gothamcity has asked for the wisdom of the Perl Monks concerning the following question:

Hi, I have a text file that has several columns of data that are tab separated. I'm trying to use command-line perl to read the file, make a substitution, and write the modified version. However, when I do so, the output file is blank.

Here is my command: 
perl -i.bak -F\t -lane "$F[4] =~ s/.*y.*/yes/" results.txt

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Here's a segment of my data (results.txt): 
name1 name2 NA NA ye[s]* no OK
name1 name2 NA NA <ls>yes no OK
name1 name2 NA NA <ls>yes<ln> no OK

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Basically, I'm trying to clean up the fifth column so that it is only 'yes' without extra information.

Any help is greatly appreciated.

Replies are listed 'Best First'.
Re: In-place edit on tabular data not working
by Corion (Patriarch) on Oct 18, 2010 at 18:05 UTC

    See perlrun on -n. If you use the -n switch, you need to do the printing yourself. Did you maybe want the -p switch?

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      Thanks, I got it to print out using -p, but the substitution is not being printed in the output (parsed) file.

      I know that the regex is working because when i run the following it prints the substituted words. 
      perl -F\t -lape "$F[4] =~ s/.*y.*/yes/; print $F[4], \"\n\";" results.
+txt

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      Not sure why the substituted words aren't being written into the file. Any thoughts? Thanks!
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Re: In-place edit on tabular data not working
by toolic (Bishop) on Oct 18, 2010 at 18:19 UTC
    All you did was change one element of the @F array. You did not change $_. You need to reconstruct $_, and as Corion said, use -p instead of -n: 
    perl -i.bak -lape '$F[4] =~ s/.*y.*/yes/; $_ = join "\t", @F' results.
+txt

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    -p says to use print, which prints $_ by default.

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      Thank you so much! Reconstructing the $_ for worked. I see the mistake I was making: the modified $F4 wasn't making it back to $_ for output.
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