Hi eventual,
On Friday 27 May 2011 11:18:01 eventual wrote:
> Hi,
> I have an array , @datas, and each element within @datas is a string that's
> made up of 6 digits with spaces in between like this “1 2 3 4 5 6”, so the
> array look like this @datas = ('1 2 3 4 5 6', '1 2 9 10 11 12', '1 2 3 4 5
> 8', '1 2 3 4 5 9' , '6 7 8 9 10 11'); Now I wish to compare each element
> of @datas with the rest of the elements in @datas in such a way that if 5
> of the digits match, to take note of the matching indices, and so the
> script I wrote is appended below. However, the script below takes a long
> time to run if the datas at @datas are huge( eg 30,000 elements). I then
> wonder is there a way to rewrite the script so that the script can run
> faster. Thanks
>
> ###### script below #######################
>
> #!/usr/bin/perl
> use strict;
>
> my @matched_location = ();
> my @datas = ('1 2 3 4 5 6', '1 2 9 10 11 12', '1 2 3 4 5 8', '1 2 3 4 5 9'
> , '6 7 8 9 10 11');
> my $iteration_counter = -1;
> foreach (@datas){
> $iteration_counter++;
> my $reference = $_;
>
> my $second_iteration_counter = -1;
> my $string = '';
> foreach (@datas){
> $second_iteration_counter++;
> my @individual_digits = split / /,$_;
>
> my $ctr = 0;
> foreach(@individual_digits){
> if($reference =~/^$_ | $_ | $_$/){
> $ctr++;
> }
> }
> if ($ctr >= 5){
> $string = $string . "$second_iteration_counter ";
> }
> }
> $matched_location[$iteration_counter] = $string;
> }
>
> my $ctr = -1;
> foreach(@matched_location){
> $ctr++;
> print "Index $ctr of \@matched_location = $_\n";
> }
>
First of all, you should add "use warnings;" to your code. Then you should get
rid of the implicit $_ as loop iterator because it's easy to break. For more
information see:
http://perl-begin.org/tutorials/bad-elements/
Other than that - you should use a better algorithm. One option would be to
sort the integers and then use a diff/merge-like algorithm:
http://en.wikipedia.org/wiki/Merge_algorithm
A different way would be to use a hash to count the number of times each
number occured in the two sets, and then see how many of them got a value of 2
(indicating they are in both sets).
But at the moment, everything is very inefficient there.
Regards,
Shlomi Fish
--
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