Optimization code. One line takses 86% of working time.

조회 수: 1 (최근 30일)
Jan Kowal
Jan Kowal 2016년 5월 13일
댓글: Ahmet Cecen 2016년 5월 14일
Hello Everyone! I am developing a Clasificator for hand written signs and digits using Logical Regression and Softmax function.
I have a problem with one line in one function because it takes too much time.
function [grad] = logistic_cost_function(xTrain, yTrain, w)
%xTrain = [30134 x 11760]
%xTrain = [30134 x 1]
%w = [36 x 11760]
N = size(xTrain,1);
K = 36;
grad = zeros(size(w));
ARG = w * xTrain';
for n = 1 : N
xRow = xTrain(n,:);
Probability = softmax(ARG(:,n));
label_n = yTrain(n);
for k = 1 : K
IND = label_n == k;
prob_k = Probability(k);
grad(k,:) = grad(k,:) + (IND - prob_k) * xRow;
end
end
grad = -grad/N;
end
It is the time counted on only 2 epochs. My point is to count a few thousands of epchos.
Does anyone have any idea how to improve an algorithm?
Bet Regards
  댓글 수: 3
Walter Roberson
Walter Roberson 2016년 5월 14일
IND is a logical value, a 0 or 1. Are you sure you want to be subtracting prob_k from that? That would be likely to give you a negative value from the subtraction for the cases where the two are not equal.
Ahmet Cecen
Ahmet Cecen 2016년 5월 14일
I believe that might be the point. Don't think of it as a negative probability, think of IND as an indicator function. You either subtract expectation by the probability of hitting the case, or add expectation by the probability of not hitting the case. I may be wrong though as I have no idea what he is actually trying to do.

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답변 (1개)

Ahmet Cecen
Ahmet Cecen 2016년 5월 14일
편집: Ahmet Cecen 2016년 5월 14일
for k = 1 : K
IND = label_n == k;
prob_k = Probability(k);
grad(k,:) = grad(k,:) + (IND - prob_k) * xRow;
end
Here is a way out of this loop (well hopefully, as I can't actually run the code to verify):
IND = zeros(K,1); IND(label_n)=1;
prob_k = Probability;
% (IND - prob_k) IS A COLUMN VECTOR, SO OUTER PRODUCT.
grad = grad + (IND - prob_k)*xRow;
Hmm... turns out no need for bsxfun.
  댓글 수: 4
John D'Errico
John D'Errico 2016년 5월 14일
Ahmet - bsxfun is not needed here because * between a column and row vector is well defined as a matrix product already. That outer product has always worked.
Ahmet Cecen
Ahmet Cecen 2016년 5월 14일
Yeah, in a previous version of my answer I didn't see that operation was an outer product for some reason and used a weird combination of bsxfun and repmat. Sometimes when I focus on trying to understand other people's code the simple stuff manages to elude me.

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