First of all, you need to know why the extra filename is added at the bottom. The Perl script splits your input line due to the "-a" option, and the tokens are filled in a system-defined array called "@F". Therefore, $F[3] is the 4th element of the array (it starts with index 0). And $F[4] is the 5th element.

Now, my hunch is that you have an extra blank line at the end of your file "3.dat". You can check this by using "cat -n 3.dat". In the output below, I see line number 7 but no data in there which means it is a blank line.

Code:

$ 
$ 
$ cat -n 3.dat
     1    # Rg          Mass            Density       Rcm-x            Rcm-y           Area             Rsph/sqrt(2)
     2    # ==          ====            =======       =====            =====           ====             ============
     3    0.086692       2850.681081    345.545735    607.726550       255.130096      138.0069921579   1.35210017578
     4    0.834407      22025.882038     40.423520    986.900503       548.502799      863.9181962071   5.10301645927
     5    0.731624       6960.747103    645.505466    457.952424       691.618854      820.1901098519   5.88622435063
     6    0.511055      17352.058392    426.119742    800.869647       671.749259      752.8314025622   3.07408757877
     7    
$ 
$

The Perl one-liner splits this blank line faithfully and assigns null strings to $F[3] and $F[4]. And the print command prints nulls for $F[3] and $F[4] and the filename for $ARGV. See below -

Code:

$ 
$ perl -lane '$.>2 && print "$F[3] $F[4] $ARGV"; close ARGV  if eof' 3.dat
607.726550 255.130096 3.dat
986.900503 548.502799 3.dat
457.952424 691.618854 3.dat
800.869647 671.749259 3.dat
  3.dat
$

I've added one more condition before printing - check if the line is not a blank line (or a line that contains nothing but whitespaces). The updated Perl one-liner is as follows -

Code:

$ 
$ 
$ perl -lane 'print "$F[3] $F[4] $ARGV" if ($.>2 and !/^\s*$/); close  ARGV if eof' 3.dat
607.726550 255.130096 3.dat
986.900503 548.502799 3.dat
457.952424 691.618854 3.dat
800.869647 671.749259 3.dat
$ 
$

In my earlier post, I had specified the regular expression "file?.dat" for this. That was because the test files (for your problem) in my system were named "file1.dat", "file2.dat" and "file3.dat".

If you can come up with a regular expression for all your "dat" files that your shell understands, then you can put that instead.

Note that your shell probably won't expand "0.dat .. 82.dat" to an array of 83 elements - 0.dat, 1.dat, 2.dat, ... 82.dat. If all your data files have the extension ".dat" then you can use "*.dat", as shown below -

[code]
$
$
$ # display all ".dat" test files in my system
$ ls -1 *.dat
0.dat
1.dat
2.dat
3.dat
$
$
$ # now feed all of these files to the Perl one-liner
$ perl -lane 'print "$F[3] $F[4] $ARGV" if ($.>2 and !/^\s*$/); close ARGV if eof' *.dat
4.556663 238.047684 0.dat
77.605812 268.392334 0.dat
152.255829 256.117035 0.dat
161.680557 252.263550 1.dat
164.127808 242.561691 1.dat
182.847580 246.922028 1.dat
269.891937 255.048630 2.dat
270.984741 269.790710 2.dat
283.803314 246.089340 2.dat
288.413605 107.780937 2.dat
607.726550 255.130096 3.dat
986.900503 548.502799 3.dat
457.952424 691.618854 3.dat
800.869647 671.749259 3.dat
$
$
[quote]

Maybe you want more control than that. Say, for example, you have data in files "0.dat" through "82.dat". But you also have files like so - "abc.dat", "zzz.dat", "myfile.dat" which you do not want to process.

The one-liner above will actually feed the files "abc.dat", "zzz.dat" and "myfile.dat" because you are simply specifying a wildcard character "*" for whatever is there to the left of the dot i.e. the base file name.

One way out is "brace expansion". See if your shell supports this - if it does, it will expand numbers within braces like so -

Code:

$ 
$ echo {1..25}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
$ 
$

And you can use something similar for your files -

Code:

perl -lane 'print "$F[3] $F[4] $ARGV" if ($.>2 and !/^\s*$/); close  ARGV if eof' {0..82}.dat

Of course, I don't have all 83 test files in my system, so I can show you the test for 0.dat, 1.dat, 2.dat, 3.dat -

Code:

$ 
$ perl -lane 'print "$F[3] $F[4] $ARGV" if ($.>2 and !/^\s*$/); close  ARGV if eof' {0..3}.dat
4.556663 238.047684 0.dat
77.605812 268.392334 0.dat
152.255829 256.117035 0.dat
161.680557 252.263550 1.dat
164.127808 242.561691 1.dat
182.847580 246.922028 1.dat
269.891937 255.048630 2.dat
270.984741 269.790710 2.dat
283.803314 246.089340 2.dat
288.413605 107.780937 2.dat
607.726550 255.130096 3.dat
986.900503 548.502799 3.dat
457.952424 691.618854 3.dat
800.869647 671.749259 3.dat
$

Yes, it possible in Perl without renaming all the files. The scalar variable "$ARGV" simply holds the current file name, i.e. the name of the file that is being processed right now. It's value can be modified to remove ".dat" from the end. Here's the updated one-liner -

Code:

$ 
$ 
$ # Remove ".dat" from the end of $ARGV before printing the line
$ perl -lane '$ARGV=~s/.dat$//; print "$F[3] $F[4] $ARGV" if ($.>2  and !/^\s*$/); close ARGV if eof' *.dat
4.556663 238.047684 0
77.605812 268.392334 0
152.255829 256.117035 0
161.680557 252.263550 1
164.127808 242.561691 1
182.847580 246.922028 1
269.891937 255.048630 2
270.984741 269.790710 2
283.803314 246.089340 2
288.413605 107.780937 2
607.726550 255.130096 3
986.900503 548.502799 3
457.952424 691.618854 3
800.869647 671.749259 3
$ 
$ # Same one-liner as above, except that the file names are fed via the  "brace expansion" feature of the shell
$ perl -lane '$ARGV=~s/.dat$//; print "$F[3] $F[4] $ARGV" if ($.>2  and !/^\s*$/); close ARGV if eof' {0..3}.dat
4.556663 238.047684 0
77.605812 268.392334 0
152.255829 256.117035 0
161.680557 252.263550 1
164.127808 242.561691 1
182.847580 246.922028 1
269.891937 255.048630 2
270.984741 269.790710 2
283.803314 246.089340 2
288.413605 107.780937 2
607.726550 255.130096 3
986.900503 548.502799 3
457.952424 691.618854 3
800.869647 671.749259 3
$ 
$

HTH,
tyler_durden