CODEWITHSUNDEEP
unixawksedline
DrewC
DrewC

Reputation: 9

Sed/Awk deleting selective lines

This may be a simple question, but I am struggling to find a solution, could anyone provide a simple UNIX sed or awk command to delete every 3rd and 4th line of a file starting from say the 5th line.

Upvotes: 0

Views: 2016

Answers (4)

glenn jackman
glenn jackman

Reputation: 247200

an awk solution, where you pass your numbers as parameters:

awk -v start=5 -v d1=3 -v d2=4 '
    {n = NR - start}
    n > 0 && (n % d1 == 0 || n % d2 == 0) {next} 
    1
'

This skips lines 8, 11, 14, ... and 9, 13, 17, ...

Upvotes: 2

paxdiablo
paxdiablo

Reputation: 882596

Assuming I've understood the requirements, you can use awk as follows:

pax$ echo '1
2
3
4
5
6
7
8
9
10
11
12
13
14
15' | awk '{if (((NR - 5) % 4 < 2) || (NR < 5)) {print}}'
1
2
3
4
5
6
9
10
13
14

As you can see, this uses NR (the line number) to decide whether or not to print the line.

Starting from line 5, it prints two then skips two, ad infinitum. Below 5, it prints everything.

If I've misunderstood exactly which lines you want skipped, the theory is still sound, you just have to adjust the if statement.

Upvotes: 2

btilly
btilly

Reputation: 46507

How about Perl?

perl -pi -e 'print if $. < 5 or 1 == $. % 4 or 2 = $. % 4' your_file

Upvotes: 0

Hasturkun
Hasturkun

Reputation: 36422

If you're using GNU sed, you can use the FIRST~STEP address format, eg.

sed -e '6~3 d' -e '8~4 d' file

deletes the 6th line, then every 3rd, and deletes the 8th line, then every 4th.

Upvotes: 2

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