extract multiple items on single line using grep/sed/perl

Question

I have a massive text file a bit like this:

=?accession=P12345;=?position=999;
=?accession=Q19283;=?position=777;
=?accession=A918282;=?position=888;

and I would like to extract terms between accession= and ;, and then also between position= and ;

so that I get:

P12345 999
Q19283 777
A918282 888

The strings I need to grep between do get more complicated, so I imagine a hardcoded solution.

I know I can take the "grep between two strings" approach:

grep -Po 'accession= \K.*(?= ;)'

but I don't know how to get subsequent extractions from the same line of the input to also appear on the same line as the output.

I really don't mind how this is done as long as I can call it from a linux command line.

Thanks.

Answer 1

1. You can update your grep expression like this.

   grep -oP "(accession=\K\w+)|(position=\K\d+)" file
   

   Output:

   P12345
   999
   Q19283
   777
   A918282
   888
   

   To format it the way you want, use paste :

   grep -oP "(accession=\K\w+)|(position=\K\d+)" file | paste -d ' ' - -
   

   Output:

   P12345 999
   Q19283 777
   A918282 888
   

2. Another simple awk solution :

   awk -F"=|;" '{print $3, $6}' file
   

   Output:

   P12345 999
   Q19283 777
   A918282 888
   

Answer 2

This awk should work:

awk -F ';' '{gsub(/=[^=]*=/, ""); $1=$1} 1' file

P12345 999
Q19283 777
A918282 888

Answer 3

sed -r 's/.*accession=([^;]*);.*position=([^;]*).*/\1 \2/' textfile

Answer 4

This perl one-liner

perl -wnE'say join " ", /(?:accession|position)=([^;]+)/g' input.txt

prints the desired output.