Regular expression problems

rattytatty has asked for the wisdom of the Perl Monks concerning the following question:

I have a problem with a pattern match. I want to print what the match is whether it be by using $1 or assigning the match to another variable. It seems to be taking a scalar of the value instead of the value and I don't know why unless perl 5.10.1 is an issue

my $procount2 = 1;

foreach my $readingframe2(@protein){
    my $protein;
    print "string is $readingframe2\n";
    
    my $fada;
    if($readingframe2 =~ /MHGR/){
        print "$1\n";
        ($protein)= $readingframe2 =~ /MHGR/;
        $fada = length($protein);
        print "motif is $protein\n";
        if($fada >=1){
           print "length of motif is $fada\n";
           print OUT1 "Protein $procount2\n$protein\n";
           $procount2++;
        }
    }
}
[download]

Here is what comes up in the console:

$ perl readingframe.pl
Enter file name
t.txt
Enter output file name
o.txt
string is MHGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRD_MHGRRRRRRRRRRRRRRRRRRRRRRR
+RRRRRRD_
Use of uninitialized value $1 in concatenation (.) or string at readin
+gframe.pl line 68, <> line 2.

motif is 1
length of motif is 1
string is CMAAAAAAAAAAAAAAAAAAAAAAAAAAAAAVTECMAAAAAAAAAAAAAAAAAAAAAAAA
+AAAAAVT
string is AWPPPPPPPPPPPPPPPPPPPPPPPPPPPPP_LNAWPPPPPPPPPPPPPPPPPPPPPPPP
+PPPPP_L
[download]

Comment on Regular expression problems Select or Download Code

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Re: Regular expression problems by brx (Pilgrim) on Apr 23, 2012 at 16:10 UTC
`if($readingframe2 =~ /MHGR/){ print "$1\n"; ($protein)= $readingframe2 =~ /MHGR/;` [download] $1 contains what you capture with parens: no parens here in regex, so nothing in $1. What would you like to find in $prottein ? `example: $ perl -e '$s="pilpoil";if ($s=~/p(.)lp(.)il/) { print "h".$1."h".$2;} +' hiho` [download]	[reply] [d/l] [select]
Re^2: Regular expression problems by rattytatty (Initiate) on Apr 24, 2012 at 08:22 UTC
Thanks for the help. I used this as an example but I will be using it to find an ORF (A Start codon'M', followed by any of the rest of the Amino Acids until a Stop codon'_' I think that will be: `=~/(M)[GAVLIFWPSTCYNQDEKRH]+(_)/` or `=~/(M[GAVLIFWPSTCYNQDEKRH]+_)/` I'm not sure but I will try both. Thanks again.	[reply] [d/l] [select]
Re^3: Regular expression problems by brx (Pilgrim) on Apr 24, 2012 at 14:16 UTC
You can want to add 'M' and '_' in parens to capture in one shot or concatanate 'M'.$1.'_' depending of what you want to do. One way to do it: `#!/usr/bin/perl -w use strict; my $seqnum=1; while (my $seq=<DATA>) { chomp $seq; print "sequence #",$seqnum++,":\n"; while ($seq =~ /M([GAVLIFWPSTCYNQDEKRH]+)_/g) { print "\t",$1,"\n"; # or: print "\t","M${1}_","\n"; } } __DATA__ MHGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRD_MHGRRRRRRRRD_ CMAAAAAAAAAAAAAAAAAAAAAAAAAAAAAVTECMAAAAAAAAAAAAAAAAAAAAAAAAAAAAAVT AWPPPPPPPPPPPPPPPPPPPPPPPPPPPPP_LNAWPPPPPPPPPPPPPPPPPPPPPPPPPPPPP_L FOOBARMNOTTHISONE_XYZMTHISYES_MTHISNOT_` [download]	[reply] [d/l]
Re: Regular expression problems by choroba (Cardinal) on Apr 23, 2012 at 16:14 UTC
If you want to populate $1, you have to use parentheses: `$readingframe =~ /(MHGR)/; $protein = $1;` [download] or you can assign it directly: `($protein) = $readingframe =~ /(MHGR)/;` [download] In your case, though, you can just output MHGR, as there is nothing dynamic in your regular expression. See also Regexp Quote Like Operators, perlretut, perlre.	[reply] [d/l] [select]
Re: Regular expression problems by pvaldes (Chaplain) on Apr 23, 2012 at 16:21 UTC
Use of uninitialized value $1... Because, $1 is in fact undefined in your example. `my $procount2 = 1; foreach my $readingframe2(@protein){ my $protein; print "string is $_\n"; if(/MHGR/){ print "$1\n","motif is $_\n"; ##### <----- if(length($_)>=1){ print "length of motif is length($_)\n"; print OUT1 "Protein $procount2\n$_\n"; $procount2++; } } }` [download]	[reply] [d/l]