faster winner takes all

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

[Walter Roberson]

Your code does not match your written requirements for the situation where S(r,1) is 0.75 exactly (which _is_ possible in binary floating point.)

[Dimitris]

Correct, it is just that 0.75 is so arbitrary that it does not matter which way is goes. You are correct mathematically though.

[Walter Roberson]

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 .

Answer 1

[Andrei Bobrov]

H = S > .75;

[Walter Roberson]

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.

[Dimitris]

Many thanks.