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Home/ Questions/Q 717977
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T05:28:26+00:00 2026-05-14T05:28:26+00:00

I have a data that looks like this: foo foo scaffold_7 1 4845 6422

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I have a data that looks like this:

foo foo      scaffold_7      1 4845 6422 4845
bar bar      scaffold_7      -1 14689 16310 16310

What I want to do is to process the above lines
where I just want to print column 1,2,3, 7 and one more column after 7th.
But with condition when printing column 7 onwards.

Below is my awk script:

awk '{ 
        if ($4=="+") { {end=$6-$5}{print $1 "\t" $2 "\t" $3 "\t" $4 "\t" $7 "\t" end+$7} } 
        else 
            {end=$6-$5}{print $1 "\t" $2 "\t" $3 "\t" $4 "\t" $7-end "\t" $7} 
    }'  

But why it doesn’t achieve the desired result like this?

foo foo    scaffold_7      1       4845    6422
bar bar   scaffold_7      -1      14689   16310

Note that the arithmetic (e.g. $7-end or end+$7) is a must. So we can’t just swap column
from input file. Furthermore this AWK will be inside a bash script.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T05:28:27+00:00Added an answer on May 14, 2026 at 5:28 am

    Try:

    awk 'BEGIN {OFS = "\t"} { 
        end = $6 - $5
        if ($4 >= 0) {
            print $1, $2, $3, $4, $7, end + $7
        } else {
            print $1, $2, $3, $4, $7 - end, $7
        }
    }'
    

    I’ve never used braces within braces and prefer the semi-colon separators for this purpose. I’m pretty certain the else clause is bound only to {end=$6-$5} in your example, not to both the braced entries. In that case, {print $1 "\t" $2 "\t" $3 "\t" $4 "\t" $7-end "\t" $7} will be executed for all lines.

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