Re: Element Count from Array to HoH

I think that last; bit is messing your results.. Do you really want to break the moment you found a match? In this case, you'll find yourself for half of the array, and your neighbour for the other half - sounds confusing to me..

Also, from your example it's not clear what you are counting - if we had a couple of rows like "I1 TTTT", and "I2 TTAT", currently you'd increase {TTTT}{I1}, when you see "I2 TTAT" - is that what you want? Perhaps you ought to be increasing {TTAT}{I1}, ie {$item2}{$tid}, or possibly {TTTT}{I2}?

Last question, can you have repeated elements in your array, and if so, what should be the behaviour?

Ok, now I'll suggest a rewrite, based on my assumptions what this should be doing, but don't be too disappointed if they were wrong. I prefer returning a new ref out of the sub, rather than passing one in:

my $transaction_map_mismatch = get_trans_array_mismatch(\@to_proc,$d);

print Dumper $transaction_map_mismatch;

#--------Sub------------
sub get_trans_array_mismatch {
    my $transaction_array   = shift;
    my $d                   = shift;

    my $trans_map_ref = {map {(split)[1]=>{}} @$transaction_array};

    foreach (@{$transaction_array}) {
        my ($tid, $item) = split;

        foreach (grep {hd($item,$_) <= $d} keys %$trans_map_ref) {
            $trans_map_ref->{$_}{$tid}++;
        }
    }
    return $trans_map_ref;
}
[download]

Not efficient, in the slightest, but I think it does the job, or at least matches your desired output

Comment on Re: Element Count from Array to HoH Select or Download Code

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Re^2: Element Count from Array to HoH by monkfan (Curate) on May 26, 2005 at 07:15 UTC
Hi Ivancho, Q: Perhaps you ought to be increasing {TTAT}{I1}, ie {$item2}{$tid} A: I can't do that, cause it will miss counting "TTTT". if what you mean is changing them to this: `$$transaction_map_ref{$item2}{$tid}++;` [download] Q:or possibly {TTTT}{I2}? A:I don't get what you mean by that. What's the difference with the above statement. Sorry..I'm rather slow here.. Q:can you have repeated elements in your array, and if so, what should be the behaviour? A: Yes if that's the case, as long as condition "hd<=$d", they should be counted. I hope that clarifies the problem. Regards, Edward	[reply] [d/l]
Re^3: Element Count from Array to HoH by ivancho (Hermit) on May 26, 2005 at 07:56 UTC
Yeah, my bad, I only meant {TTTT}{I2} - which is what my code does, I shouldn't have asked about {TTAT}{I1}.. My understanding is as follows. You get a pair "TTAT", "I2", and you say to yourself 'But this TTAT might actually be TTTT - so let's increase {TTTT}{I2} for good measure.. We'll increase {TTAT}{I2} too, when we get there...' If this is what you want to do, then my snippet should work fine.. The `{map {(split)...` line saves the unique 4 letter codes we have. Then for each line '$tid $item', `grep {hd($item,$_) <= $d} keys %$trans_map_ref` [download] gives us which of our 4 letter codes $item might be. For each of them, we increase their {$tid} field. I think the problem here is that your function tries to do everything at once.. Maybe you can split it in 2 pieces - first find all pairs of neighbours between the 4 letter codes, then go through your array and for every item with a tid, also increase this tid for it's neighbours.. hopefully this isn't just more mud...	[reply] [d/l] [select]