Issues using an efficient positivity constraint
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Martin Pott
am 24 Nov. 2014
Beantwortet: Martin Pott
am 26 Nov. 2014
I want to use the following command to obtain only values of 0 or higher for the values in the 600x495 weight matrix p_w:
for j = 1:size(p_w,1)
for i = 1:size(p_w,2)
p_w(j,i) = max(0,p_w(j,i))/nansum(max(0,p_w(j,:)));
end
end
However, when running the optimization it takes a very long time (probably because of the loops), so I have tried to get rid of the loops and came up with the following:
temp = sum(max(0,p_w),2);
p_w_plus = max(0,p_w)./repmat(temp, 1, 495);
However, the optimized values in this case are very different from the function with the loops ,and they are also so extreme that they seem erroneous, so I am probably making a mistake somewhere in transforming the code.
Hopefully someone can tell me what the mistake is or suggest another way to make the first code more efficient.
Thanks in advance,
Martin
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Thorsten
am 24 Nov. 2014
Bearbeitet: Thorsten
am 24 Nov. 2014
I see. Than you can use your second code or (only one max, probably faster):
temp = max(0,p_w);
p_w_plus = temp./repmat(sum(temp, 2), 1, size(p_w, 2));
Your first solution with the loops is wrong, as I have explained in my first answer. You can check this using
sum(p_w, 2)
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Thorsten
am 24 Nov. 2014
Bearbeitet: Thorsten
am 24 Nov. 2014
To obtain values of 0 or higher, of course you can simply use
p_w = abs(p_w);
or
p_w = max(0, p_w);
Obviously to try to meet another constrain, maybe that the rows contain only positive values that sum to 1? You can achieve this with your second code. The reason why this "optimized" code gives different results than your loop is that you change p_w within the loop, so nansum(max(0, p_w(j,:)) will give different results depending on i.
So far I am not sure what additional constrains your p_w has to meet. Please expand on that.
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