[bash] transfer last line to first place

Alexos · 01-30-2012, 05:00 AM

I want to transfer last line in the file to the first place, without processing all file. Is it possible?
For example if I have

Code:

Line 1...
Line 2...
....
Line n-1...
Line n...

I want to receive

Code:

Line n...
Line 1...
Line 2...
....
Line n-1...

Or if it is impossible, how can I get

Code:

Line n...
Line 2...
...
Line n-1...

colucix · 01-30-2012, 05:45 AM

Maybe using something like

Code:

cat <(tail -n1 file) <(head -n-1 file)

this requires redirecting the output to a temporary file and rename it later. How many lines are in the file?

Alexos · 01-30-2012, 06:30 AM

Quote:

Originally Posted by colucix

How many lines are in the file?

unknown

Alexos · 01-30-2012, 07:10 AM

OK, solved)

Code:

size=`wc -l file.in | awk '{print $1}'`
let size--
cat <(tail -1 file.in) <(head -$size file.in) > file.out

colucix · 01-30-2012, 08:45 AM

Nice! Actually the size calculation is not necessary since option -n of the head command accepts negative numbers to print all but the last N lines of each file. Specifying

Code:

head -n-1

as in my example, will exclude the last line of the file independently from the total length. My concern about the number of lines was related only to performance.

Alexos · 01-31-2012, 04:46 AM

Great! thanks. but I found easier method again) without creating new files

Code:

total=`tail -n 1 "file"`
sed -i -e '$d' -e "1c\\$total\n" "file"

firstfire · 01-31-2012, 06:31 AM

Hi.

Here is even simpler solution:

Code:

$ cat infile2.txt 
Line 1...
Line 2...
....
Line n-1...
Line n...
$ echo -e '$m0\nw' | ed infile2.txt 
47
47
$ cat infile2.txt 
Line n...
Line 1...
Line 2...
....
Line n-1...