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Re: finding matches in the same array

by polettix (Vicar)
on Apr 13, 2005 at 17:02 UTC ( [id://447507]=note: print w/replies, xml ) Need Help??


in reply to finding matches in the same array

If you have to evaluate the average for all sequences, you can build the following hash: 
my %occurrences;
push @{$occurrences{$sequence[$_]}}, $numbers[$_]
   foreach (0 .. $#sequence);

[download]
At this point, each entry in the hash has a "sequence" for key and a reference to an array containing the "numbers" for that sequence as value - computing the average should be pretty easy, e.g.: 
my $sum;
$sum += $_ foreach (@{$occurrences{'atcg'}});
my $average = $sum / scalar(@{$occurrences{'atcg'}});

[download]
I leave the union of the two snippets as an exercise, just in case it's an homework :)

Update: fixed a typo in first snipped, thanks to Postular Postulant.

Flavio (perl -e "print(scalar(reverse('ti.xittelop@oivalf')))")

Don't fool yourself.

Replies are listed 'Best First'.
Re^2: finding matches in the same array
by bageler (Hermit) on Apr 13, 2005 at 18:07 UTC
    let's reduce the loops by calculating the average as we move along: 
    my %oc;
for (0 .. $#sequence) {
  $oc{$sequence[$_]}[0] =
    ($oc{$sequence[$_]}[0]+$numbers[$_])/(++$oc{$sequence[$_]}[1]);
}
foreach my $seq (keys %oc) {
  print "The average for $seq is ".$oc{$seq}[0].$/;
}

    [download]
[reply]
[d/l]
      This does not seem to be an average :) First item is divided by 1, second divided by 2, third by 3... - maybe it's better to make the division at the end: 
      my %oc;
for (0 .. $#sequence) {
  $oc{$sequence[$_]}[0] += $numbers[$_];
  ++$oc{$sequence[$_]}[1];
}
foreach my $seq (keys %oc) {
  print "The average for $seq is ".($oc{$seq}[0] / $oc{$seq}[1]).$/;
}

      [download]

      Flavio (perl -e "print(scalar(reverse('ti.xittelop@oivalf')))")

      Don't fool yourself.
[reply]
[d/l]

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