如何使用SED或AWK删除除第一行以外的奇数行

Question

I have the following file

# header1 header2
zzzz yyyy
1
kkkkk wwww
2

What I want to do is to remove odd lines except the header yielding:

# header1 header2
zzzz yyyy
kkkkk wwww

I tried this but it removes the header too

awk 'NR%2==0'

What's the right way to do it?

Answer 1

awk 'NR==1 || NR%2==0'

If the record number is 1 or is even, print it.

awk 'NR % 2 == 0 || NR == 1'

Reversing the comparisons might be marginally faster. The difference probably isn't measurable. (And the choice of spacing is essentially immaterial too.)

Answer 2

Works on GNU sed

sed '3~2d' ip.txt 

This deletes line numbers starting from 3rd line and then +2,+4,+6, etc

Example:

$ seq 10 | sed '3~2d'
1
2
4
6
8
10

Answer 3

You just need

awk 'NR==1 || NR%2==0' file

• This keeps the header part of the file intact and applies the rule NR%2==0 , which is true only for even lines(starting from the header) in which case it is printed.

Another variant of the same above answer

awk 'NR==1 || !(NR%2)' file

• For even lines (NR%2) becomes 0 and negation of that becomes a true condition to print the line

Answer 4

sed '1!{N;P;d}'

1! On lines other than the first (the default behavior echoes the first line)
N append the next line to the current line
P print only the first of the two
d delete them both.

Answer 5

This might work for you (GNU sed):

sed '1b;n;d' file

But:

sed '3~2d' file

Is far neater.