Grep column value in same line and print line

December 21, 2023

I have this 5-columns file :

m64071_220512_054244/46858502/ccs TCTACACGACGCTCTTCCGATCTTATTGGGCACGGTGTCGCCATCTGATCGTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTCGAGGTTTGCAGCTATTTTATTTACAAGTATACATTTAACACAATGAAATAAACACTGATATACTGAAGCCTAGTTAATAGTAGTGTAACAATATGCATCATTTTGATGATTACATTATTTTAAACAACAAACTACACTGAAAAATTAATGCCGATAAAATTCTTGGTCATAATATTAAGAAATACAATATATAAATTGAAAATATGATTGCTTAAAATTTGAAAATGGAAGTGAACTCATTTGGACAGACTCAGAGTTAACATAATCTGAAGGGAGGGGAGCTCTGACCCAAATGATATCTTTCAGGTTAACAGAAGAAAAAAGAAGCATAGTTTATCTTCAAGGAGAACGGGCAGTTTGCTTCTTCAGGTA fwd pet047-9952 TATTGGGCACGGTGTC
m64071_220512_054244/52233509/ccs AGCTTTTTTGGAATCTTCTGCTAAAGAAAATCAGACTGCTGTGGATGTTTTTCGAAGGATAATTTTGGAGGCAGAAAAAATGGACGGGGCAGCTTCACAAGGCAAGTCTTCATGCTCGGTGATGTGATTCTGCTGCAAAGCCTGAGGACACTGGGAATATATTCTACCTGAAGAAGCAAACTGCCCGTTCTCCTTGAAGATAAACTATGCTTCTTTTTTCTTCTGTTAACCTGAAAGATATCATTTGGGTCAGAGCTCCCCTCCCTTCAGATTATGTTAACTCTGAGTCTGTCCAAATGAGTTCACTTCCATTTTCAAATTTTAAGCAATCATATTTTCAATTTATATATTGTATTTCTTAATATTATGACCAAGAATTTTATCGGCATTAATTTTTCAGTGTAGTTTGTTGTTTAAAATAATGTAATCATCAAAATGATGCATATTGTTACACTACTATTAACTAGGCTTCAGTATATCAGTGTTTATTTCATTGTGTTAAATGTATACTTGTAAATAAAATAGCTGCAAACCTCGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACGATCAGATGGCGACACCGTGCCCAATAAGATCGGAAGAGCGTCGTGTAGA rev pet047-9952 GACACCGTGCCCAATA
m64071_220512_054244/91226755/ccs TCTACACGACGCTCTTCCGATCTTATTGGGCACGGTGTCGCCATCTGATCGTTTTTTTTTTTTTTTTTTTTTTTTTTTTTCGAGGTTTGCAGCTATTTTATTTACAAGTATACATTTAACACAATGAAATAAACACTGATATACTGAAGCCTAGTTAATAGTAGTGTAACAATATGCATCATTTTGATGATTACATTATTTTAAACAACAAACTACACTGAAAAATTAATGCCGATAAAATTCTTGGTCATAATATTAAGAAATACAATATATAAATTGAAAATATGATTGCTTAAAATTTGAAAATGGAAGTGAACTCATTTGGACAGACTCAGAGTTAACATAATCTGAAGGGAGGGGAGCTCTGACCCAAATGATATCTTTCAGGTTAACAGAAGAAAAAAGAAGCATAGTTTATCTTCAAGGAGAACGGGCAGTTTGCTTCTTCAGGTAGAATATATTCCCAGTGTCCTCAGGCTTTGCAGCAGAATCACATCACCGAGCATGAAGACTTGCCTTGTGAAGCTGCCCCGTCCATTTTTTCTGCCTCCAA fwd pet047-9952 TATTGGGCACGGTGTC

For each line, I need to grep the last column value $5 in the second field $2.
Then, I need to print the same line with an additional $6 column with the grep result with a condition : if ($3 == rev) , the $6 is grep result + 12 characters after grep or if ($3 == fwd) the grep result + 12 characters before grep.

awk '$2~/$5/ {match($0, /$5/); if ($4=="rev") print substr($0, RSTART +12, RLENGTH + 12); else print substr($0, RSTART + 0, RLENGTH + 12) ;}' file

$5 values are necessary 16 characters and the pattern I look for is always 12 characters. Then, my $6 output is 28 characters.

Expected output :

m64071_220512_054244/46858502/ccs TCTACACGACGCTCTTCCGATCTTATTGGGCACGGTGTCGCCATCTGATCGTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTCGAGGTTTGCAGCTATTTTATTTACAAGTATACATTTAACACAATGAAATAAACACTGATATACTGAAGCCTAGTTAATAGTAGTGTAACAATATGCATCATTTTGATGATTACATTATTTTAAACAACAAACTACACTGAAAAATTAATGCCGATAAAATTCTTGGTCATAATATTAAGAAATACAATATATAAATTGAAAATATGATTGCTTAAAATTTGAAAATGGAAGTGAACTCATTTGGACAGACTCAGAGTTAACATAATCTGAAGGGAGGGGAGCTCTGACCCAAATGATATCTTTCAGGTTAACAGAAGAAAAAAGAAGCATAGTTTATCTTCAAGGAGAACGGGCAGTTTGCTTCTTCAGGTA fwd pet047-9952 TATTGGGCACGGTGTC TATTGGGCACGGTGTCGCCATCTGATCG
m64071_220512_054244/52233509/ccs AGCTTTTTTGGAATCTTCTGCTAAAGAAAATCAGACTGCTGTGGATGTTTTTCGAAGGATAATTTTGGAGGCAGAAAAAATGGACGGGGCAGCTTCACAAGGCAAGTCTTCATGCTCGGTGATGTGATTCTGCTGCAAAGCCTGAGGACACTGGGAATATATTCTACCTGAAGAAGCAAACTGCCCGTTCTCCTTGAAGATAAACTATGCTTCTTTTTTCTTCTGTTAACCTGAAAGATATCATTTGGGTCAGAGCTCCCCTCCCTTCAGATTATGTTAACTCTGAGTCTGTCCAAATGAGTTCACTTCCATTTTCAAATTTTAAGCAATCATATTTTCAATTTATATATTGTATTTCTTAATATTATGACCAAGAATTTTATCGGCATTAATTTTTCAGTGTAGTTTGTTGTTTAAAATAATGTAATCATCAAAATGATGCATATTGTTACACTACTATTAACTAGGCTTCAGTATATCAGTGTTTATTTCATTGTGTTAAATGTATACTTGTAAATAAAATAGCTGCAAACCTCGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACGATCAGATGGCGACACCGTGCCCAATAAGATCGGAAGAGCGTCGTGTAGA rev pet047-9952 GACACCGTGCCCAATA CGATCAGATGGCGACACCGTGCCCAATA
m64071_220512_054244/91226755/ccs TCTACACGACGCTCTTCCGATCTTATTGGGCACGGTGTCGCCATCTGATCGTTTTTTTTTTTTTTTTTTTTTTTTTTTTTCGAGGTTTGCAGCTATTTTATTTACAAGTATACATTTAACACAATGAAATAAACACTGATATACTGAAGCCTAGTTAATAGTAGTGTAACAATATGCATCATTTTGATGATTACATTATTTTAAACAACAAACTACACTGAAAAATTAATGCCGATAAAATTCTTGGTCATAATATTAAGAAATACAATATATAAATTGAAAATATGATTGCTTAAAATTTGAAAATGGAAGTGAACTCATTTGGACAGACTCAGAGTTAACATAATCTGAAGGGAGGGGAGCTCTGACCCAAATGATATCTTTCAGGTTAACAGAAGAAAAAAGAAGCATAGTTTATCTTCAAGGAGAACGGGCAGTTTGCTTCTTCAGGTAGAATATATTCCCAGTGTCCTCAGGCTTTGCAGCAGAATCACATCACCGAGCATGAAGACTTGCCTTGTGAAGCTGCCCCGTCCATTTTTTCTGCCTCCAA fwd pet047-9952 TATTGGGCACGGTGTC TATTGGGCACGGTGTCGCCATCTGATCG

But I do not get what I want.

>Solution :

You can use the below script to achieve your result:

awk '{
    idx = index($2, $5);
    if (idx != 0) {
        if ($3 == "rev") {
            substr_28 = substr($2, idx - 12, 28);
        } else {
            substr_28 = substr($2, idx, 28);
        }
        print $0, substr_28;
    }
}' your_file_containing_inp

Basically, the script:

first finds the index of the $5 value in the second field. If the value is found (idx != 0).

a. it checks the value of the third field.
b. If it’s "rev", it extracts the 28-character substring starting 12 characters before the index.
c. else, it extracts the substring starting at the index.

CODE DEMO