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Theres no reason your awk script awk '{print $0," "}' in > out should not work. I see that your lines are 80 char long. Is it possible there's line wrapping occurring in your terminal/editor?

Hi,
you can try as here:
$ cat -A inp.file
$
2012 0811 1223 15.2 L 38.393 46.806 9.0 Teh 78 0.5 6.5LTeh 1 $
GAP=74 ...

use tr and then try awk something like this

$ tr -s '\n' <file | awk '$0=$0" "'

With awk
awk '/^?\?\?/ {$0="#"} 1'

Hi Jotne,
I don't see the need for the backslashes? Why not just:
awk '$1=="???"{$0="#"}1' in-file > out-file
or
awk '/^???/{$0="#"}1' in-file > out-file

Hi saeed.soltani,
This can also...

Yes, obviously. I obviously didn't get enough sleep before sending that message.

For anyone who hasn't followed this discussion, the awk command:
/^???/{$0="#"}
matches any line with zero or...

@don: Since awk uses ERE when using matching, ? has a special meaning and needs to be escaped, except for the first ? since it follows a leading caret sign and is therefore not special..

$ sed "s/^???.*/#/" input
TAB 20120201 21:02:02
MRD 20121301 22:21:01
#
SAS 20123010 12:25:41
MMT 20120102 11:01:12
POP 20120306 02:01:01
#

Is a 3d ellipse a 2d ellipse on a plane in 3D? Standard rotation, translation and scaling would return you to a unit ellipse in the xy z=0 plane. And vice-versa, in reverse order.

awk '/^#/{print;c=0}c<2&&!/^#/{print;++c;}' filename

roughly:
awk '
# c print line counter
# d skip line flag
# e skipped line counter
{if (++c % 6 && !d) { # mod test for c line count (not 0) and skip line flag set to 0
e=d=0; print ...

Unless I'm missing something, the above scripts seem overly complex. Try:
awk 'NR % 8 == 1 && NR > 1 {
# Print an empty line separator between sets of 8 input lines.
printf("\n")...

try:
awk '{if (++c % 6 && !d) {e=d=0; print } else {if (++e % 4) {d=1; c=0} else {print "\n"$0; d=e=0}}}' input

awk 'BEGIN{L=1}{if(!(L>5&&L<9)) print; L++; if(L==9) {print " "; L=1;} } ' filename

awk '{if(NR>1){if(sqrt(($6-s)*($6-s)) > 1 && sqrt(($7-p)*($7-p)) > 1 && sqrt(($9-q)*($9-q)) > 5){print "###"}}}{
s=$6;p=$7;q=$9}1' file

You could use "cut" to extract a certain range of characters from each line:

cut -cnn-mm /path/to/infile

will extract columns nn to mm as string and display this at <stdout>. You can even...

Or, since you know the field widths and want to print the entire lines:
awk 'substr($0,119,4)=="ISC "' input >output

FIELDWIDTHS is a GNU extension.

What's your system?

awk 'FNR==NR{a[$1]=$0;next}{split(a[$1],M,FS);FV=((M[6]-$6) >=0) ? (M[6]-$6) : ($6-M[6]);
SV=((M[3]-$3) >=0) ? (M[3]-$3) : ($3-M[3]); TV=((M[4]-$4) >=0) ? (M[4]-$4) : ($4-M[4]);
if(FV < 5 && SV < 1...

Try: join -j1 file1 file2 | awk 'sqrt(($6-$11)^2)<5&&sqrt(($3-$8)^2)<1&&sqrt(($4-$9)^2)<1'Note that it will not print column 1 from file 2...

This will print 3rd record, not 2nd, you realize that, right?

You can make it simpler:
awk 'NR==2' RS=### infile

{if(NR==3) print}

agama said he found a # by itself on a line as the section separator. When I copied the sample input and fed it through od -cb, I found that the separator line contained the octal byte values 343,...

How abut this ?
#!/usr/bin/perl

$/="\n\n";

while (<DATA>) {
chomp;
if ( ((split))[11] > 3 ) {
print ;
}
}



__DATA__
2011 0701 2015 21.2 L 37.692 46.202 18.0 Teh 4 0.3 2.1 LTeh 1...

I think this does what you might be looking for:


awk '
$1 == "#" { if( snarf ) print " "; snarf = 0; next; } # turn off section capture, write a trailing blank line
snarf ||...