Is there a faster way to remove a line (given a line number) from a file?

What you could do to avoid writing a copy of the file is to write the file over itself like:

{
  sed "$l1,$l2 d" < file
  perl -le 'truncate STDOUT, tell STDOUT'
} 1<> file

Dangerous as you've no backup copy there.

Or avoiding sed, stealing part of manatwork's idea:

{
  head -n "$(($l1 - 1))"
  head -n "$(($l2 - $l1 + 1))" > /dev/null
  cat
  perl -le 'truncate STDOUT, tell STDOUT'
} < file 1<> file

That could still be improved because you're overwriting the first l1 - 1 lines over themselves while you don't need to, but avoiding it would mean a bit more involved programming, and for instance do everything in perl which may end up less efficient:

perl -ne 'BEGIN{($l1,$l2) = ($ENV{"l1"}, $ENV{"l2"})}
    if ($. == $l1) {$s = tell(STDIN) - length; next}
    if ($. == $l2) {seek STDOUT, $s, 0; $/ = \32768; next}
    if ($. > $l2) {print}
    END {truncate STDOUT, tell STDOUT}' < file 1<> file

Some timings for removing lines 1000000 to 1000050 from the output of seq 1e7:

• sed -i "$l1,$l2 d" file: 16.2s
• 1st solution: 1.25s
• 2nd solution: 0.057s
• 3rd solution: 0.48s

They all work on the same principle: we open two file descriptors to the file, one in read-only mode (0) using < file short for 0< file and one in read-write mode (1) using 1<> file (<> file would be 0<> file). Those file descriptors point to two open file descriptions that will have each a current cursor position within the file associated with them.

In the second solution for instance, the first head -n "$(($l1 - 1))" will read $l1 - 1 lines worth of data from fd 0 and write that data to fd 1. So at the end of that command, the cursor on both open file descriptions associated with fds 0 and 1 will be at the start of the $l1th line.

Then, in head -n "$(($l2 - $l1 + 1))" > /dev/null, head will read $l2 - $l1 + 1 lines from the same open file description through its fd 0 which is still associated to it, so the cursor on fd 0 will move to the beginning of the line after the $l2 one.

But its fd 1 has been redirected to /dev/null, so upon writing to fd 1, it will not move the cursor in the open file description pointed to by {...}'s fd 1.

So, upon starting cat, the cursor on the open file description pointed to by fd 0 will be at the start of the next line after $l2, while the cursor on fd 1 will still be at the beginning of the $l1th line. Or said otherwise, that second head will have skipped those lines to remove on input but not on output. Now cat will overwrite the $l1th line with the next line after $l2 and so on.

cat will return when it reaches the end of file on fd 0. But fd 1 will point to somewhere in the file that has not been overwritten yet. That part has to go away, it corresponds to the space occupied by the deleted lines now shifted to the end of the file. What we need is to truncate the file at the exact location where that fd 1 points to now.

That's done with the ftruncate system call. Unfortunately, there's no standard Unix utility to do that, so we resort on perl. tell STDOUT gives us the current cursor position associated with fd 1. And we truncate the file at that offset using perl's interface to the ftruncate system call: truncate.

In the third solution, we replace the writing to fd 1 of the first head command with one lseek system call.

Using sed is a good approach: It is clear, it streams the file (no problem with long files), and can easily be generalized to do more. But if you want a simple way to edit the file in-place, the easiest thing is to use ed or ex:

(echo 10,31d; echo wq) | ed input.txt

A better approach, guaranteed to work with files of unlimited size (and for lines as long as your RAM allows) is the following perl one-liner which edits the file in place:

perl -n -i -e 'print if $. < 10 || $. > 31' input.txt

Explanation:

-n: Apply the script to each line. Produce no other output.
-i: Edit the file in-place (use -i.bck to make a backup).
-e ...: Print each line, except lines 10 to 31.