Entropy: Difference between revisions
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fq' c = map (\x -> x / (sum c)) c |
fq' c = map (\x -> x / (sum c)) c |
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</lang> |
</lang> |
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=={{header|Perl 6}}== |
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<lang Perl 6> |
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sub entropy(@a) { |
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my %count; %count{$_}++ for @a; |
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my @p = %count.values »/» @a.elems; |
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-log(2) R/ [+] map { $_ * log $_ }, @p; |
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} |
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say entropy "1223334444".comb;</lang> |
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Revision as of 20:39, 21 February 2013
You are encouraged to solve this task according to the task description, using any language you may know.
Calculate the entropy (shannon entropy) of a given input sequence. Use "1223334444" as an example sequence. The result should be around 1.84644.
Burlesque
<lang burlesque> blsq ) "1223334444"F:u[vv^^{1\/?/2\/LG}m[?*++ 1.8464393446710157 </lang>
Haskell
<lang haskell> import Data.List
main = print $ entropy "1223334444"
entropy s =
sum . map lg' . fq' . map (fromIntegral.length) . group . sort $ s
where lg' c = (c * ) . logBase 2 $ 1.0 / c
fq' c = map (\x -> x / (sum c)) c
</lang>
Perl 6
<lang Perl 6> sub entropy(@a) {
my %count; %count{$_}++ for @a;
my @p = %count.values »/» @a.elems;
-log(2) R/ [+] map { $_ * log $_ }, @p;
}
say entropy "1223334444".comb;</lang>