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Answered
Charlis
Asked: 2022-08-25T08:22:32+00:00

awk select rows that has column 1 partially matching column 2

I have an “example_file” like below,

a2022.1 a2022.1 80
a2022.1 a2022.2 90
a2022.1 a2023.1 80
a2022.2 a2022.2 90
a2022.2 a2023.1 40
a2022.3 a2022.1 50
b20225.1 a2022.1 80

and I would like to select lines which has the third column > 80 and also has the 1st column (before the dot) different from the 2nd column (before the dot).

So the desired result would be

a2022.1 a2023.1 80
b20225.1 a2022.1 80

So I already managed to exclude the lines with 3rd column <80 and with the same 1st and 2nd column by using the line below

awk '($3 > 80 && $1!=$2)' example_file

and I tried awk '($3 > 80 && $1!~$2)' example_file, but it does not exclude the line

a2022.1 a2022.2 90

I thought it could be easy, but I can’t figure it out. Could somebody help? Thanks!

awk
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1 Answer

  1. Best Answer

    You may use this awk:

    awk -F '[.[:blank:]]' '$NF >= 80 && $1 != $3' file

a2022.1 a2023.1 80
b20225.1 a2022.1 80

    Here -F '[.[:blank:]]' sets input field separator to a dot or whitespace character. This allows us to use parts of first column as $1 and $2 and second column as $3 and $4.

    Alternatively using gnu-awk:

    awk '$3 >= 80 && 
    gensub(/\..*/, "", "1", $1) != gensub(/\..*/, "", "1", $2)' file

a2022.1 a2023.1 80
b20225.1 a2022.1 80
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