Extracting specific data from text file in Perl

Question

I am new to Perl and am trying to extract specific data from a file, which looks like this:

 Print of   9 heaviest strained elements:    


   Element no   Max strain 
      20004         9.6 % 
      20013         0.5 % 
      11189         0.1 % 
      20207         0.1 % 
      11157         0.1 % 
      11183         0.0 % 
      10665         0.0 % 
      20182         0.0 % 
      11160         0.0 % 


 ==================================================

I would like to extract the element numbers only (20004, 20013 etc.) and write these to a new file. The reading of the file should end as soon as the line (=========) is reached, as there are more element numbers with the same heading later on in the file. Hope that makes sense. Any advice much appreciated!

I now have this code, which gives me a list of the numbers, maximum 10 in a row:

my $StrainOut = "PFP_elem"."_$loadComb"."_"."$i";
open DATAOUT, ">$StrainOut" or die "can't open $StrainOut";  # Open the file for writing.

open my $in, '<', "$POSTout" or die "Unable to open file: $!\n";
my $count = 0;

 while(my $line = <$in>) {
  last if $line =~ / ={10}\s*/;
  if ($line =~ /% *$/) {
    my @columns = split "         ", $line;
    $count++;
    if($count % 10 == 0) {
      print DATAOUT "$columns[1]\n";
    }
    else {
      print DATAOUT "$columns[1] ";
    }      
  }
}
close (DATAOUT);
close $in;

What needs changing is the "my @columns = split..." line. At the moment it splits up the $line scalar whenever it has '9 spaces'. As the number of digits of the element numbers can vary, this is a poor way of extracting the data. Is it possible to just read from left to right, omitting all spaces and recording numbers only until the numbers are followed by more spaces (that way the percentage value is ignored)?

this seems to work: my @columns = split(/\s+/,$line);

charles hendry
– charles hendry

2012-02-02 16:18:35 +00:00
Commented Feb 2, 2012 at 16:18 — charles hendry
– charles hendry, Commented Feb 2, 2012 at 16:18

choroba · Accepted Answer · 2012-02-02 11:34:39Z

1

#!/usr/bin/perl
use strict;
use warnings;

while (<>) {                        # read the file line by line
    if (/% *$/) {                   # if the line ends in a percent sign
        my @columns = split;        # create columns
        print $columns[0], "\n";    # print the first one
    }
    last if /={10}/;                # end of processing
}

answered Feb 2, 2012 at 11:34

choroba

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Comments

Birei · Accepted Answer · 2012-02-02 14:06:05Z

1

A one-liner using flip-flop:

perl -ne '
  if ( m/\A\s*(?i)element\s+no/ .. ($end = /\A\s*=+\s*\Z/) ) {
    printf qq[$1\n] if m/\A\s*(\d+)/;
    exit 0 if $end
  }
' infile

Result:

answered Feb 2, 2012 at 14:06

Birei

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Comments

Alien Life Form · Accepted Answer · 2012-02-02 11:36:07Z

0

#!/usr/bin/perl
use strict;
use warnings;

while (my $f= shift) {
   open(F, $f) or (warn("While opening $f: $!", next);
   my foundstart=0;
  while(<F>) {
     ($foundstart++, next) if /^\s#Element/;
     last if /\s*=+/;
     print $_ if $foundstart;
  }
  $foundstart=0;
  close(F);
}

answered Feb 2, 2012 at 11:36

Alien Life Form

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1 Comment

Birei Over a year ago

It has compile errors. 1.- There is a miss of a parentheses in the warn instruction. 2.- Declare foundstart variable as a scalar with $ and in the next regex I think there is a typo with # instead *. It then prints numbers but percentages too in my test.

flesk · Accepted Answer · 2012-02-02 11:41:44Z

0

#!/usr/bin/perl
use strict;
use warnings;

open my $rh, '<', 'input.txt' or die "Unable to open file: $!\n";
open my $wh, '>', 'output.txt' or die "Unable to open file: $!\n";

while (my $line = <$rh>) {        
    last if $line =~ /^ ={50}/;
    next unless $line =~ /^ {6}(\d+)/;
    print $wh "$1\n";
}

close $wh;

answered Feb 2, 2012 at 11:41

flesk

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eyevan · Accepted Answer · 2012-02-02 12:01:40Z

0

You could do it by running this one-liner in a command shell.

On *nix:

cat in_file.txt | perl -ne 'print "$1\n" if ( m/\s*(\d+)\s*\d+\.\d+/ )' > out_file.txt

On Windows:

type in_file.txt | perl -ne "print qq{$1\n} if ( m/\s*(\d+)\s*\d+\.\d+/ )" > out_file.txt

answered Feb 2, 2012 at 12:01

eyevan

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2 Comments

flesk Over a year ago

He wants to stop reading from file when the line with the equals signs are reached though.

JRFerguson Over a year ago

The cat is a waste of a process. The -n switch causes an iteration over the file names in @ARGV as if you had written LINE: while (<>) { ...

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