Re: opening many files

[jwkrahn]

UNTESTED, but this may work:

@ARGV = glob 'datafile*';

open my $OUTPUT, '>', 'output_file' or die "Cannot open 'output_file' 
+because: $!";

while ( <> ) {
    if ( $. == 1 ) {
        print $OUTPUT $ARGV =~ /(\d+)$/, " has";
        }
    print $OUTPUT " ", ( split )[ 2 ];
    if ( eof ) {
        print $OUTPUT " $. columns\n";
        close ARGV;
        }
    }

[wanttoprogram]

thank you for reply. it is working file except i want 948 rows and 200 columns from 200 files. the program is giving me 200 rows and 948 columns. thank you.

[jwkrahn]

@ARGV = glob 'datafile*';

open my $OUTPUT, '>', 'output_file' or die "Cannot open 'output_file' 
+because: $!";

my %data;
while ( <> ) {
    next unless /^(.+) (\S+)$/;
    push @{ $data{ $1 } }, $2;
    }

for my $key ( sort keys %data ) {
    print join ' ', $key, @{ $data{ $key } }, scalar @{ $data{ $key } 
+}, "columns\n";
    }

[wanttoprogram]

thank you. the code is working fine. could you please help me understand the code. i am a beginner in perl and had hard time understanding the hi-fi code. also could you please explain how sort is working in your code. what changes do i have to make to code to get all the data with out any sorting. thank you.

[wanttoprogram]

thank you. the code is working fine. could you please explain how sort is working in your code. what changes do i have to make to code to get all the data with out any sorting. thank you

[wanttoprogram]

@ARGV = glob 'datafile*';

open my $OUTPUT, '>', 'output_file' or die "Cannot open 'output_file' 
+because: $!";

my %data;
while ( <> ) {
    next unless /^(.+) (\S+)$/;
    push @{ $data{ $1 } }, $2;
    }

for my $key ( sort keys %data ) {
    print join ' ', $key, @{ $data{ $key } }, scalar @{ $data{ $key } 
+}, "columns\n";
    }

======================== The program you have given works fantastic. But I have to deal with one more thing...Here it is. I have files by name datafile1,datafile2,datafile3,datafile4.........datafile200 in a directory. But these files are stored in a a different pattern ...datafile1,datafile10,datafile11,datafile12 and so on. The above program is reading and writing in the same order. I would like to pick data from in the an order from datafile1, 2, 3 to data200. I guess I need to sort. Any suggestions and additions are appreciated. Thank you.

[aaron_baugher]

That's because glob returns the results of datafile* interpolated in the order a shell would return them, so datafile11 comes before datafile2, as you've seen. A quick and dirty solution, if you know you're only dealing with up to 3 digits:

@ARGV = glob 'datafile? datafile?? datafile???';

A more general solution that'll sort on any number of digits:

@ARGV = sort { my $an = substr $a, 8; my $bn = substr $b, 8; $an <=> $
+bn } glob 'datafile*';

[Anonymous Monk]

http://learn.perl.org/faq/perlfaq4.html#How-do-I-sort-an-array-by-anything-

http://perldoc.perl.org/perlfaq4.html#How-do-I-sort-an-array-by-%28anything%29%3f

Sorting ip addresses quickly