Move specific line from a position to another

Using ed (the line-editor that sed and grep are derived from):

printf '%s\n' '11m0' 'w bonjour2' 'q' | ed -s bonjour

This applies the editing command 11m0 to the file which moves line 11 to before the first line. It then writes the resulting document to the file bonjour2 and quits.

Alternatively:

printf '%s\n' '11m0' ',p' 'Q' | ed -s bonjour >bonjour2

... which instead of writing to a specific file using ed commands just prints the whole file to standard output. The result is then redirected to a new filename. The ,p command (short for 1,$p) will output the whole document to standard output and Q will force quit (even though the document has changed).

To make the change in the document in-place (i.e. change the original file), write the result back to the file itself:

printf '%s\n' '11m0' 'wq' | ed -s bonjour

Example run of one of the above variations:

$ printf '%s\n' '11m0' ',p' 'Q' | ed -s bonjour >bonjour2
$ cat bonjour2
French: Bonjour
English: Hello
Turkish: Marhaba
Italian: Ciao
German: Hallo
Spanish: Hola
Latin: Salve
Greek: chai-ray
Welsh: Helo
Finnish: Hei
Breton: Demat

To always move the last line to the top, use $m0 in place of 11m0. To always move the line that starts with the string French:, use /^French:/m0.

If I understand correctly, you just want to move the 11th line to the first line.

So if your original looked like

Original line 1
Original line 2
...
Original line 18
Original line 19
Original line 20

Then the result would be

Original line 11
Original line 1
Original line 2
...
Original line 9
Original line 10
Original line 12
Original line 13
...
Original line 19
Original line 20

This can be achieved with two passes through the file, but the first pass can be short-cutted so it's not too horrendous

( sed -n '11 {p;q}' file ; sed 11d file ) > newfile

Basically the first sed command only prints line 11 and then quits. The second sed command prints every line except line 11.

The result is that, effectively, line 11 is moved to the front of the new file.

After posting my original answer below, I realized now that having an array was overkill and all you really need is:

awk '
NR < 11 { buf = buf $0 ORS; next }
{
    print
    printf "%s", buf
    buf = ""
}
' file > new_file

e.g.:

$ seq 15 | awk '
NR < 11 { buf = buf $0 ORS; next }
{
    print
    printf "%s", buf
    buf = ""
}
'
11
1
2
3
4
5
6
7
8
9
10
12
13
14
15

Original answer using an array instead of a scalar variable for the buffer:

awk '
NR < 11 { buf[++bufSz] = $0; next }
{
    print
    for (i=1; i<=bufSz; i++) {
        print buf[i]
    }
    bufSz = 0
}
' file > new_file

e.g.:

$ seq 15 | awk '
NR < 11 { buf[++bufSz] = $0; next }
{
    print
    for (i=1; i<=bufSz; i++) {
        print buf[i]
    }
    bufSz = 0
}
'
11
1
2
3
4
5
6
7
8
9
10
12
13
14
15

With POSIX sed:

$ sed -e '1,11!b' -e '11!H;1h;11!d;G' file

Using perl:

$ perl -lne '
   push(@A,$_),next if $. < 11;
   print for $_, splice @A;
' file

Using bash:

$ (head -n 11 - | tee log | tail -n 1 -; head -n 10 log; cat -;) < file

A simpler sed than the ones posted (POSIX, no labels):

seq 20 | sed '1,10{H;1h;d;};11G'

That's: on lines 1 to 10 fill the hold space with those lines (don't print). On line 11, get the hold space back, and let sed do the default from then on: just print each line. Note that after line 11 the script is a no-op, just leave sed do the default, maybe faster.

In awk:

awk 'NR <11{ hold=hold RS $0 }
     NR==11{ print $0, hold  }
     NR >11
    ' file

As one liner:

$ seq 15 | awk 'NR<11{hold=hold RS $0}NR==11{print $0,hold}NR>11'
11 
1
2
3
4
5
6
7
8
9
10
12
13
14
15