Question

I have this string in my data file (dat.txt) =>

00000
10101

I want to replace the 1st row, second and fourth column, I tried this:

$>sed -i "/^1/s/0/1/2" dat.txt
$>sed -i "/^1/s/0/1/4" dat.txt

But when I reload the dat.txt, it does not change anything. Is my command wrong?

Answer 1

For a start, unless that first line is your zeroth row, you’ll be changing the second (the one that begins with 1).

And that 2 and 4 don’t change the second/fourth column, they change the second/fourth occurrence of the search pattern.

See the following transcript:

$ cat dat.txt
00000
10101

$ sed -i "/^1/s/0/1/2" dat.txt ; cat dat.txt
00000
10111

$ sed -i "/^1/s/0/1/4" dat.txt ; cat dat.txt
00000
10111

You can see that the first sed changed the second occurrence of 0 to 1, and that was in the fourth column. The second sed changed nothing since you asked it to do it to the fourth occurrence and there aren’t four 0 characters there.

If you want to change specific columns, you can use capturing patterns, as in the following transcript:

$ cat dat.txt
00000
10101

$ sed -i "/^1/s/^\(.\)0/\11/" dat.txt ; cat dat.txt
00000
11101

$ sed -i "/^1/s/^\(...\)0/\11/" dat.txt ; cat dat.txt
00000
11111

The parentheses capture the pattern so that \1 in the replacement text can be used to refer to that captured text. Because you’re talking about a small number of characters before the one you want to change, you can use ... as a pattern. If you wanted to change (for example) the 85^th column, you’d probably be better off with something like:

sed -i "/^1/s/^\(.\{84\}\)0/\11/" dat.txt ; cat dat.txt