Word frequency: Difference between revisions
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Dim Dict As Object |
Dim Dict As Object |
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Dim Book As String, temp As String |
Dim Book As String, temp As String |
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Dim K |
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Dim T# |
Dim T# |
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T = Timer |
T = Timer |
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FillDictionary Dict, arr |
FillDictionary Dict, arr |
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Erase arr |
Erase arr |
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| ⚫ | |||
| ⚫ | |||
SortDictByFreq Dict, arr |
SortDictByFreq Dict, arr |
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DisplayTheTopMostUsedWords arr, 10 |
DisplayTheTopMostUsedWords arr, 10 |
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| ⚫ | |||
| ⚫ | |||
Debug.Print "" |
Debug.Print "" |
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Debug.Print "Optionally : " |
Debug.Print "Optionally : " |
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Debug.Print "Frequency of the word DISASTER : " & DisplayFrequencyOf("DISASTER", Dict) |
Debug.Print "Frequency of the word DISASTER : " & DisplayFrequencyOf("DISASTER", Dict) |
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Debug.Print "Frequency of the word ROSETTA_CODE : " & DisplayFrequencyOf("ROSETTA_CODE", Dict) |
Debug.Print "Frequency of the word ROSETTA_CODE : " & DisplayFrequencyOf("ROSETTA_CODE", Dict) |
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Debug.Print "-------------------------" |
Debug.Print "-------------------------" |
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Debug.Print "Execution Time : " & Format(Timer - T, "0.000") & " sec." |
Debug.Print "Execution Time : " & Format(Timer - T, "0.000") & " sec." |
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Revision as of 08:14, 27 March 2018
- 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
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])
D
<lang D>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);
}
}</lang>
- 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#
<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)
Factor
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. <lang factor> USING: ascii io math.statistics prettyprint sequences splitting ; IN: rosetta-code.word-count
lines " " join " .,?!:;()\"-" split harvest [ >lower ] map sorted-histogram reverse 10 head . </lang>
- Output:
{
{ "the" 41021 }
{ "of" 19945 }
{ "and" 14938 }
{ "a" 14522 }
{ "to" 13938 }
{ "in" 11201 }
{ "he" 9600 }
{ "was" 8618 }
{ "that" 7822 }
{ "it" 6532 }
}
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
Objeck
<lang objeck>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{$count}"->PrintLine();
index += 1;
};
};
}
}</lang>
Output:
Rank Word Frequency ==== ==== ==== 1 the 41036 2 of 19946 3 and 14940 4 a 14589 5 to 13939 6 in 11204 7 he 9645 8 was 8619 9 that 7922 10 it 6659
Perl 6
Note: much of the following exposition is no longer critical to the task as the requirements have been updated, but is left here for historical and informational reasons.
This is slightly trickier than it appears initially. The task specifically states: "A word is a sequence of one or more contiguous letters", so contractions and hyphenated words are broken up. Initially we might reach for a regex matcher like /\w+/ , but \w includes underscore, which is not a letter but a punctuation connector; and this text is full of underscores since that is how Project Gutenberg texts denote italicized text. The underscores are not actually parts of the words though, they are markup.
We might try /A-Za-z/ as a matcher but this text is bursting with French words containing various accented glyphs. Those are letters, so words will be incorrectly split up; (Misérables will be counted as 'mis' and 'rables', probably not what we want.)
Actually, in this case /A-Za-z/ returns very nearly the correct answer. Unfortunately, the name "Alèthe" appears once (only once!) in the text, gets incorrectly split into Al & the, and incorrectly reports 41089 occurrences of "the". The text has several words like "Panathenæa", "ça", "aérostiers" and "Keksekça" so the counts for 'a' are off too. The other 8 of the top 10 are "correct" using /A-Za-z/, but it is mostly by accident.
A more accurate regex matcher would be some kind of Unicode aware /\w/ minus underscore. It may also be useful, depending on your requirements, to recognize contractions with embedded apostrophes, hyphenated words, and hyphenated words broken across lines.
Here is a sample that shows the result when using various different matchers. <lang perl6>sub MAIN ($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];
}
}</lang>
- 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
<lang Phix>?"loading..." constant subs = "\t\r\n_.,\"\'!;:?][()|=<>#/*{}+@%&$",
reps = repeat(' ',length(subs)),
fn = open("135-0.txt","r")
string text = lower(substitute_all(get_text(fn),subs,reps)) close(fn) sequence words = append(sort(split(text,no_empty:=true)),"") constant wf = new_dict() string last = words[1] integer count = 1 for i=2 to length(words) do
if words[i]!=last then
setd({count,last},0,wf)
count = 0
last = words[i]
end if
count += 1
end for count = 10 function visitor(object key, object /*data*/, object /*user_data*/)
?key count -= 1 return count>0
end function traverse_dict(routine_id("visitor"),0,wf,true)</lang>
- Output:
loading...
{40743,"the"}
{19925,"of"}
{14881,"and"}
{14474,"a"}
{13704,"to"}
{11174,"in"}
{9623,"he"}
{8613,"was"}
{7867,"that"}
{6612,"it"}
PicoLisp
<lang PicoLisp>(setq *Delim " ^I^J^M-_.,\"'*[]?!&@#$%^\(\):;") (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)) )</lang>
- Output:
"the" 41088 "of" 19949 "and" 14942 "a" 14545 "to" 13950 "in" 11214 "he" 9647 "was" 8620 "that" 7924 "it" 6661
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 (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. <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 /* " " " " " " */ @.=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; @.z=@.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; k=@.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 _</lang>
- 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
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). <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 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'</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
Ring
<lang ring>
- project : Word count
- date : 2017/11/30
- author : Gal Zsolt (~ calmosoft ~)
- email : <calmosoft@gmail.com>
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]
</lang> Output:
the 41089 of 19949 and 14942 a 14608 to 13951 in 11214 he 9648 was 8621 that 7924 it 6661
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
Scala
Featuring online remote file as input
<lang Scala>import scala.io.Source
object WordCount extends App {
val (url, header) = ("http://www.gutenberg.org/files/135/135-0.txt",
"Rank Word Frequency\n==== ======== ======")
def wordCnt = Source.fromURL(url).
getLines().filter(!_.isEmpty).
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]")
}</lang>
- 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
<lang ruby>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}"
}</lang>
- Output:
the -> 41088 of -> 19949 and -> 14942 a -> 14596 to -> 13951 in -> 11214 he -> 9648 was -> 8621 that -> 7924 it -> 6661
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
This is derived from Doug McIlroy's original 6-line note in the ACM article cited in the task. <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
VBA
In order to use it, you have to adapt the PATHFILE Const.
<lang vb> 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</lang>
- 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
<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