删除交替行的最后3个字符

Question

I need to remove the last 3 characters in alternating lines. I have

 36960
32768
40800
16384
22656
4096

I need to have output like

 36960
32
40800
16
22656
4

Could someone please help?

Answer 1

An awk solution:

~$ cat file
 36960
32768
40800
16384
22656
4096
~$ awk '{if (NR%2){print $0}else{print substr($0, 0, length($0)-3)}}' file
 36960
32
40800
16
22656
4

Answer 2

Preceding a "sed" command with a "step" amount will do what you want. This looks like X~Y, which will match every Y'th line starting with line X.

sed '2~2 s/...$//g'

Answer 3

Divide the number by 1000 and store the answer into an integer.

As such:

int a = 32768/1000;

The value stored in a will be 32

For the Alternate lines:

Use a loop that increments by 2 (assuming you store those numbers in an array)

for(a=0;a<6;a=a+2)
{ 
    place code described above here; 
}

Answer 4

Here's an awk program that does it:

if (NR % 2 == 0) {
  print gensub(/(.*).../, "\\1", "g"); 
}
else
{
 print $0
}

If I run it like this I get the shown output:

:> awk '{ if (NR % 2 == 0) { print gensub(/(.*).../, "\\1", "g")} else { print $0 }}' << EOL
>  36960
> 32768
> 40800
> 16384
> 22656
> 4096
> EOL
 36960
32
40800
16
22656
4