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Nested Cycle - Statistic measure

by remluvr (Sexton)
on Mar 12, 2012 at 15:55 UTC ( #959138=perlquestion: print w/ replies, xml ) Need Help??
remluvr has asked for the wisdom of the Perl Monks concerning the following question:

Hi everyone.
My question is the following. I have the following structure:

1 beast-n into transform-v 356.9551 2 beast-n obj kill-v 266.2511 3 beast-n obj see-v 252.3623 4 beast-n prd become-v 250.9534 5 beast-n obj tame-v 224.6948 6 beast-n into turn-v 191.9883 7 beast-n obj call-v 171.4000 8 beast-n sbj_intr devour-v 165.3228 9 beast-n obj hunt-v 155.7637 10 beast-n obj fight-v 150.4370 11 beast-n obj slay-v 150.3982 1 frog-n obj find-v 322.5589 2 frog-n into turn-v 307.3012 3 frog-n sbj_intr jump-v 235.0503 4 frog-n coord-1 toad-n 207.3611 5 frog-n obj see-v 207.2610 6 frog-n obj eat-v 204.1762 7 frog-n obj kill-v 64.6689

Using these data I need to implement a statistical measure to check the relevance of a given semantic relation in regards to the names it occur with.
Apart from the list above, I have two words as input. (sticking to the previous example, let's say they are beast-n and frog-n. If a given feature occurring with the first word also occurs whit the second, I have to compute Precision of the feature in regard to the first word. If I am at rank 1, and I found a feature that occurs also with the second word, my precision is 1, because it's computed as found_relevant_feat/rank_found. In the example above, the only feat that occurs both with beast and frog is kill-v. My precision would then be 1(which is the number of found_rel_feat until that rank)/2 which is the rank in which it occurs.
Also, I have to found the rank in which the given feature has been founded with word2. (in this case 7) When I am done with this I also need to know the total number of occurrence of the first word and the total number of occurrence of the second word. (given the example before, it would be 11 for beast-n and 7 for frog-n).

my ($prop,$rank, $score); my ($prop2,$rank2, $score2); while (my($name1,$aref) = each %matrice ) { my $num=0; foreach my $item (@$aref){ $count_feat_trovate=0; ($prop,$rank, $score) = split(',',$item); my $lastrank=&lastrank2($name2,$prop); while (my($name2,$aref2) = each %matrice) { foreach my $item1 (@$aref2){ ($prop2,$rank2, $score2) = split(',',$item1); if($prop eq $prop2){ $count_feat_trovate++; #number of feat found $rank_trovato_2=$rank2; #Rank of which the feat has been f +ound in regards to the second element } } } $last_rank_2=$rank2; } $last_rank_1=$rank; }

I guess I just made a lot of mess without achieving anything. Any ideas on how to retrieve the needed data?
Thanks
Giulia

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Re: Nested Cycle - Statistic measure
by Marshall (Prior) on Mar 12, 2012 at 17:14 UTC
    Excuse me for my mis-understanding. Can you clarify? What is special about "kill-v"?
    #!/usr/bin/perl -w use strict; use Data::Dumper; use Data::Dump qw (pp); my %animals; my %verbs; # Your say, "In the example above, the only feat that occurs # both with beast and frog is kill-v". # # I don't see it # Obviously there is something that I don't understand... # # this simple code gets multiple things in common # I don't understand how to filter out "see-v" and "turn-v" # Prints: # turn-v: beast-n frog-n # see-v: beast-n frog-n # kill-v: beast-n frog-n while (<DATA>) { my ($animal, $verb) = (split(' ', $_))[1,3]; $animals{$animal}++; push @{$verbs{$verb}}, $animal; } foreach my $verb (keys %verbs) { if (@{$verbs{$verb}} > 1) { print "$verb: @{$verbs{$verb}}", "\n"; } } __DATA__ 1 beast-n into transform-v 356.9551 2 beast-n obj kill-v 266.2511 3 beast-n obj see-v 252.3623 4 beast-n prd become-v 250.9534 5 beast-n obj tame-v 224.6948 6 beast-n into turn-v 191.9883 7 beast-n obj call-v 171.4000 8 beast-n sbj_intr devour-v 165.3228 9 beast-n obj hunt-v 155.7637 10 beast-n obj fight-v 150.4370 11 beast-n obj slay-v 150.3982 1 frog-n obj find-v 322.5589 2 frog-n into turn-v 307.3012 3 frog-n sbj_intr jump-v 235.0503 4 frog-n coord-1 toad-n 207.3611 5 frog-n obj see-v 207.2610 6 frog-n obj eat-v 204.1762 7 frog-n obj kill-v 64.6689

      Kill-v is the only feature that occur both with beast-n and frog-n. I have been given two words, (frog-n and beast-n, for example) and I have to take care only of features that occur with both of the words I have.
      I hope this clarify it a little
      Thanks again Giulia

        Kill-v is the only feature that occur both with beast-n and frog-n.

        Sorry, no. I can see that's not right even by eye-balling your data.

        And Marshall's code above (did you actually bother to run it?) proves that your claim is wrong!

        What haven't you told us about turn-v and see-v ??

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