in reply to Hash of Arrays and File Operations

I would like to know how can I do the sum of the rows without taking the the values of the same id present in other columns of the file. 
use strict;
use warnings;
use List::Util qw(sum);

my %data;
while (<DATA>) {
    next if /^id/;
    my ($id, @cols) = split;
    push @{ $data{$id} }, \@cols;
}

for my $id (sort keys %data) {
    for my $aref (@{ $data{$id} }) {
        print "$id ", sum(@{$aref}), "\n";
    }
}

__DATA__
id   col1   col2    col3
12   100    12        196        
12   120    15        190
13   90     190        200
13   70       20       20 
13   101    340        25 
14   100    123        19 
15   80     389        39

[download]
prints: 
12 308
12 325
13 480
13 110
13 466
14 242
15 508

[download]

Update:

1. I would like to know that how will I do the translation/ substitution operation on the data stored in Hash of Array data structures.
In general, you will loop through your data structure using for, but, depending on exactly what your need to modify, you might be able to use map. Of course, if possible, it would be best to modify before you load it into your data structure in the first place.

Replies are listed 'Best First'.
Re^2: Hash of Arrays and File Operations
by snape (Pilgrim) on Jan 28, 2010 at 19:20 UTC

    I am trying to use map but it doesn't seem to be working. I tried doing it as : 

    @cols[1 .. $#cols] = map(tr /[1,2]/[0,1]/, @cols)

    [download]

    Also, do u think doing (i)split and getting the values as $id and @cols; then doing (ii)join to get the cols as scalar values so that we can do translate and then do (iii) split again to get the array value so that I can store it in the hash of arrays a good idea ?? for eg: 

    my %data;
while (<DATA>) {
    next if /^id/;
    my ($id, @cols) = split; ## first split
    my $cols = join('',@cols); ## Join 
    @cols = split('',$cols); ## and then split
    push @{ $data{$id} }, \@cols;
}

    [download]

    I am having second thoughts about this method as it may lead to slower process or it might not be a good programming practice. Please Advise.

[reply]
[d/l]
[select]
      @cols[1 .. $#cols] = map(tr /[1,2]/[0,1]/, @cols)

      [download]
      I'm no tr expert, but isn't tr /[1,2]/[0,1]/ more simply written as tr/12/01/?

      The following will replace all 1's with 0's, and all 2's with 1's, in all elements of the @cols array. It modifies the array in-place. 

      map { tr/12/01/ } @cols;

      [download]

      Update: I concur with AnonyMonk's hint that it is better written as: 

      tr/12/01/ for @cols;

      [download]
[reply]
[d/l]
[select]
        tr/12/91/ for @cols[1.. $#cols];
[reply]
[d/l]

In Section Seekers of Perl Wisdom