How to process every other line in bash

Question

I would like to print odd lines (1,3,5,7..) without any change, but even lines (2,4,6,8) process with pipeline beginning with grep. I would like to write everything to new file (odd lines without any change and new values for even lines).

I know how to print every other line in awk:

awk ' NR % 2 == 1 { print; } NR % 2 ==0 {print; }' file.fasta

However, for even lines, I dont want to use {print; } but I want to use my grep pipeline instead.

An advice will be appreciated. Thanks a lot.

Answer 1

If you're planning to do a simple grep, you can do away with the additional step and do the filtering within awk itself, e.g.:

awk 'NR % 2 {print} !(NR % 2) && /pattern/ {print}' file.fasta

However, if you intend to do a lot more then, as chepner already pointed out, you can indeed pipe from inside awk. For example:

awk 'NR % 2 {print} !(NR % 2) {print | "grep pattern | rev" }' file.fasta

That opens a pipe to the command "grep pattern | rev" (note the surrounding quotes) and redirects the print output to it. Do note that the output in this case may not be as you might expect; you will end up with all odd lines being output first followed by the output of the piped command (which consumes the even lines).

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(In response to your comments) to count the number of chars in each even line, try:

awk 'NR % 2 {print} !(NR % 2) {print length($0)}' file.fasta