awk remove line feed


 
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# 1  
Old 06-29-2011
awk remove line feed

Hi,

I've this file:

1,
2,
3,
4,
5,
6,

I need to remove the line feed LF every 3 row.

1,2,3,
4,5,6,


Thanks in advance,

Alfredo
# 2  
Old 06-29-2011
Seems like a simpler job for paste than awk:

Code:
$ paste -d"\0" - - - < file
1,2,3,
4,5,6,

An awk one anyway:
Code:
$ awk '{printf} !NR%3{print ""}' file
1,2,3,
4,5,6,

This User Gave Thanks to Scott For This Post:
# 3  
Old 06-29-2011
some possibilities:

Code:
#  nawk 'NR%3!=0{printf "%s", $0;next}1' infile
1,2,3,
4,5,6,

#  nawk 'ORS=(FNR%3)?FS:RS' infile
1, 2, 3,
4, 5, 6,

#  paste - - - < infile
1,      2,      3,
4,      5,      6,

#  paste -d" " - - - < infile
1, 2, 3,
4, 5, 6,

cat infile | xargs -n3
1, 2, 3,
4, 5, 6,

HTH

---------- Post updated at 12:18 PM ---------- Previous update was at 12:13 PM ----------

nice sneaky use of printf by scottn there that I've never really appreciated....learn something new every day :-)
# 4  
Old 06-29-2011
Quote:
Originally Posted by scottn
Code:
$ awk '{printf} !NR%3{print ""}' file
1,2,3,
4,5,6,

Using printf without any argument is dangerous.
If $0 contains format sequence like %s or %d, awk will failed.
An example:
Code:
$ cat in.txt
aaa
bbb
__ %s __
ddd
$ awk '{printf; print "."}' in.txt
aaa.
bbb.
awk: not enough arguments in printf(__ %s __) or sprintf(__ %s __)
 record number 3
$

Allways specify an argument to printf.
Replace :
Code:
printf ;

with
Code:
printf "";

Jean-Pierre.
# 5  
Old 06-29-2011
With sed..
Code:
sed 'N;N;s/\n//g' inputfile > outfile

# 6  
Old 06-29-2011
@aigles: Good point. I was just being lazy. It should, anyway be {printf "%s", $0}

The sed solution doesn't work so well if there is not an exact multiple of (in this case) three lines.
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