faster winner takes all
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I 've made an mplementation of the winner takes all rule (WTA) using a for loop (code bellow). But now I want to "harden" ~8 million values and the loop is very slow (several hours). Any idea for a faster WTA implementation? In detail, if we have a matrix with values in [0,1] we need to replace them with 1 for > 0.75 or with 0 for < 0.75.
function H = wta(S)
% Winner Takes All (WTA) rule
% returns the hard values for S
%
[row,col] = size(S);
for r = 1:row
if S(r,1) > 0.75
H(r,1) = 1;
else
H(r,1) = 0;
end
end
3 件のコメント
Walter Roberson
2011 年 11 月 6 日
For 0.75 exactly, as you do not define any replacement conditions, the implication would be that those locations should not be replaced, leaving them 0.75 .
採用された回答
Andrei Bobrov
2011 年 11 月 6 日
H = S > .75;
2 件のコメント
Walter Roberson
2011 年 11 月 6 日
H = S(:,1) > 0.75;
would be a closer match to the code. The difference would be important if S is two-dimensional or S is a row vector.
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