topic Re: help with script in Operating System - HP-UX

help with script

George_216 — Tue, 28 Jun 2005 02:25:07 GMT

Im trying to check a file with 1000s of lines all the columns and rows should be

0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00

Is there any way to run a script on this file and report any lines with any column not equal to "0.00000000E+00"

Any help would be appreciated. attached is a sample text file.

Re: help with script

Bharat Katkar — Tue, 28 Jun 2005 02:31:14 GMT

Hi,
See if this works for you:

# cat filename | xargs | grep -v "0.00000000E+00"

Regards,

Re: help with script

George_216 — Tue, 28 Jun 2005 02:50:27 GMT

im sorry, it didn,t help

Re: help with script

Sudeesh — Tue, 28 Jun 2005 02:53:38 GMT

grep -v "0.00000000E+00"
will work for you. You can write a script like this:

for files in /home/testdir/*
do
echo $files
grep -v "0.00000000E+00" $files
echo "****************************"
done

Sudeesh

Re: help with script

George_216 — Tue, 28 Jun 2005 02:56:25 GMT

your solutions will exclude all lines with 0.00000000E+00.

for example It will exclude this line as well:
0.00000000E+00 4444444444 0.00000000E+00

Re: help with script

Stuart Browne — Tue, 28 Jun 2005 08:44:43 GMT

If every line will have 5 entries on it (like in the example), then a slight expansion on the above suggestions:

grep -v "0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00 0.00000000E+00" file

As this will only exclude lines which are all 0'd.

Now, that being said, if there is a variable number of entries per line, then a quick awk:

awk '{ count = 1;while (count < NF) { if ($count != "0.00000000E+00") { print; next } ; count++ } }' < file

Anyway, just some ideas. ;)

Note: with the above 'awk', if you want to only print out the different actual fields (instead of the entire line), change the 'print; next' to 'printf( "%s\n", $count)' .