in reply to Convert a string into a hash

The code below demonstrates that $x (scalar), @x (list), %x (hash) are different things.
Your instinct that anything with Perl using array indices is probably wrong, is correct. 
#!/usr/bin/perl -w
use strict;
use Data::Dumper;

my $tokens = "32,15,4,72,13,28,14";
my @tokens = (split/,/,$tokens);
my %tokens = map {$_ => 1}@tokens;

print Dumper (\%tokens);

__END__

Prints:
$VAR1 = {
          '4' => 1,
          '32' => 1,
          '28' => 1,
          '72' => 1,
          '13' => 1,
          '14' => 1,
          '15' => 1
        };

[download]
Reconsidering your requirements, and I will say that this is bizarre: 
#!/usr/bin/perl -w
use strict;
use Data::Dumper;

my $tokens = "32,15,4,72,13,28,14";
my @tokens = (split/,/,$tokens);
my $i=0;
my %tokens = map {$_ => $i++}@tokens;

print Dumper (\%tokens);
__END__
Prints:
VAR1 = {
          '4' => 2,
          '32' => 0,
          '28' => 5,
          '72' => 3,
          '13' => 4,
          '14' => 6,
          '15' => 1
        };

[download]
Now again of course since this initialization, why would you need the scalar string at all? 
my @tokens = qw (32 15 4 72 13 28 14);
my $i=0;
my %tokens = map {$_ => $i++}@tokens;

[download]
yields the same result as above.

Ok, I am going to go "crazy" here and ask why you want a hash in the first place? I am at a loss the see the usefulness of a hash here. Why do you think that you need it?

Replies are listed 'Best First'.
Re^2: Convert a string into a hash
by vitoco (Hermit) on Aug 15, 2009 at 04:26 UTC
    Ok, I am going to go "crazy" here and ask why you want a hash in the first place?

    I just want to iterate the elements of another list of tokens, but in the same order them appear in the main list of tokens. Simplified:

    my $tokens = "32,15,4,72,13,28,14";
my %code = do { my $i = 0; map { $_ => $i++ } split /,/, $tokens };

my $list = "4,13,15";
my $ok;
for my $token (sort { $code{$a} <=> $code{$b} } split /,/, $list) {
  print "$token ";
  # more code with $token...
  last if $ok;
}

__END__
15 4 13

    [download]

    Obviously, both $tokens and $list are dynamically loaded from somewhere else...

[reply]
[d/l]
      This is a straight-forward implementation.
      Post again if I didn't get it right.... 
      #!/usr/bin/perl -w
use strict;

my $tokens = "32,15,4,72,13,28,14";
my @tokens = split (/,/, $tokens);

my $list = "4,13,15";
my @list = split (/,/,$list);
my %list = map {$_ => 1}@list;

my @ordered_nums = grep{$list{$_}}@tokens; 
print "@ordered_nums\n";

__END__
PRINTS:
15 4 13

      [download]
[reply]
[d/l]

        That was the other aproach I thought, but the requirement changed to pick only the best token from $list, which is loaded many times based on some external events, while %code is initialized once.

        Based on the previous, I decided to keep that part, and now I want to rewrite:

        my $best = 14; # actually, last value from $tokens.
for my $b (split /,/, $list) {
  $best = $b if $code{$b} < $code{$best};
}

        [download]

        Update: syntax error, extra ")"...

[reply]
[d/l]
      Ooops, mess up and double post...sorry.... This is a straight-forward implementation.
      Post again if I didn't get it right.... 
      #!/usr/bin/perl -w
use strict;

my $tokens = "32,15,4,72,13,28,14";
my @tokens = split (/,/, $tokens);

my $list = "4,13,15";
my @list = split (/,/,$list);
my %list = map {$_ => 1}@list;

my @ordered_nums = grep{$list{$_}}@tokens; 
print "@ordered_nums\n";

__END__
PRINTS:
15 4 13

      [download]
      Update: I would perhaps replace "list" with "list_order". Matter of choice.
[reply]
[d/l]

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