CODEWITHSUNDEEP
bashawk
Andy K
Andy K

Reputation: 5044

bash - how to delete every nth line of a file

I have this 

line1
line2
line3
line4
line5
line6
line7
line8
line9
line10
line11
line12
line13
line14
line15

My aim is to have 

line1
line3
line4
line6
line7
line9
line10
line12
line13
line15

I tried the code below this but it is not doing what I want. i'm not sure what I'm doing wrong.

awk 'NR == 1 || NR % 3 != 0' file

Any clues on the why?

Upvotes: 3

Views: 4246

Answers (5)

Adam
Adam

Reputation: 18807

You can just use GNU sed:

 sed -i '2~3d' test.txt

You can also remove the -i in case you would not want to save the result.

Extract for sed man page:

first~step Match every step'th line starting with line first. For example, sed -n 1~2p will print all the odd-numbered lines in the input stream, and the address 2~5 will match every fifth line, starting with the second.

Upvotes: 8

Thor
Thor

Reputation: 47089

Here is one way of doing it with portable sed.

To delete every 3rd line:

sed 'n; n; d' infile

To delete every 3rd line, starting with line 2:

sed '1n; 2d; n; n; d' infile

Output:

line1
line3
line4
line6
line7
line9
line10
line12
line13
line15

Upvotes: 2

Ben Grimm
Ben Grimm

Reputation: 4371

You had the right idea using % 3 - just off by one since you want to remove starting at line 2 rather than 3. The test for NR == 1 is superfluous, falling within the modulus test, but it wasn't causing any trouble:

$ awk '(NR+1) % 3' file
line1
line3
line4
line6
line7
line9
line10
line12
line13
line15

Upvotes: 8

Wintermute
Wintermute

Reputation: 44023

To ignore the lines like in your example output, use

awk 'NR % 3 != 2' file

Upvotes: 3

anubhava
anubhava

Reputation: 784898

You can use:

awk '(NR+1)%3' file
line1
line3
line4
line6
line7
line9
line10
line12
line13
line15

This will skip every 3rd line starting from line #2. So line # 2, 5, 8, 11, 14 will be skipped.

Upvotes: 2

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