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Full Discussion: join command
Top Forums Shell Programming and Scripting join command Post 302368545 by Dedalus on Thursday 5th of November 2009 04:03:14 AM
Old 11-05-2009
join command

Hi

my problem is:
I've 2 files like:

file1:
Quote:
1 3.791406e-03 65388.3
6 8.568031e-01 1.47768e+07
17 1.323488e-01 2.28255e+06
47 5.928275e-03 102242
50 1.037607e-03 17895
89 9.085876e-05 1566.99
file2
Quote:
1 3.839507e-03 66217.9
6 8.563433e-01 1.47689e+07
10 0.000000e+00 0
17 0.000000e+00 0
..so many zeros
47 5.863369e-03 101122
50 1.020259e-03 17595.9
..so many zeros
89 0.0000000e+00 0
..so many zeros
I'd like to get file_out:
Quote:
 
1 3.791406e-03 65388.3 3.839507e-03 66217.9
6 8.568031e-01 1.47768e+07 8.563433e-01 1.47689e+07
17 1.323488e-01 2.28255e+06 0.000000e+00 0
47 5.928275e-03 102242 5.863369e-03 101122
50 1.037607e-03 17895 1.020259e-03 17595.9
89 9.085876e-05 1566.99 0.0000000e+00 0
I mean if the field is different from zero in the first file i want take that field also in the second(no matter if in the second is zero).

I tried with the join command but it doesn't work always and i don't understand why.
Can anyone help please?

thanks in advance

D
 

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SHELL-QUOTE(1p) 					User Contributed Perl Documentation					   SHELL-QUOTE(1p)

NAME
shell-quote - quote arguments for safe use, unmodified in a shell command SYNOPSIS
shell-quote [switch]... arg... DESCRIPTION
shell-quote lets you pass arbitrary strings through the shell so that they won't be changed by the shell. This lets you process commands or files with embedded white space or shell globbing characters safely. Here are a few examples. EXAMPLES
ssh preserving args When running a remote command with ssh, ssh doesn't preserve the separate arguments it receives. It just joins them with spaces and passes them to "$SHELL -c". This doesn't work as intended: ssh host touch 'hi there' # fails It creates 2 files, hi and there. Instead, do this: cmd=`shell-quote touch 'hi there'` ssh host "$cmd" This gives you just 1 file, hi there. process find output It's not ordinarily possible to process an arbitrary list of files output by find with a shell script. Anything you put in $IFS to split up the output could legitimately be in a file's name. Here's how you can do it using shell-quote: eval set -- `find -type f -print0 | xargs -0 shell-quote --` debug shell scripts shell-quote is better than echo for debugging shell scripts. debug() { [ -z "$debug" ] || shell-quote "debug:" "$@" } With echo you can't tell the difference between "debug 'foo bar'" and "debug foo bar", but with shell-quote you can. save a command for later shell-quote can be used to build up a shell command to run later. Say you want the user to be able to give you switches for a command you're going to run. If you don't want the switches to be re-evaluated by the shell (which is usually a good idea, else there are things the user can't pass through), you can do something like this: user_switches= while [ $# != 0 ] do case x$1 in x--pass-through) [ $# -gt 1 ] || die "need an argument for $1" user_switches="$user_switches "`shell-quote -- "$2"` shift;; # process other switches esac shift done # later eval "shell-quote some-command $user_switches my args" OPTIONS
--debug Turn debugging on. --help Show the usage message and die. --version Show the version number and exit. AVAILABILITY
The code is licensed under the GNU GPL. Check http://www.argon.org/~roderick/ or CPAN for updated versions. AUTHOR
Roderick Schertler <roderick@argon.org> perl v5.8.4 2005-05-03 SHELL-QUOTE(1p)
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