Entropy: Difference between revisions
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my %count; %count{$_}++ for @a; |
my %count; %count{$_}++ for @a; |
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my @p = %count.values »/» @a.elems; |
my @p = %count.values »/» @a.elems; |
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- |
-[+] map { $_ * log $_ }, @p; |
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} |
} |
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say entropy "1223334444".comb;</lang> |
say log(2) R/ entropy "1223334444".comb;</lang> |
Revision as of 21:06, 21 February 2013
Entropy
You are encouraged to solve this task according to the task description, using any language you may know.
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 bit.
Burlesque
<lang burlesque> blsq ) "1223334444"F:u[vv^^{1\/?/2\/LG}m[?*++ 1.8464393446710157 </lang>
D
<lang d>import std.stdio, std.algorithm, std.math;
double entropy(T)(T[] s) /*pure nothrow*/ if (__traits(compiles, sort(s))) {
return s .sort() .group .map!(g => g[1] / cast(double)s.length) .map!(p => -p * log2(p)) .reduce!q{a + b};
}
void main() {
"1223334444"d.dup.entropy.writeln;
}</lang>
- Output:
1.84644
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
<lang Perl>sub entropy {
my %count; $count{$_}++ for @_; my @p = map $_/@_, values %count; my $entropy = 0; $entropy += - $_ * log $_ for @p; $entropy / log 2
}
print entropy split //, "1223334444";</lang>
Perl 6
<lang Perl 6>sub entropy(@a) {
my %count; %count{$_}++ for @a; my @p = %count.values »/» @a.elems; -[+] map { $_ * log $_ }, @p;
}
say log(2) R/ entropy "1223334444".comb;</lang>