?: 0) + 1
}
def top = (mapCounts.sort { a, b -> b.value <=> a.value }.collect{ it })[0..<n]
println "Rank Word Frequency\n==== ==== ========="
(0..<n).each { printf ("%4d %-4s %9d\n", it+1, top[it].key, top[it].value) }
}

Test:

def rawText = "http://www.gutenberg.org/files/135/135-0.txt".toURL().text
topWordCounts(rawText, 10)

Output:

Rank Word Frequency
==== ==== =========
   1 the      41036
   2 of       19946
   3 and      14940
   4 a        14589
   5 to       13939
   6 in       11204
   7 he        9645
   8 was       8619
   9 that      7922
  10 it        6659

Haskell[edit]

Translation of: Clojure
module Main where
 
import Control.Category -- (>>>)
import Data.Char -- toLower, isSpace
import Data.List -- sortBy, (Foldable(foldl')), filter
import Data.Ord -- Down
import System.IO -- stdin, ReadMode, openFile, hClose
import System.Environment -- getArgs
 
-- containers
import Data.Map.Strict (Map)
import qualified Data.Map.Strict as M
import qualified Data.IntMap.Strict as IM
 
-- text
import Data.Text (Text)
import qualified Data.Text as T
import qualified Data.Text.IO as T
 
frequencies :: Ord a => [a] -> Map a Integer
frequencies = foldl' (\m k -> M.insertWith (+) k 1 m) M.empty
{-# SPECIALIZE frequencies :: [Text] -> Map Text Integer #-}
 
main :: IO ()
main = do
args <- getArgs
(n,hand,filep) <- case length args of
0 -> return (10,stdin,False)
1 -> return (read $ head args,stdin,False)
_ -> let (ns:fp:_) = args
in fmap (\h -> (read ns,h,True)) (openFile fp ReadMode)
T.hGetContents hand >>=
(T.map toLower
>>> T.split isSpace
>>> filter (not <<< T.null)
>>> frequencies
>>> M.toList
>>> sortBy (comparing (Down <<< snd)) -- sort the opposite way
>>> take n
>>> print)
when filep (hClose hand)
Output:
$ ./word_count 10 < ~/doc/les_miserables*
[("the",40368),("of",19863),("and",14470),("a",14277),("to",13587),("in",11019),("he",9212),("was",8346),("that",7251),("his",6414)]

Java[edit]

Translation of: Kotlin
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.stream.Collectors;
 
public class WordCount {
public static void main(String[] args) throws IOException {
Path path = Paths.get("135-0.txt");
byte[] bytes = Files.readAllBytes(path);
String text = new String(bytes);
text = text.toLowerCase();
 
Pattern r = Pattern.compile("\\p{javaLowerCase}+");
Matcher matcher = r.matcher(text);
Map<String, Integer> freq = new HashMap<>();
while (matcher.find()) {
String word = matcher.group();
Integer current = freq.getOrDefault(word, 0);
freq.put(word, current + 1);
}
 
List<Map.Entry<String, Integer>> entries = freq.entrySet()
.stream()
.sorted((i1, i2) -> Integer.compare(i2.getValue(), i1.getValue()))
.limit(10)
.collect(Collectors.toList());
 
System.out.println("Rank Word Frequency");
System.out.println("==== ==== =========");
int rank = 1;
for (Map.Entry<String, Integer> entry : entries) {
String word = entry.getKey();
Integer count = entry.getValue();
System.out.printf("%2d  %-4s  %5d\n", rank++, word, count);
}
}
}
Output:
Rank  Word  Frequency
====  ====  =========
 1    the     41088
 2    of      19949
 3    and     14942
 4    a       14596
 5    to      13951
 6    in      11214
 7    he       9648
 8    was      8621
 9    that     7924
10    it       6661

Julia[edit]

Works with: Julia version 1.0
 
using FreqTables
 
txt = read("les-mis.txt", String)
words = split(replace(txt, r"\P{L}"i => " "))
table = sort(freqtable(words); rev=true)
println(table[1:10])
Output:
Dim1   │
───────┼──────
"the"  │ 36671
"of"   │ 19618
"and"  │ 14081
"to"   │ 13541
"a"    │ 13529
"in"   │ 10265
"was"  │  8545
"that" │  7326
"he"   │  6816
"had"  │  6140

Kotlin[edit]

The author of the Perl 6 entry has given a good account of the difficulties with this task and, in the absence of any clarification on the various issues, I've followed a similar 'literal' approach.

So, after first converting the text to lower case, I've assumed that a word is any sequence of one or more lower-case Unicode letters and obtained the same results as the Perl 6 version.

There is no change in the results if the numerals 0-9 are also regarded as letters.

// version 1.1.3
 
import java.io.File
 
fun main(args: Array<String>) {
val text = File("135-0.txt").readText().toLowerCase()
val r = Regex("""\p{javaLowerCase}+""")
val matches = r.findAll(text)
val wordGroups = matches.map { it.value }
.groupBy { it }
.map { Pair(it.key, it.value.size) }
.sortedByDescending { it.second }
.take(10)
println("Rank Word Frequency")
println("==== ==== =========")
var rank = 1
for ((word, freq) in wordGroups)
System.out.printf("%2d  %-4s  %5d\n", rank++, word, freq)
}
Output:
Rank  Word  Frequency
====  ====  =========
 1    the     41088
 2    of      19949
 3    and     14942
 4    a       14596
 5    to      13951
 6    in      11214
 7    he       9648
 8    was      8621
 9    that     7924
10    it       6661

Objeck[edit]

use System.IO.File;
use Collection;
use RegEx;
 
class Rosetta {
function : Main(args : String[]) ~ Nil {
if(args->Size() <> 1) {
return;
};
 
input := FileReader->ReadFile(args[0]);
filter := RegEx->New("\\w+");
words := filter->Find(input);
 
word_counts := StringMap->New();
each(i : words) {
word := words->Get(i)->As(String);
if(word <> Nil & word->Size() > 0) {
word := word->ToLower();
if(word_counts->Has(word)) {
count := word_counts->Find(word)->As(IntHolder);
count->Set(count->Get() + 1);
}
else {
word_counts->Insert(word, IntHolder->New(1));
};
};
};
 
count_words := IntMap->New();
words := word_counts->GetKeys();
each(i : words) {
word := words->Get(i)->As(String);
count := word_counts->Find(word)->As(IntHolder);
count_words->Insert(count->Get(), word);
};
 
counts := count_words->GetKeys();
counts->Sort();
 
index := 1;
"Rank\tWord\tFrequency"->PrintLine();
"====\t====\t===="->PrintLine();
for(i := count_words->Size() - 1; i >= 0; i -= 1;) {
if(count_words->Size() - 10 <= i) {
count := counts->Get(i);
word := count_words->Find(count)->As(String);
"{$index}\t{$word}\t{$count}"->PrintLine();
index += 1;
};
};
}
}

Output:

Rank    Word    Frequency
====    ====    ====
1       the     41036
2       of      19946
3       and     14940
4       a       14589
5       to      13939
6       in      11204
7       he      9645
8       was     8619
9       that    7922
10      it      6659

Perl[edit]

Translation of: Perl 6
$top = 10;
 
open $fh, "<", '135-0.txt';
($text = join '', <$fh>) =~ tr/A-Z/a-z/;
 
@matcher = (
qr/[a-z]+/, # simple 7-bit ASCII
qr/\w+/, # word characters with underscore
qr/[a-z0-9]+/, # word characters without underscore
);
 
for $reg (@matcher) {
print "\nTop $top using regex: " . $reg . "\n";
@matches = $text =~ /$reg/g;
my %words;
for $w (@matches) { $words{$w}++ };
$c = 0;
for $w ( sort { $words{$b} <=> $words{$a} } keys %words ) {
printf "%-7s %6d\n", $w, $words{$w};
last if ++$c >= $top;
}
}
Output:
Top 10 using regex: (?^:[a-z]+)
the      41089
of       19949
and      14942
a        14608
to       13951
in       11214
he        9648
was       8621
that      7924
it        6661

Top 10 using regex: (?^:\w+)
the      41036
of       19946
and      14940
a        14589
to       13939
in       11204
he        9645
was       8619
that      7922
it        6659

Top 10 using regex: (?^:[a-z0-9]+)
the      41089
of       19949
and      14942
a        14608
to       13951
in       11214
he        9648
was       8621
that      7924
it        6661

Perl 6[edit]

Works with: Rakudo version 2017.07

Note: much of the following exposition is no longer critical to the task as the requirements have been updated, but is left here for historical and informational reasons.

This is slightly trickier than it appears initially. The task specifically states: "A word is a sequence of one or more contiguous letters", so contractions and hyphenated words are broken up. Initially we might reach for a regex matcher like /\w+/ , but \w includes underscore, which is not a letter but a punctuation connector; and this text is full of underscores since that is how Project Gutenberg texts denote italicized text. The underscores are not actually parts of the words though, they are markup.

We might try /A-Za-z/ as a matcher but this text is bursting with French words containing various accented glyphs. Those are letters, so words will be incorrectly split up; (Misérables will be counted as 'mis' and 'rables', probably not what we want.)

Actually, in this case /A-Za-z/ returns very nearly the correct answer. Unfortunately, the name "Alèthe" appears once (only once!) in the text, gets incorrectly split into Al & the, and incorrectly reports 41089 occurrences of "the". The text has several words like "Panathenæa", "ça", "aérostiers" and "Keksekça" so the counts for 'a' are off too. The other 8 of the top 10 are "correct" using /A-Za-z/, but it is mostly by accident.

A more accurate regex matcher would be some kind of Unicode aware /\w/ minus underscore. It may also be useful, depending on your requirements, to recognize contractions with embedded apostrophes, hyphenated words, and hyphenated words broken across lines.

Here is a sample that shows the result when using various different matchers.

sub MAIN ($filename, $top = 10) {
my $file = $filename.IO.slurp.lc.subst(/ (<[\w]-[_]>'-')\n(<[\w]-[_]>) /, {$0 ~ $1}, :g );
my @matcher = (
rx/ <[a..z]>+ /, # simple 7-bit ASCII
rx/ \w+ /, # word characters with underscore
rx/ <[\w]-[_]>+ /, # word characters without underscore
rx/ <[\w]-[_]>+[["'"|'-'|"'-"]<[\w]-[_]>+]* / # word characters without underscore but with hyphens and contractions
);
for @matcher -> $reg {
say "\nTop $top using regex: ", $reg.perl;
.put for $file.comb( $reg ).Bag.sort(-*.value)[^$top];
}
}
Output:

Passing in the file name and 10:

Top 10 using regex: rx/ <[a..z]>+ /
the	41089
of	19949
and	14942
a	14608
to	13951
in	11214
he	9648
was	8621
that	7924
it	6661

Top 10 using regex: rx/ \w+ /
the	41035
of	19946
and	14940
a	14577
to	13939
in	11204
he	9645
was	8619
that	7922
it	6659

Top 10 using regex: rx/ <[\w]-[_]>+ /
the	41088
of	19949
and	14942
a	14596
to	13951
in	11214
he	9648
was	8621
that	7924
it	6661

Top 10 using regex: rx/ <[\w]-[_]>+[["'"|'-'|"'-"]<[\w]-[_]>+]* /
the	41081
of	19930
and	14934
a	14587
to	13735
in	11204
he	9607
was	8620
that	7825
it	6535

Phix[edit]

?"loading..."
constant subs = "\t\r\n_.,\"\'!;:?][()|=<>#/*{}[email protected]%&$",
reps = repeat(' ',length(subs)),
fn = open("135-0.txt","r")
string text = lower(substitute_all(get_text(fn),subs,reps))
close(fn)
sequence words = append(sort(split(text,no_empty:=true)),"")
constant wf = new_dict()
string last = words[1]
integer count = 1
for i=2 to length(words) do
if words[i]!=last then
setd({count,last},0,wf)
count = 0
last = words[i]
end if
count += 1
end for
count = 10
function visitor(object key, object /*data*/, object /*user_data*/)
 ?key
count -= 1
return count>0
end function
traverse_dict(routine_id("visitor"),0,wf,true)
Output:
loading...
{40743,"the"}
{19925,"of"}
{14881,"and"}
{14474,"a"}
{13704,"to"}
{11174,"in"}
{9623,"he"}
{8613,"was"}
{7867,"that"}
{6612,"it"}

PicoLisp[edit]

(setq *Delim " ^I^J^M-_.,\"'*[]?!&@#$%^\(\):;")
(setq *Skip (chop *Delim))
 
(de word+ NIL
(prog1
(lowc (till *Delim T))
(while (member (peek) *Skip) (char)) ) )
 
(off B)
(in "135-0.txt"
(until (eof)
(let W (word+)
(if (idx 'B W T) (inc (car @)) (set W 1)) ) ) )
(for L (head 10 (flip (by val sort (idx 'B))))
(println L (val L)) )
Output:
"the" 41088
"of" 19949
"and" 14942
"a" 14545
"to" 13950
"in" 11214
"he" 9647
"was" 8620
"that" 7924
"it" 6661

Python[edit]

Python2.7[edit]

import collections
import re
import string
import sys
 
def main():
counter = collections.Counter(re.findall(r"\w+",open(sys.argv[1]).read().lower()))
print counter.most_common(int(sys.argv[2]))
 
if __name__ == "__main__":
main()
Output:
$ python wordcount.py 135-0.txt 10
[('the', 41036), ('of', 19946), ('and', 14940), ('a', 14589), ('to', 13939),
 ('in', 11204), ('he', 9645), ('was', 8619), ('that', 7922), ('it', 6659)]

Python3.6[edit]

from collections import Counter
from re import findall
 
les_mis_file = 'les_mis_135-0.txt'
 
def _count_words(fname):
with open(fname) as f:
text = f.read()
words = findall(r'\w+', text.lower())
return Counter(words)
 
def most_common_words_in_file(fname, n):
counts = _count_words(fname)
for word, count in [['WORD', 'COUNT']] + counts.most_common(n):
print(f'{word:>10} {count:>6}')
 
 
if __name__ == "__main__":
n = int(input('How many?: '))
most_common_words_in_file(les_mis_file, n)
Output:
How many?: 10
      WORD  COUNT
       the  41036
        of  19946
       and  14940
         a  14586
        to  13939
        in  11204
        he   9645
       was   8619
      that   7922
        it   6659

R[edit]

I chose to remove apostrophes only if they're followed by an s (so "mom" and "mom's" will show up as the same word but "they" and "they're" won't). I also chose not to remove hyphens.

 
wordcount<-function(file,n){
punctuation=c("`","~","!","@","#","$","%","^","&","*","(",")","_","+","=","{","[","}","]","|","\\",":",";","\"","<",",",">",".","?","/","'s")
wordlist=scan(file,what=character())
wordlist=tolower(wordlist)
for(i in 1:length(punctuation)){
wordlist=gsub(punctuation[i],"",wordlist,fixed=T)
}
df=data.frame("Word"=sort(unique(wordlist)),"Count"=rep(0,length(unique(wordlist))))
for(i in 1:length(unique(wordlist))){
df[i,2]=length(which(wordlist==df[i,1]))
}
df=df[order(df[,2],decreasing = T),]
row.names(df)=1:nrow(df)
return(df[1:n,])
}
 
Output:
> wordcount("MobyDick.txt",10)
Read 212793 items
   Word Count
1   the 14346
2    of  6590
3   and  6340
4     a  4611
5    to  4572
6    in  4130
7  that  2903
8   his  2516
9    it  2308
10    i  1845

Racket[edit]

#lang racket
 
(define (all-words f (case-fold string-downcase))
(map case-fold (regexp-match* #px"\\w+" (file->string f))))
 
(define (l.|l| l) (cons (car l) (length l)))
 
(define (counts l (>? >)) (sort (map l.|l| (group-by values l)) >? #:key cdr))
 
(module+ main
(take (counts (all-words "data/les-mis.txt")) 10))
Output:
'(("the" . 41036)
  ("of" . 19946)
  ("and" . 14940)
  ("a" . 14589)
  ("to" . 13939)
  ("in" . 11204)
  ("he" . 9645)
  ("was" . 8619)
  ("that" . 7922)
  ("it" . 6659))

REXX[edit]

version 1[edit]

This REXX version doesn't need to sort the list of words.

Currently, this version recognizes all the accented (non-Latin) accented letters that are present in the text (file) that is specified to be used   (and some other non-Latin letters as well).   This means that the word     Alèthe     is treated as one word, not as two words     Al  the     (and not thereby adding two words).

This version also supports words that contain embedded apostrophes ( ' )     [that is, within a word, but not those words that start or end with an apostrophe; for those words, the apostrophe is elided].

Thus,   it's   is counted separately from   it   or   its.

Since REXX doesn't support UTF-8 encodings, code was added to this REXX version to support the accented letters in the mandated input file.

/*REXX pgm displays top 10 words in a file (includes foreign letters),  case is ignored.*/
parse arg fID top . /*obtain optional arguments from the CL*/
if fID=='' | fID=="," then fID= 'les_mes.TXT' /*None specified? Then use the default.*/
if top=='' | top=="," then top= 10 /* " " " " " " */
@.=0; c=0; abcL="abcdefghijklmnopqrstuvwxyz'" /*initialize word list, count; alphabet*/
q= "'"; abcU= abcL; upper abcU /*define uppercase version of alphabet*/
totW=0; accL= 'üéâÄàÅÇêëèïîìéæôÖòûùÿáíóúÑ' /* " " of some accented chrs*/
accU= 'ÜéâäàåçêëèïîìÉÆôöòûùÿáíóúñ' /* " lowercase accented characters.*/
accG= 'αßΓπΣσµτΦΘΩδφε' /* " some upper/lower Greek letters*/
a=abcL || abcL ||accL ||accL || accG /* " char string of after letters.*/
b=abcL || abcU ||accL ||accU || accG || xrange() /* " char string of before " */
x= 'Çà åÅ çÇ êÉ ëÉ áà óâ ªæ ºç ¿è ⌐é ¬ê ½ë «î »ï ▒ñ ┤ô ╣ù ╗û ╝ü' /*list of 16-bit chars.*/
xs= words(x) /*num. " " " */
!.= /*define the original word instances. */
do #=0 while lines(fID)\==0; $=linein(fID) /*loop whilst there are lines in file. */
if pos('├', $)\==0 then do k=1 for xs; _=word(x, k) /*any 16-bit chars? */
$=changestr('├'left(_, 1), $, right(_, 1) ) /*convert.*/
end /*k*/
$=translate( $, a, b) /*remove superfluous blanks in the line*/
do while $\=''; parse var $ z $ /*now, process each word in the $ list.*/
parse var z z1 2 zr '' -1 zL /*extract: first, middle, & last char.*/
if z1==q then do; z=zr; if z=='' then iterate; end /*starts with apostrophe? */
if zL==q then z=strip(left(z, length(z) - 1)) /*ends " " */
if z=='' then iterate /*if Z is now null, skip.*/
if @.z==0 then do; c=c+1; !.c=z; end /*bump word count; assign word to array*/
totW=totW + 1; @.[email protected].z + 1 /*bump total words & count of the word.*/
end /*while*/
end /*#*/
say commas(totW) ' words found ('commas(c) "unique) in " commas(#),
' records read from file: ' fID; say
say right('word', 40) " " center(' rank ', 6) " count " /*display title for output*/
say right('════', 40) " " center('══════', 6) " ═══════" /* " title separator.*/
tops=1
do until otops==tops | tops>top /*process enough words to satisfy TOP.*/
WL=; mk=0; otops=tops /*initialize the word list (to a NULL).*/
do n=1 for c; z=!.n; [email protected].z /*process the list of words in the file*/
if k==mk then WL=WL z /*handle cases of tied number of words.*/
if k> mk then do; mk=k; WL=z; end /*this word count is the current max. */
end /*n*/
wr=max( length(' rank '), length(top) ) /*find the maximum length of the rank #*/
do d=1 for words(WL); _=word(WL, d) /*process all words in the word list. */
if d==1 then w=max(10, length(@._) ) /*use length of the first number used. */
say right(@._, 40) right(commas(tops), wr) right(commas(@._), w)
@._= -1 /*nullify word count for next go around*/
end /*d*/ /* [↑] this allows a non-sorted list. */
tops=tops + words(WL) /*correctly handle any tied rankings.*/
end /*until*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n, #, "M")
e=verify(n, #'0', , verify(n, #"0.", 'M') ) - 4
do j=e to b by -3; _=insert(',', _, j); end /*j*/; return _
output   when using the default inputs:
574,122  words found  (23,414 unique)  in  67,663  records read from file:  les_mes.TXT

                                    word    rank    count
                                    ════   ══════  ═══════
                                     the      1     41,088
                                      of      2     19,949
                                     and      3     14,942
                                       a      4     14,595
                                      to      5     13,950
                                      in      6     11,214
                                      he      7      9,607
                                     was      8      8,620
                                    that      9      7,826
                                      it     10      6,535

To see a list of the top 1,000 words that show (among other things) words like   it's   and other accented words, see the discussion page.

version 2[edit]

Inspired by version 1 and adapted for ooRexx. It ignores all characters other than a-z and A-Z (which are translated to a-z).

/*REXX program   reads  and  displays  a  count  of words a file.  Word case is ignored.*/
Call time 'R'
abc='abcdefghijklmnopqrstuvwxyz'
abcABC=abc||translate(abc)
parse arg fID_top /*obtain optional arguments from the CL*/
Parse Var fid_top fid ',' top
if fID=='' then fID= 'mis.TXT' /* Use default if not specified */
if top=='' then top= 10 /* Use default if not specified */
occ.=0 /* occurrences of word (stem) in file */
wn=0
Do While lines(fid)>0 /*loop whilst there are lines in file. */
line=linein(fID)
line=translate(line,abc||abc,abcABC||xrange('00'x,'ff'x)) /*use only lowercase letters*/
Do While line<>''
Parse Var line word line /* take a word */
If occ.word=0 Then Do /* not yet in word list */
wn=wn+1
word.wn=word
End
occ.word=occ.word+1
End
End
Say 'We found' wn 'different words'
say right('word',40) ' rank count ' /* header */
say right('----',40) '------ -------' /* separator. */
tops=0
Do Until tops>=top | tops>=wn /*process enough words to satisfy TOP.*/
max_occ=0
tl='' /*initialize (possibly) a list of words*/
Do wi=1 To wn /*process the list of words in the file*/
word=word.wi /* take a word from the list */
Select
When occ.word>max_occ Then Do /* most occurrences so far */
tl=word /* candidate for output */
max_occ=occ.word /* current maximum occurrences */
End
When occ.word=max_occ Then Do /* tied */
tl=tl word /* add to output candidate */
End
Otherwise /* no candidate (yet) */
Nop
End
End
do d=1 for words(tl)
word=word(tl,d)
say right(word,40) right(tops+1,4) right(occ.word,8)
occ.word=0 /*nullify this word count for next time*/
End
tops=tops+words(tl) /*correctly handle the tied rankings. */
end
Say time('E') 'seconds elapsed'
Output:
We found 22820 different words
                                    word  rank   count
                                    ---- ------ -------
                                     the    1    41089
                                      of    2    19949
                                     and    3    14942
                                       a    4    14608
                                      to    5    13951
                                      in    6    11214
                                      he    7     9648
                                     was    8     8621
                                    that    9     7924
                                      it   10     6661
1.750000 seconds elapsed

Ring[edit]

 
# project : Word count
 
fp = fopen("Miserables.txt","r")
str = fread(fp, getFileSize(fp))
fclose(fp)
 
mis =substr(str, " ", nl)
mis = lower(mis)
mis = str2list(mis)
count = list(len(mis))
ready = []
for n = 1 to len(mis)
flag = 0
for m = 1 to len(mis)
if mis[n] = mis[m] and n != m
for p = 1 to len(ready)
if m = ready[p]
flag = 1
ok
next
if flag = 0
count[n] = count[n] + 1
ok
ok
next
if flag = 0
add(ready, n)
ok
next
for n = 1 to len(count)
for m = n + 1 to len(count)
if count[m] > count[n]
temp = count[n]
count[n] = count[m]
count[m] = temp
temp = mis[n]
mis[n] = mis[m]
mis[m] = temp
ok
next
next
for n = 1 to 10
see mis[n] + " " + (count[n] + 1) + nl
next
 
func getFileSize fp
c_filestart = 0
c_fileend = 2
fseek(fp,0,c_fileend)
nfilesize = ftell(fp)
fseek(fp,0,c_filestart)
return nfilesize
 
func swap(a, b)
temp = a
a = b
b = temp
return [a, b]
 

Output:

the	41089
of	19949
and	14942
a	14608
to	13951
in	11214
he	9648
was	8621
that	7924
it	6661

Ruby[edit]

 
class String
def wc
n = Hash.new(0)
downcase.scan(/[A-Za--ÿ]+/) { |g| n[g] += 1 }
n.sort{|n,g| n[1]<=>g[1]}
end
end
 
open('135-0.txt') { |n| n.read.wc[-10,10].each{|n| puts n[0].to_s+"->"+n[1].to_s} }
 
Output:
it->6661
that->7924
was->8621
he->9648
in->11214
to->13951
a->14596
and->14942
of->19949
the->41088

Rust[edit]

use std::cmp::Reverse;
use std::collections::HashMap;
use std::fs::File;
use std::io::{BufRead, BufReader};
 
extern crate regex;
use regex::Regex;
 
fn word_count(file: File, n: usize) {
let word_regex = Regex::new("(?i)[a-z']+").unwrap();
 
let mut words = HashMap::new();
for line in BufReader::new(file).lines() {
word_regex
.find_iter(&line.expect("Read error"))
.map(|m| m.as_str())
.for_each(|word| {
*words.entry(word.to_lowercase()).or_insert(0) += 1;
});
}
 
let mut words: Vec<_> = words.iter().collect();
words.sort_unstable_by_key(|&(word, count)| (Reverse(count), word));
 
for (word, count) in words.iter().take(n) {
println!("{:8} {:>8}", word, count);
}
}
 
fn main() {
word_count(File::open("135-0.txt").expect("File open error"), 10)
}
Output:
the         41083
of          19948
and         14941
a           14604
to          13951
in          11212
he           9604
was          8621
that         7824
it           6534

Scala[edit]

Featuring online remote file as input[edit]

Output:

Best seen running in your browser Scastie (remote JVM).

import scala.io.Source
 
object WordCount extends App {
 
val url = "http://www.gutenberg.org/files/135/135-0.txt"
val header = "Rank Word Frequency\n==== ======== ======"
 
def wordCnt =
Source.fromURL(url).getLines()
.filter(_.nonEmpty)
.flatMap(_.split("""\W+""")).toSeq
.groupBy(_.toLowerCase())
.mapValues(_.size).toSeq
.sortWith { case ((_, v0), (_, v1)) => v0 > v1 }
.take(10).zipWithIndex
 
println(header)
wordCnt.foreach {
case ((word, count), rank) => println(f"${rank + 1}%4d $word%-8s $count%6d")
}
 
println(s"\nSuccessfully completed without errors. [total ${scala.compat.Platform.currentTime - executionStart} ms]")
 
}
Output:
Rank Word  Frequency
==== ======== ======
   1 the       41036
   2 of        19946
   3 and       14940
   4 a         14589
   5 to        13939
   6 in        11204
   7 he         9645
   8 was        8619
   9 that       7922
  10 it         6659

Successfully completed without errors. [total 4528 ms]

Sidef[edit]

var count = Hash()
var file = File(ARGV[0] \\ '135-0.txt')
 
file.open_r.each { |line|
line.lc.scan(/[\pL]+/).each { |word|
count{word} := 0 ++
}
}
 
var top = count.sort_by {|_,v| v }.last(10).flip
 
top.each { |pair|
say "#{pair.key}\t-> #{pair.value}"
}
Output:
the	-> 41088
of	-> 19949
and	-> 14942
a	-> 14596
to	-> 13951
in	-> 11214
he	-> 9648
was	-> 8621
that	-> 7924
it	-> 6661

Simula[edit]

COMMENT COMPILE WITH
$ cim -m64 word-count.sim
;
BEGIN
 
COMMENT ----- CLASSES FOR GENERAL USE ;
 
 ! ABSTRACT HASH KEY TYPE ;
CLASS HASHKEY;
VIRTUAL:
PROCEDURE HASH IS
INTEGER PROCEDURE HASH;;
PROCEDURE EQUALTO IS
BOOLEAN PROCEDURE EQUALTO(K); REF(HASHKEY) K;;
BEGIN
END HASHKEY;
 
 ! ABSTRACT HASH VALUE TYPE ;
CLASS HASHVAL;
BEGIN
 ! THERE IS NOTHING REQUIRED FOR THE VALUE TYPE ;
END HASHVAL;
 
CLASS HASHMAP;
BEGIN
CLASS INNERHASHMAP(N); INTEGER N;
BEGIN
 
INTEGER PROCEDURE INDEX(K); REF(HASHKEY) K;
BEGIN
INTEGER I;
IF K == NONE THEN
ERROR("HASHMAP.INDEX: NONE IS NOT A VALID KEY");
I := MOD(K.HASH,N);
LOOP:
IF KEYTABLE(I) == NONE OR ELSE KEYTABLE(I).EQUALTO(K) THEN
INDEX := I
ELSE BEGIN
I := IF I+1 = N THEN 0 ELSE I+1;
GO TO LOOP;
END;
END INDEX;
 
 ! PUT SOMETHING IN ;
PROCEDURE PUT(K,V); REF(HASHKEY) K; REF(HASHVAL) V;
BEGIN
INTEGER I;
IF V == NONE THEN
ERROR("HASHMAP.PUT: NONE IS NOT A VALID VALUE");
I := INDEX(K);
IF KEYTABLE(I) == NONE THEN BEGIN
IF SIZE = N THEN
ERROR("HASHMAP.PUT: TABLE FILLED COMPLETELY");
KEYTABLE(I) :- K;
VALTABLE(I) :- V;
SIZE := SIZE+1;
END ELSE
VALTABLE(I) :- V;
END PUT;
 
 ! GET SOMETHING OUT ;
REF(HASHVAL) PROCEDURE GET(K); REF(HASHKEY) K;
BEGIN
INTEGER I;
IF K == NONE THEN
ERROR("HASHMAP.GET: NONE IS NOT A VALID KEY");
I := INDEX(K);
IF KEYTABLE(I) == NONE THEN
GET :- NONE ! ERROR("HASHMAP.GET: KEY NOT FOUND");
ELSE
GET :- VALTABLE(I);
END GET;
 
PROCEDURE CLEAR;
BEGIN
INTEGER I;
FOR I := 0 STEP 1 UNTIL N-1 DO BEGIN
KEYTABLE(I) :- NONE;
VALTABLE(I) :- NONE;
END;
SIZE := 0;
END CLEAR;
 
 ! DATA MEMBERS OF CLASS HASHMAP ;
REF(HASHKEY) ARRAY KEYTABLE(0:N-1);
REF(HASHVAL) ARRAY VALTABLE(0:N-1);
INTEGER SIZE;
 
END INNERHASHMAP;
 
PROCEDURE PUT(K,V); REF(HASHKEY) K; REF(HASHVAL) V;
BEGIN
IF IMAP.SIZE >= 0.75 * IMAP.N THEN
BEGIN
COMMENT RESIZE HASHMAP ;
REF(INNERHASHMAP) NEWIMAP;
REF(ITERATOR) IT;
NEWIMAP :- NEW INNERHASHMAP(2 * IMAP.N);
IT :- NEW ITERATOR(THIS HASHMAP);
WHILE IT.MORE DO
BEGIN
REF(HASHKEY) KEY;
KEY :- IT.NEXT;
NEWIMAP.PUT(KEY, IMAP.GET(KEY));
END;
IMAP.CLEAR;
IMAP :- NEWIMAP;
END;
IMAP.PUT(K, V);
END;
 
REF(HASHVAL) PROCEDURE GET(K); REF(HASHKEY) K;
GET :- IMAP.GET(K);
 
PROCEDURE CLEAR;
IMAP.CLEAR;
 
INTEGER PROCEDURE SIZE;
SIZE := IMAP.SIZE;
 
REF(INNERHASHMAP) IMAP;
 
IMAP :- NEW INNERHASHMAP(16);
END HASHMAP;
 
CLASS ITERATOR(H); REF(HASHMAP) H;
BEGIN
INTEGER POS,KEYCOUNT;
 
BOOLEAN PROCEDURE MORE;
MORE := KEYCOUNT < H.SIZE;
 
REF(HASHKEY) PROCEDURE NEXT;
BEGIN
INSPECT H DO
INSPECT IMAP DO
BEGIN
WHILE KEYTABLE(POS) == NONE DO
POS := POS+1;
NEXT :- KEYTABLE(POS);
KEYCOUNT := KEYCOUNT+1;
POS := POS+1;
END;
END NEXT;
 
END ITERATOR;
 
COMMENT ----- PROBLEM SPECIFIC CLASSES ;
 
HASHKEY CLASS TEXTHASHKEY(T); VALUE T; TEXT T;
BEGIN
INTEGER PROCEDURE HASH;
BEGIN
INTEGER I;
T.SETPOS(1);
WHILE T.MORE DO
I := 31*I+RANK(T.GETCHAR);
HASH := I;
END HASH;
BOOLEAN PROCEDURE EQUALTO(K); REF(HASHKEY) K;
EQUALTO := T = K QUA TEXTHASHKEY.T;
END TEXTHASHKEY;
 
HASHVAL CLASS COUNTER;
BEGIN
INTEGER COUNT;
END COUNTER;
 
REF(INFILE) INF;
REF(HASHMAP) MAP;
REF(TEXTHASHKEY) KEY;
REF(COUNTER) VAL;
REF(ITERATOR) IT;
TEXT LINE, WORD;
INTEGER I, J, MAXCOUNT, LINENO;
INTEGER ARRAY MAXCOUNTS(1:10);
REF(TEXTHASHKEY) ARRAY MAXWORDS(1:10);
 
WORD :- BLANKS(1000);
MAP :- NEW HASHMAP;
 
COMMENT MAP WORDS TO COUNTERS ;
 
INF :- NEW INFILE("135-0.txt");
INF.OPEN(BLANKS(4096));
WHILE NOT INF.LASTITEM DO
BEGIN
BOOLEAN INWORD;
 
PROCEDURE SAVE;
BEGIN
IF WORD.POS > 1 THEN
BEGIN
KEY :- NEW TEXTHASHKEY(WORD.SUB(1, WORD.POS - 1));
VAL :- MAP.GET(KEY);
IF VAL == NONE THEN
BEGIN
VAL :- NEW COUNTER;
MAP.PUT(KEY, VAL);
END;
VAL.COUNT := VAL.COUNT + 1;
WORD := " ";
WORD.SETPOS(1);
END;
END SAVE;
 
LINENO := LINENO + 1;
LINE :- COPY(INF.IMAGE).STRIP; INF.INIMAGE;
 
COMMENT SEARCH WORDS IN LINE ;
COMMENT A WORD IS ANY SEQUENCE OF LETTERS ;
 
INWORD := FALSE;
LINE.SETPOS(1);
WHILE LINE.MORE DO
BEGIN
CHARACTER CH;
CH := LINE.GETCHAR;
IF CH >= 'a' AND CH <= 'z' THEN
CH := CHAR(RANK(CH) - RANK('a') + RANK('A'));
IF CH >= 'A' AND CH <= 'Z' THEN
BEGIN
IF NOT INWORD THEN
BEGIN
SAVE;
INWORD := TRUE;
END;
WORD.PUTCHAR(CH);
END ELSE
BEGIN
IF INWORD THEN
BEGIN
SAVE;
INWORD := FALSE;
END;
END;
END;
SAVE; COMMENT LAST WORD ;
END;
INF.CLOSE;
 
COMMENT FIND 10 MOST COMMON WORDS ;
 
IT :- NEW ITERATOR(MAP);
WHILE IT.MORE DO
BEGIN
KEY :- IT.NEXT;
VAL :- MAP.GET(KEY);
FOR I := 1 STEP 1 UNTIL 10 DO
BEGIN
IF VAL.COUNT >= MAXCOUNTS(I) THEN
BEGIN
FOR J := 10 STEP -1 UNTIL I + 1 DO
BEGIN
MAXCOUNTS(J) := MAXCOUNTS(J - 1);
MAXWORDS(J) :- MAXWORDS(J - 1);
END;
MAXCOUNTS(I) := VAL.COUNT;
MAXWORDS(I) :- KEY;
GO TO BREAK;
END;
END;
BREAK:
END;
 
COMMENT OUTPUT 10 MOST COMMON WORDS ;
 
FOR I := 1 STEP 1 UNTIL 10 DO
BEGIN
IF MAXWORDS(I) =/= NONE THEN
BEGIN
OUTINT(MAXCOUNTS(I), 10);
OUTTEXT(" ");
OUTTEXT(MAXWORDS(I) QUA TEXTHASHKEY.T);
OUTIMAGE;
END;
END;
 
END
 
Output:
     41089 THE
     19949 OF
     14942 AND
     14608 A
     13951 TO
     11214 IN
      9648 HE
      8621 WAS
      7924 THAT
      6661 IT

6 garbage collection(s) in 0.2 seconds.

UNIX Shell[edit]

Works with: Bash
Works with: zsh

This is derived from Doug McIlroy's original 6-line note in the ACM article cited in the task.

#!/bin/sh
cat ${1} | tr -cs A-Za-z '\n' | tr A-Z a-z | sort | uniq -c | sort -rn | sed ${2}q


Output:
$ ./wordcount.sh 135-0.txt 10 
41089 the
19949 of
14942 and
14608 a
13951 to
11214 in
9648 he
8621 was
7924 that
6661 it

VBA[edit]

In order to use it, you have to adapt the PATHFILE Const.

 
Option Explicit
 
Private Const PATHFILE As String = "C:\HOME\VBA\ROSETTA"
 
Sub Main()
Dim arr
Dim Dict As Object
Dim Book As String, temp As String
Dim T#
T = Timer
Book = ExtractTxt(PATHFILE & "\les miserables.txt")
temp = RemovePunctuation(Book)
temp = UCase(temp)
arr = Split(temp, " ")
Set Dict = CreateObject("Scripting.Dictionary")
FillDictionary Dict, arr
Erase arr
SortDictByFreq Dict, arr
DisplayTheTopMostUsedWords arr, 10
 
Debug.Print "Words different in this book : " & Dict.Count
Debug.Print "-------------------------"
Debug.Print ""
Debug.Print "Optionally : "
Debug.Print "Frequency of the word MISERABLE : " & DisplayFrequencyOf("MISERABLE", Dict)
Debug.Print "Frequency of the word DISASTER : " & DisplayFrequencyOf("DISASTER", Dict)
Debug.Print "Frequency of the word ROSETTA_CODE : " & DisplayFrequencyOf("ROSETTA_CODE", Dict)
Debug.Print "-------------------------"
Debug.Print "Execution Time : " & Format(Timer - T, "0.000") & " sec."
End Sub
 
Private Function ExtractTxt(strFile As String) As String
'http://rosettacode.org/wiki/File_input/output#VBA
Dim i As Integer
i = FreeFile
Open strFile For Input As #i
ExtractTxt = Input(LOF(1), #i)
Close #i
End Function
 
Private Function RemovePunctuation(strBook As String) As String
Dim T, i As Integer, temp As String
Const PUNCT As String = """,;:!?."
T = Split(StrConv(PUNCT, vbUnicode), Chr(0))
temp = strBook
For i = LBound(T) To UBound(T) - 1
temp = Replace(temp, T(i), " ")
Next
temp = Replace(temp, "--", " ")
temp = Replace(temp, "...", " ")
temp = Replace(temp, vbCrLf, " ")
RemovePunctuation = Replace(temp, " ", " ")
End Function
 
Private Sub FillDictionary(d As Object, a As Variant)
Dim L As Long
For L = LBound(a) To UBound(a)
If a(L) <> "" Then _
d(a(L)) = d(a(L)) + 1
Next
End Sub
 
Private Sub SortDictByFreq(d As Object, myArr As Variant)
Dim K
Dim L As Long
ReDim myArr(1 To d.Count, 1 To 2)
For Each K In d.keys
L = L + 1
myArr(L, 1) = K
myArr(L, 2) = CLng(d(K))
Next
SortArray myArr, LBound(myArr), UBound(myArr), 2
End Sub
 
Private Sub SortArray(a, Le As Long, Ri As Long, Col As Long)
Dim ref As Long, L As Long, r As Long, temp As Variant
ref = a((Le + Ri) \ 2, Col)
L = Le
r = Ri
Do
Do While a(L, Col) < ref
L = L + 1
Loop
Do While ref < a(r, Col)
r = r - 1
Loop
If L <= r Then
temp = a(L, 1)
a(L, 1) = a(r, 1)
a(r, 1) = temp
temp = a(L, 2)
a(L, 2) = a(r, 2)
a(r, 2) = temp
L = L + 1
r = r - 1
End If
Loop While L <= r
If L < Ri Then SortArray a, L, Ri, Col
If Le < r Then SortArray a, Le, r, Col
End Sub
 
Private Sub DisplayTheTopMostUsedWords(arr As Variant, Nb As Long)
Dim L As Long, i As Integer
i = 1
Debug.Print "Rank Word Frequency"
Debug.Print "==== ======= ========="
For L = UBound(arr) To UBound(arr) - Nb + 1 Step -1
Debug.Print Left(CStr(i) & " ", 5) & Left(arr(L, 1) & " ", 8) & " " & Format(arr(L, 2), "0 000")
i = i + 1
Next
End Sub
 
Private Function DisplayFrequencyOf(Word As String, d As Object) As Long
If d.Exists(Word) Then _
DisplayFrequencyOf = d(Word)
End Function
Output:
Words different in this book : 25884
-------------------------
Rank Word    Frequency
==== ======= =========
1    THE      40 831
2    OF       19 807
3    AND      14 860
4    A        14 453
5    TO       13 641
6    IN       11 133
7    HE       9 598
8    WAS      8 617
9    THAT     7 807
10   IT       6 517

Optionally : 
Frequency of the word MISERABLE : 35
Frequency of the word DISASTER : 12
Frequency of the word ROSETTA_CODE : 0
-------------------------
Execution Time : 7,785 sec.

zkl[edit]

fname,count := vm.arglist;	// grab cammand line args
 
// words may have leading or trailing "_", ie "the" and "_the"
File(fname).pump(Void,"toLower", // read the file line by line and hash words
RegExp("[a-z]+").pump.fp1(Dictionary().incV)) // line-->(word:count,..)
.toList().copy().sort(fcn(a,b){ b[1]<a[1] })[0,count.toInt()] // hash-->list
.pump(String,Void.Xplode,"%s,%s\n".fmt).println();
Output:
$ zkl bbb ~/Documents/Les\ Miserables.txt 10
the,41089
of,19949
and,14942
a,14608
to,13951
in,11214
he,9648
was,8621
that,7924
it,6661