我有运行的Matlab代码,但我想通过删除for循环使其更快,并且基本上只使用矩阵进行相同的计算(我的数据集非常大,所以我需要这种优化):
Matrices dimensions: x(N,D), m(K,D), p(K), fL(N,K), maxf(N), maxfL(N), z(N,K).
Also p, maxf, maxfL are row vectors
代码1:
f = zeros(N,K);
maxf = zeros(1,N);
for n=1:N
for k=1:K
% here i had a loop for d dimension but made it more efficient like this:
f2 = x(n,:) * log(m(k,:))' + (1 - x(n,:)) * log(1 - m(k,:))';
f(n,k) = log(p(k)) + f2;
end
maxf(n) = max(f(n,:));
f(n,:) = f(n,:) - maxf(n);
end
代码2:
for k=1:K
sum2 = sum(z(:,k));
p(k)= sum2/N;
for d=1:D
% here i had a n loop for sum1 and made it like this:
sum1 = z(:,k)' * x(:,d);
m(k,d) = sum1/sum2;
end
end
代码3:
L_new = 0;
for n=1:N
suma = sum(fL(n,:));
L_new = L_new + maxfL(n) + log(suma);
end
现在,我将在工作量中使用以下提供的答案来总结一些平均执行时间(结果以秒为单位)(N = 1000,K = 2,D = 784):
CodeNumber Execution Time
1 3.98(for_loops), 1.01(Divakar), 0.5(Nishant)
2 0.2(for_loops), 0.40-0.42(Divakar-2 approaches), 0.13(Nishant)
3 0.03(for_loops), 0.0026(Divakar), 0.0024(Nishant)
感谢您的回答!
对于代码1:
f2 = x*log(m)' + (1-x)*(log(1-m))' ;
f = f2 + ones(N,1)*log(p);
maxf = max(f');
f = f - maxf'*ones(1,K);
对于代码2:
sum2 = sum(z);
sum1 = z'*x;
temp = sum2'*ones(1,D);
m = sum1./temp;
p = sum2/N;
对于代码3:
L_new = sum(log(sum(fL'))) + sum(maxfL');
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