Joining 3 lines at a time - Shell Programming and Scripting

Sabari Nath S · #1 (**permalink**) 12-09-2005

Hi,

I have a file which has the contents as below :

07:38:36 EST
date 20041117
07:39:06 EST
07:00:29 EDT
date 20050504
07:25:16 EDT
07:00:40 EDT
date 20050505
07:23:12 EDT

I need to delete the new line character from all lines except 3rd,6th,9th etc. lines so that i get the output as below :

07:38:36 EST date 20041117 07:39:06 EST
07:00:29 EDT date 20050504 07:25:16 EDT
07:00:40 EDT date 20050505 07:23:12 EDT

How can I achieve this ?? I tried some combinations of sed which I know but no success ...... for eg: sed 'N;s/\n/ /' can join pairs of lines ... but not 3 lines at a time .. I am searching for some kind of one liner like that .. not a script ..

Can anyone pls help ??

mona · #2 (**permalink**) 12-09-2005

I think the below awk command will help you

Code:

awk 'NR % 3 !=0 {printf $0;printf " "} NR % 3 ==0 {print " "}' join_three_lines.dat

vino · #3 (**permalink**) 12-09-2005

And a sed solution

Code:

[/tmp]$ echo "07:38:36 EST
date 20041117
07:39:06 EST
07:00:29 EDT
date 20050504
07:25:16 EDT
07:00:40 EDT
date 20050505
07:23:12 EDT
" | sed -e 'N;N;s/\n/ /g'
07:38:36 EST date 20041117 07:39:06 EST
07:00:29 EDT date 20050504 07:25:16 EDT
07:00:40 EDT date 20050505 07:23:12 EDT

Sabari Nath S · #4 (**permalink**) 12-09-2005

Hi,

This gives the result like this :

07:38:36 EST date 20041117
07:00:29 EDT date 20050504
07:00:40 EDT date 20050505

Not showing the last time ....

I got it done through another way by appending a | to lines other than 3,6,9 etc and then using sed like this ...

cat filename | awk '(NR%3!="0") {print $0,"|"} (NR%3=="0") {print $0}' t1 | sed -e :a -e '/|$/N; s/|\n//; ta'

Not sure of any other better way to do this ...

vino · #5 (**permalink**) 12-09-2005

That awk statment should be

Code:

 awk '($NR+1) % 3 !=0 {printf $0; printf " "} NR % 3 == 0 {print " "}'

Code:

[/tmp]$ echo "07:38:36 EST
date 20041117
07:39:06 EST
07:00:29 EDT
date 20050504
07:25:16 EDT
07:00:40 EDT
date 20050505
07:23:12 EDT
" | awk '($NR+1) % 3 !=0 {printf $0; printf " "} NR % 3 == 0 {print " "}'
07:38:36 EST date 20041117 07:39:06 EST  
07:00:29 EDT date 20050504 07:25:16 EDT  
07:00:40 EDT date 20050505 07:23:12 EDT

mona · #6 (**permalink**) 12-09-2005

Sorry. This will give you the expected result

awk '{printf $0;printf " "}NR % 3 ==0 {print " "}' join_three_lines.dat

Sabari Nath S · #7 (**permalink**) 12-09-2005

Great Vino .. that worked fine ... One more doubt .... can u tell me whats the meaning of N in this sed command and how does that work here ?

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