> -----Original Message-----
> From: amit hetawal [mailto:[EMAIL PROTECTED]
> Sent: Sunday, April 23, 2006 7:23 AM
> To: [EMAIL PROTECTED]; [email protected]
> Subject: Re: unigrams,bigrams,trigrams?
>
> hello ,
> Thanks for the help .
> But my problem is that i need to get 3 different hashes for all uni bi
> tri grams ..
> coz later in my code i have to calculate the value as
>
> trigram{hello how are} /bigram{hello how}
I think perhaps instead of 3 different hashes, you should have a different
data structure.
I was thinking of a 3 dimensional hash with each key being a word, but that
would be hard to save the counts. Perhaps a hash w/ an array where the
first element is the count and the second is a hash. Where the second hash
is similar to the first in that the keys are words and values are arrays,
the first value the count the second a hash for trigrams. Using this method
you could also count out all your uni/bi/trigrams at the same time.
For example you would do this to get the count for selected items:
print "unigrams\n";
print "hello: ", $WordCount{hello}[0], "\n";
print "how: ", $WordCount{how}[0] , "\n";
print "bigrams\n";
print "Hello how: ", $WordCount{hello}[1]->{how}[0], "\n";
print "trigrams\n";
print "hello how are: ", $WordCount{hello}[1]->{how}[1]->{are}[0], "\n";
This way to "relate them" you just need to get the keys leading up to the
trigram. It'd be pretty easy to write a function something like:
Bigram_Divides_Trigram('hello', 'how', 'are') that returns a number.
I leave that as an exercise for the reader. :-D
-Wayne
PS. this solution is only tractable since you're only going up to trigram.
If this is a simple problem to try and answer a larger more general question
that data structure would become hopelessly complex (in my opinion).
--
CAM Automation Programmer
Unicircuit Inc.
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