Re: recursively count the words occurrence in the text files
On Thu, 30 Dec 2010 10:34:59 -0800, S Mathias wrote: I just can't google for it: (...) Just out of curiosity... why not? :-? http://www.google.com/search?hl=encomplete=0biw=1280bih=839sa=Xei=WpodTYuiIImi8QORlqXrBQved=0CBYQvwUoAQq=bash+count+word+occurrences+in+filesspell=1 Greetings, -- Camaleón -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of unsubscribe. Trouble? Contact listmas...@lists.debian.org Archive: http://lists.debian.org/pan.2010.12.31.08.59...@gmail.com
Re: recursively count the words occurrence in the text files
Hi, S Mathias wrote: I just can't google for it: I'm searching for a bash one liner (awk, perl, or anything) for this: there are text files, in several directories: mkdir one mkdir two mkdir three echo word1 word2 word3 one/asf.txt echo word2 word4, word5 one/asfcxv saf.txt echo word1. word2 one/dsgsdg.txt echo word6, word3! two/sdgsd dsf.txt echo word6 two/ergd.txt echo asdf, word2 three/werdf.txt echo word7, word8 word9 word10 three/qwerb erfsdgdsg.txt echo word4 word3 three/web erg as.txt so it does the magic* recursively: $ SOMEMAGIC output.txt cat output.txt asdf 1 word1 2 word2 4 word3 3 word4 2 word5 1 word6 2 word7 1 word8 1 word9 1 word10 1 $ find . ! -name output.txt -type f|xargs cat | \ tr -s ' .,' '\n\n\n'|sort -u | \ while read x;do find . ! -name output.txt -type f| \ xargs cat|echo $x: `grep -c $x `;done | \ tee output.txt That was fun... of course, it was a one liner, but I added the formatting above for the email ;-) Cheers -- Kind Regards AndrewM Andrew McGlashan Broadband Solutions now including VoIP -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of unsubscribe. Trouble? Contact listmas...@lists.debian.org Archive: http://lists.debian.org/4d1db31e.1070...@affinityvision.com.au
recursively count the words occurrence in the text files
I just can't google for it: I'm searching for a bash one liner (awk, perl, or anything) for this: there are text files, in several directories: mkdir one mkdir two mkdir three echo word1 word2 word3 one/asf.txt echo word2 word4, word5 one/asfcxv saf.txt echo word1. word2 one/dsgsdg.txt echo word6, word3! two/sdgsd dsf.txt echo word6 two/ergd.txt echo asdf, word2 three/werdf.txt echo word7, word8 word9 word10 three/qwerb erfsdgdsg.txt echo word4 word3 three/web erg as.txt so it does the magic* recursively: $ SOMEMAGIC output.txt cat output.txt asdf 1 word1 2 word2 4 word3 3 word4 2 word5 1 word6 2 word7 1 word8 1 word9 1 word10 1 $ *recursively count the words occurrence in the text files like: word1 2 can anyone point to a howto/link? [re: i just can't google for it :\] -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of unsubscribe. Trouble? Contact listmas...@lists.debian.org Archive: http://lists.debian.org/788911.43739...@web121407.mail.ne1.yahoo.com
Re: recursively count the words occurrence in the text files
On Thu, 30 Dec 2010 10:34:59 -0800 (PST) S Mathias smathias1...@yahoo.com wrote: I just can't google for it: I'm searching for a bash one liner (awk, perl, or anything) for this: there are text files, in several directories: mkdir one mkdir two mkdir three echo word1 word2 word3 one/asf.txt echo word2 word4, word5 one/asfcxv saf.txt echo word1. word2 one/dsgsdg.txt echo word6, word3! two/sdgsd dsf.txt echo word6 two/ergd.txt echo asdf, word2 three/werdf.txt echo word7, word8 word9 word10 three/qwerb erfsdgdsg.txt echo word4 word3 three/web erg as.txt so it does the magic* recursively: $ SOMEMAGIC output.txt cat output.txt asdf 1 word1 2 word2 4 word3 3 word4 2 word5 1 word6 2 word7 1 word8 1 word9 1 word10 1 $ *recursively count the words occurrence in the text files like: word1 2 can anyone point to a howto/link? [re: i just can't google for it :\] grep or find -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of unsubscribe. Trouble? Contact listmas...@lists.debian.org Archive: http://lists.debian.org/20101230134440.52dd8...@t61.debian-linux
Re: recursively count the words occurrence in the text files
On Thu, Dec 30, 2010 at 10:34 -0800, S Mathias wrote: I just can't google for it: I'm searching for a bash one liner (awk, perl, or anything) for this: there are text files, in several directories: […] How do you define a word? -- .''`. Wolodja Wentlandwolodja.wentl...@ed.ac.uk : :' : `. `'` 4096R/CAF14EFC `- 081C B7CD FF04 2BA9 94EA 36B2 8B7F 7D30 CAF1 4EFC signature.asc Description: Digital signature The University of Edinburgh is a charitable body, registered in Scotland, with registration number SC005336.
Re: recursively count the words occurrence in the text files
S Mathias smathias1...@yahoo.com wrote: *recursively count the words occurrence in the text files like: word1 2 can anyone point to a howto/link? [re: i just can't google for it :\] What are you actually trying to do? Neither your example nor your description makes any sense, here. Are you perhaps trying to count the number of words in a series of files? wc -l $( find * -type f ) Or something else? Chris -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of unsubscribe. Trouble? Contact listmas...@lists.debian.org Archive: http://lists.debian.org/806vu7xpr9@news.roaima.co.uk
Re: recursively count the words occurrence in the text files
On Thu, 30 Dec 2010 10:34:59 -0800 (PST),
S Mathias smathias1...@yahoo.com said:
S ...recursively count the words occurrence in the text files.
This assumes words consist of alphanumeric characters only. If that's
not the case, you'll need to change the tr/ line in the script.
me% head */*
== one/asf.txt ==
word1 word2 word3
== one/asfcxv.txt ==
word2 word4, word5
== one/dsgsdg.txt ==
word1. word2
== three/qwerbdsg.txt ==
word7, word8 word9 word10
== three/weberg.txt ==
word4 word3
== three/werdf.txt ==
asdf, word2
== two/ergd.txt ==
word6
== two/sdgsddsf.txt ==
word6, word3
me% cat words
#!/usr/bin/perl -w
use strict;
my %words;
while () {
tr/a-zA-Z0-9/ /cs;
foreach my $w (split) {
$words{$w}++;
}
}
foreach (sort keys %words) {
print $_ $words{$_}\n;
}
exit(0);
me% ./words */*
asdf 1
word1 2
word10 1
word2 4
word3 3
word4 2
word5 1
word6 2
word7 1
word8 1
word9 1
--
Karl Vogel I don't speak for the USAF or my company
Never take a beer to a job interview. --Martha Stewart's tips for Rednecks
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