Extract last n lines and prepend chars with sed one-liner

Question

Considering this file:

1 2 3 4 5

I can extract the last 3 lines with this sed one-liner:

$ /bin/sed input.txt -e ':a;$q;N;4,$D;ba'

But I would also like to add some characters (# ) in front of each line, so that the output will be like so:

# 3 # 4 # 5

How can I modify the one-liner to do that? I mean, without adding another sed invocation (with | /bin/sed -e 's/^/# /').

I would prefer to use sed only for the same reasons of the original question, i.e. I have a system with limited binaries available.

Why do you not want to use another sed invocation? I mean any issue with using sed two times? — Pandya
– Pandya, Commented Nov 11, 2021 at 18:41
Is tac available? If so, reverse the file, extract the first n lines and modify them, then reverse the result. — glenn jackman
– glenn jackman, Commented Nov 11, 2021 at 19:03

Quasímodo · Accepted Answer · 2021-11-11 18:31:27Z

4

Instead of using sed to print the last n lines use tail:

$ tail -3 input.txt | sed 's/^/# /' # 3 # 4 # 5

Also you don't need to cat the file, see Useless use of cat.

Using sed only:

sed ':a;${s/^/# /;s/\n/&# /g;q;};N;4,$D;ba' input.txt

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answered Nov 11, 2021 at 17:48

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