mjhjmtu

I am wondering whether there is a simple way to operate on certain lines with preassigned line numbers.

Let's say I want to output the 1st, 7th, 14th and 16th lines of a file, I can simply do

sed -n '1p;7p;14p;16p' input_file

but this gets more complicated when the operation is not just printing, and I don't want to write the same long command 4 times (and yes, I know I can construct this long sed command by substituting the same bash variable 4 times, but that's not ideal enough ...) , i.e.

sed -n '1long_command;7long_command;14long_command;16long_command' input_file

Is there a way to do the operation on these specific lines of my file? I am expecting something like,

sed -n '1,7,14,16p'

which certainly will not work in the current form.

Any help will be appreciated. "No, it is not possible." with explanations is also an answer that I will accept.

You can use branches:

sed '
 1b1
 7b1
 14b1
 16b1

 # for the rest to be left alone, branch off (or delete them with "d"):
 b

 :1
 long_command'

(note that you can also add some 20,25b1 line ranges, or /re/b1 to include lines that match the re).

Or you could use awk:

awk 'NR == 1 || NR == 7 || ... stuff'

Or using a hash:

awk -v l=1,7,14,16 '
 BEGINsplit(l, a, ","); for (i in a) lines[a[i]]

 NR in lines stuff'

(or BEGINlines[1]lines[7]lines[14]lines[16] if there aren't too many)

Simply invert your selection and delete it:

sed '2,6d;8,13d;15d;17,$d;long_command'

First variant:

You can use this trick:

sed -nf <(printf '%dpn' 1 7 14 16)

-f script-file - add the contents of script-file to the commands to be executed.

Testing

seq 1 20 | sed -nf <(printf '%dpn' 1 7 14 16)

Output

1
7
14
16

Second variant:

In the beginning, the auxiliary sed is used for filtering only needed lines from the file, then these lines piped to the main sed with the long command.

sed -n '1p; 7p; 14p; 16p' input.txt | sed 'long command'