imlou has asked for the wisdom of the Perl Monks concerning the following question:

If I want to search a hash by strings in an array, and if the string exists in the hash how do I increment the value of the hash. This is the code i have but its not working. 
foreach $word (@words){
        if($word eq $word_list{$word}){
        $word_list{$word}= $word_list{$word} + 1; 
        }
        else{
        $word_list{$word} = 1;
        }
}

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Replies are listed 'Best First'.
Re: hard referencing in hashes?!?
by DamnDirtyApe (Curate) on Nov 08, 2002 at 02:28 UTC

    How about something like this:

    #! /usr/bin/perl
use strict ;
use warnings ;
use Data::Dumper ;

my @match_arr  = qw( count pick words ) ;
my %match_hash = map { $_ => 1 } @match_arr ;

my %word_count ;
while ( <DATA> )
{
    my @words = split ;
    for ( @words ) { $word_count{$_}++ if $match_hash{$_} }
}

print Dumper \%word_count ;

exit ;

__DATA__
This is my data.
I will first count all the
words herein, then pick and
choose the data I want. I can pick
the words using grep.

    [download]

    _______________
    DamnDirtyApe 
    Those who know that they are profound strive for clarity. Those who
would like to seem profound to the crowd strive for obscurity.
            --Friedrich Nietzsche
[reply]
[d/l]
      Thanks! but for my assignment i only have to count all the words in a file. Thats why I used hashes....but i found the powerful use of EXISTS :P. Make life so much easier. Now my problem is sorting the results of each word by acending or decending order.
[reply]

        Look at using a Schwartzian Transform:

        my @sorted = sort { $b->[1] <=> $a->[1] }
             map  { [ $_, $word_count{$_} ] }
             keys %word_count ;

        [download]

        _______________
        DamnDirtyApe 
        Those who know that they are profound strive for clarity. Those who
would like to seem profound to the crowd strive for obscurity.
            --Friedrich Nietzsche
[reply]
[d/l]
        generally we use sort
[reply]
Re: hard referencing in hashes?!?
by jdporter (Paladin) on Nov 08, 2002 at 04:07 UTC
    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!
[reply]
[d/l]
[select]
      map { $word_list{$_}++ } @words;
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[d/l]
        $word_list{$_}++ for @words;

@sorted = sort keys %word_list;

        [download]
[reply]
[d/l]
Re: hard referencing in hashes?!?
by LEFant (Scribe) on Nov 08, 2002 at 03:19 UTC
    If I structure a solution as a c, java, or BASIC programmer might: 
    my %word_list = ( bat=> 0, cat=> 0, eat=> 0, fat=> 0 );
my @words = qw( bat bat dat fat eat eat eat fat );
my $word;

foreach $word (@words){
   if(exists $word_list{$word}){
      $word_list{$word}= $word_list{$word} + 1; 
   }
   else{
      $word_list{$word} = 1;
   }
}

foreach $word (sort keys %word_list) {
   print "$word: $word_list{$word}\n";
}

    [download]
    If I want a perlish solution I would write: 
    %word_list = ( bat=> 0, cat=> 0, eat=> 0, fat=> 0 );
my @words = qw( bat bat dat fat eat eat eat fat );
my $word;
foreach $word (@words){
   $word_list{$word}++;
}

foreach $word (sort keys %word_list) {
   print "$word: $word_list{$word}\n";
}

    [download]
    If don't need a the original word_list for displaying words that were not matched (word count is 0) but want only the counts of the words that occur in @words then an empty %word_list. 
    my %word_List = ( );

    [download]
    Using a hash and increment to count things is a fundamental technique to master early in your perl walk.

    RBL

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[d/l]
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