Your text and example differ slightly. In multiple places the text indicates word 3 comes before words 1 and 2, but your example shows word 3 coming after words 1 and 2. I am assuming you want the latter, but in reality it doesn't impact the choice of data structure.

There are a lot of ways to approach this, but I might use something like

$hash{$word1}{$word2}{$word3} = $count
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Note that I did not store the percentage, which can be calculated after all of the counts have been accumulated. If you wanted a more complicated structure to facilitate that, you could do something like

$hash{$word1}{$word2}{$word3} =
    { count => $count,
      freq  => $frequency }
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$hash{$word1}{$word2} =
    {
        total_counts => $tot_counts,
        words => {
                     $word3 => $count,
                 }
    }
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To populate the data structure, simply identify words 1, 2, and 3 and then do

$hash{$word1}{$word2}{total_counts}++;
$hash{$word1}{$word2}{words}{$word3}++;
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These structures may be able to be simplified, depending on your requirements. On the other hand, if you had a lot of data and expected the structure to become quite large, a database may be a better choice.

Note that I did not use any arrays. Since the description of the problem implied that you would be looking up data by $word, a hash seemed more appropriate. An array (or an array of hash refs) would require you to search through the contents of the array to find the $word of interest, and depending on the size of the arrays that could be a very inefficient approach.

^*If you really wanted to store the frequency of each word and you didn't need a running total, I'd suggest doing the calculation at the end and using $word3 => { count => $count, freq  => $frequency } rather than $word3 => $count, which would make adding the frequencies later a bit more straightforward.

Thank you very much for your time and thoughts. I went back after reading your excellent explination and looked over the data structures cookbook. I now feel much more informed about Perl's data structures and understand why arrays would not work well in this situation.

cheers,
-mox

I'd be inclined to use a hash containing pairs of counts and following word hashes. The following works for generating data sets of arbitary orders:

use strict;
use warnings;
use Data::Dump::Streamer;

use constant ORDER => 3;

my %chains;

while (<DATA>) {
    my @words = split;
    my @chain;
    
    for (@words) {
        push @chain, $_;
        next if @chain < ORDER;
        shift @chain if @chain > ORDER;
        
        my $root;
        
        for my $index (0 .. ORDER - 1) {
            if (defined $root) {
                ++$root->[1]{$chain[$index]}[0];
                $root = $root->[1]{$chain[$index]};
            } else {
                ++$chains{$chain[$index]}[0];
                $root = $chains{$chain[$index]};
            }
        }
    }
}

Dump (\%chains);


__DATA__
I will not eat them in a bar.
I will not eat them in a car.
I will not eat them Sam I am.
I will not eat green eggs and ham!
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Once again your time and efforts are clear and helpful. Thank you for both.

-mox

Thank you for the link. It was very useful.

-mox

One question: does the order of the pair ($word1, $word2) matter? In other words (making the same assumption as to what you mean as bobf above), is:

equivalent to:

for your purposes, or is it different?

Update: Since you say "I think I will be accessing these parings via the $word1-$word2 parings", why not use an HoH, with $word1 and $word2 stringified as the key to the outer hash, and populate it something like this:

# foreach group of three words
      $hash{$word1 . ' ' . $word2}{$word3}++;
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Or, more generally, for any number of words in your 'chain':

# foreach group of n words in array @n_words
      my $nth_word = pop @n_words;
      $hash{join ' ', @n_words}{$nth_word}++;
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And if the answer to my question in the first part of this post is that order doesn't matter, just change that last line to:

Thank you very much. I am will read up on HoH's and use your suggestion in my final solution! Looking more into whether word order is important, I have concluded that it is, and so I will be using the $word1 and $word2 stringfield as a key to the out ther hash as you suggested. Thank you again for your help!

-mox