Performing regex capture and then substitute using SED/PERL

I have a data that looks like this (let's call this file submit.txt):

dir1/pmid_5409464.txt
dir1/pmid_5788247.txt
dir1/pmid_4971884.txt

What I want to do is to perform an inline file regex change so that it results in the following

perl mycode.pl /home/neversaint/dir1/pmid_5409464.txt > /home/neversaint/dir1/pmid_5409464.output
perl mycode.pl/home/neversaint/dir1/pmid_5788247.txt > /home/neversaint/dir1/pmid_5788247.output
perl mycode.pl /home/neversaint/dir1/pmid_4971884.txt > /home/neversaint/dir1/pmid_4971884.output

Is there a SED/Perl one liner to do that?

My difficulty is in capturing the input file name and then create the output file (.output) - for each line - based on that. I'm stuck with this:

sed 's/^/perl mycode.pl \/home\/neversaint\/dir1\//g' submit.txt |
sed 's/$/ >/'

regex perl sed

Edit Report

1 Answer

You can use escaped parenthesis to capture groups, and access the groups with \1, \2, etc.

sed 's/^\(.*\).txt$/perl mycode.pl \/home\/neversaint\/\1\.txt > \/home\/neversaint\/\1.output/' submit.sh

output:

perl mycode.pl /home/neversaint/dir1/pmid_5409464.txt > /home/neversaint/dir1/pmid_5409464.output
perl mycode.pl /home/neversaint/dir1/pmid_5788247.txt > /home/neversaint/dir1/pmid_5788247.output
perl mycode.pl /home/neversaint/dir1/pmid_4971884.txt > /home/neversaint/dir1/pmid_4971884.output

edit: it doesn't look like sed has a built-in in place file editing (GNU sed has the -i option). It still possible to do but this solution just prints to standard out. You could also use a Perl one liner as shown here: sed edit file in place

answered 2013-07-31T02:39:06.873

Performing regex capture and then substitute using SED/PERL

1 Answer

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