Word frequency: Difference between revisions
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parse var z z1 2 zr '' -1 zL /*extract: first, middle, and last char*/ |
parse var z z1 2 zr '' -1 zL /*extract: first, middle, and last char*/ |
||
if z1==q then do; z=zr; if z=='' then iterate; end /*starts with apostrophe? */ |
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 zL==q then z=strip(left(z, length(z) - 1)) /*ends " " */ |
||
if z=='' then iterate /*if Z is now null, skip.*/ |
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*/ |
if @.z==0 then do; c=c+1; !.c=z; end /*bump word count; assign word to array*/ |
||
Revision as of 20:39, 24 August 2017
- 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
- 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.
- References
Clojure
<lang clojure>(defn count-words [file n]
(->> file slurp clojure.string/lower-case (re-seq #"\w+") frequencies (sort-by val >) (take n)))</lang>
- 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])
F#
<lang fsharp> 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) </lang>
- Output:
("the", 41088)
("of", 19949)
("and", 14942)
("a", 14596)
("to", 13951)
("in", 11214)
("he", 9648)
("was", 8621)
("that", 7924)
("it", 6661)
Julia
<lang julia># v0.6
using FreqTables
txt = readstring("les-mis.txt") words = split(replace(txt, r"\P{L}"i, " ")) table = sort(freqtable(words); rev=true) println(table[1:10])10-element Named Array{Int64,1}</lang>
- Output:
Dim1 │ ───────┼────── "the" │ 36671 "of" │ 19618 "and" │ 14081 "to" │ 13541 "a" │ 13529 "in" │ 10265 "was" │ 8545 "that" │ 7326 "he" │ 6816 "had" │ 6140
Kotlin
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. <lang scala>// 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)
}</lang>
- 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
Perl 6
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. The more accurate regex matcher is some kind of Unicode aware /\w/ minus underscore.
( Really, a better regex would allow for contractions and embedded apostrophes but that is beyond the scope of this task as it stands. There are words like cat-o'-nine-tails and will-o'-the-wisps in there too to make your day even more interesting. )
<lang perl6>sub MAIN ($filename, $top = 10) {
.say for ($filename.IO.slurp.lc ~~ m:g/[<[\w]-[_]>]+/)».Str.Bag.sort(-*.value)[^$top]
}</lang>
- Output:
Passing in the file name and 10:
the => 41088 of => 19949 and => 14942 a => 14596 to => 13951 in => 11214 he => 9648 was => 8621 that => 7924 it => 6661
This satisfies the task requirements as they are written, but leaves a lot to be desired. For my own amusement here is a version that recognizes contractions with embedded apostrophes, hyphenated words, and hyphenated words broken across lines. Returns the top N words and counts sorted by length with a secondary sort on frequency just to be different (and to demonstrate that it really does what is claimed.)
<lang perl6>sub MAIN ($filename, $top = 10) {
.say for ($filename.IO.slurp.lc.subst(/ (\w '-') \n ( \w ) /, {$0 ~ $1}, :g )
~~ m:g/ <[\w]-[_]>+[["'"|'-'|"'-"]<[\w]-[_]>+]* /)».Str.Bag.sort( {-$^a.key.chars, -$a.value} )[^$top];
}</lang>
- Output:
Again, passing in the same file name and 10:
police-agent-ja-vert-was-found-drowned-un-der-a-boat-of-the-pont-au-change => 1 jésus-mon-dieu-bancroche-à-bas-la-lune => 1 die-of-hunger-if-you-have-a-fire => 1 guimard-guimardini-guimardinette => 1 monsieur-i-don't-know-your-name => 1 sainte-croix-de-la-bretonnerie => 2 die-of-cold-if-you-have-bread => 1 petit-picpus-sainte-antoine => 1 saint-jacques-du-haut-pas => 7 chemin-vert-saint-antoine => 3
Python
Python2.7
<lang python>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()</lang>
- 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
<lang python>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)</lang>
- 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
Racket
<lang racket>#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))</lang>
- Output:
'(("the" . 41036)
("of" . 19946)
("and" . 14940)
("a" . 14589)
("to" . 13939)
("in" . 11204)
("he" . 9645)
("was" . 8619)
("that" . 7922)
("it" . 6659))
REXX
version 1
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 that is specified to be used (and some other non-Latin letters as well).
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. <lang rexx>/*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 /* " " " " " " */ c=0; @.=0; abcL="abcdefghijklmnopqrstuvwxyz'" /*initialize word list, count; alphabet*/ q= "'"; abcU= abcL; upper abcU /*define uppercase version of alphabet*/
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 #=1 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*/
$=space( 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, and 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*/
@@.z=z /*save the original case of the word. */
@.z=@.z + 1 /*bump the count of occurrences of word*/
end /*while*/
end /*#*/
say commas(c) ' words found in ' commas(#-1) ' records read from the file: ' fID say say right('word', 40) " " center(' rank ', 6) " count " /*display title for output*/ say right('════', 40) " " center('══════', 6) " ═══════" /* " title separator.*/
/* [↓] note the BY incr.*/
do tops=1 by 0 until otops==tops|tops>top /*process enough words to satisfy TOP.*/
tl=; mc=0; otops=tops /*initialize (possibly) a list of words*/
do n=1 for c; z=!.n; count=@.z /*process the list of words in the file*/
if count<1 then iterate /*Is count too small? Then ignore it.*/
if count==mc then tl=tl z /*handle cases of tied number of words.*/
if count> mc then do; mc=count /*this word count is the current max. */
tl=z /* " word " " " " */
end
end /*n*/
w=0 /*will be the maximum length of count. */
wr=max( length(' rank '), length(top) ) /*find the maximum length of the rank #*/
do d=1 for words(tl); _=word(tl, d) /*process each of the words in the TL. */
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)
@._=0 /*nullify this word count for next time*/
end /*d*/
tops=tops + words(tl) /*correctly handle the tied rankings. */
end /*tops*/
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 _</lang>
- output when using the default inputs:
23,414 words found in 67,663 records read from the 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
Inspired by version 1 and adapted for ooRexx. It ignores all characters other than a-z and A-Z (which are tanslated to a-z). <lang>/*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 occurences 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 candudate */
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'</lang>
- 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
Ruby
<lang ruby> class String
def wc
n = Hash.new(0)
downcase.scan(/[A-Za-zÀ-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} } </lang>
- Output:
it->6661 that->7924 was->8621 he->9648 in->11214 to->13951 a->14596 and->14942 of->19949 the->41088
Simula
<lang simula>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 </lang>
- 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
<lang bash>#!/bin/sh cat ${1} | tr -cs A-Za-z '\n' | tr A-Z a-z | sort | uniq -c | sort -rn | sed ${2}q</lang>
- 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
zkl
<lang zkl>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();</lang>
- 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