CODEWITHSUNDEEP
unix
ramee
ramee

Reputation: 47

how to print the lines of a file that are multiples of 3 like 3rd, 6th, 9th line etc

I have a file and I want only lines that are multiples of 3. Is there any UNIX command to perform this task?

Upvotes: 2

Views: 2108

Answers (2)

schesis
schesis

Reputation: 59168

With GNU sed:

sed -n 0~3p filename

You can start at a different line by changing the number before the ~, so to start with the first line instead, it'd be:

sed -n 1~3p filename

Example:

$ cat filename 
The first line
The second line
The third line
The fourth line
The fifth line
The sixth line
The seventh line
$ sed -n 0~3p filename 
The third line
The sixth line
$ sed -n 1~3p filename 
The first line
The fourth line
The seventh line

Alternatively, with a non-GNU sed like BSD sed:

$ sed -n '3,${p;n;n;}' filename 
The third line
The sixth line

Upvotes: 5

fedorqui
fedorqui

Reputation: 290035

This makes it:

awk 'NR%3==0' file

NR stands for number of record, which in this case is number of line. So the condition is "(number of line / 3) having modulus 0" === "lines that are multiple of 3".

Test

$ cat file
hello1
hello2
hello3
hello4
hello5
hello6
hello7
hello8
hello9
hello10
$ awk 'NR%3==0' file
hello3
hello6
hello9

Upvotes: 7

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