in reply to hard referencing in hashes?!?

Well, a) you can increment just about anything with the ++ operator.
And b) there's no need to see whether the element exists in the hash yet, because if you apply the ++ to a non-existent item, it treats it like it had the value of 0.

So: 

    foreach $word ( @words ) {
        $word_list{$word}++;
    }

[download]
That should do it. Perl makes the easy things easy!

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Re: Re: hard referencing in hashes?!?
by ph0enix (Friar) on Nov 08, 2002 at 11:05 UTC
    map { $word_list{$_}++ } @words;
[reply]
[d/l]
      $word_list{$_}++ for @words;

@sorted = sort keys %word_list;

      [download]
[reply]
[d/l]
        @word_list{@words}=1;
@sorted = sort keys %word_list;

        [download]

        update:
        Oh come on, that shift+9 shift+0 ... 3 extra keys man ... *me*shakes*head* ... whatever happened to lazyness ???

        :D

        ____________________________________________________
        ** The Third rule of perl club is a statement of fact: pod is sexy.

[reply]
[d/l]
        Thanks for all your help guys! I was wondering how I would be able to get word123 into my @word but not 456. So that word123 would be a word, but 456 isn't a word. Right now I have 
        my @words = map {s/[.,!\W\?]\B//g; split; } <FILE>;

        [download]
        With this i'm getting it to keep word123 but if its 39 it would remove the 3 but not the 9? Also if I add an i (case insensitive) behind the /g would that give me There = there?
[reply]
[d/l]

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