Matlab 效率 - 向量化

Question

我有这个运行的 Matlab 代码，但我想通过删除 for 循环来使其更快，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

我现在将总结在我的工作负载中使用以下提供的答案的一些平均执行时间（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！

Answer 1

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');

Answer 2

试试这些 -

代码 1：

tp1 = squeeze(sum(bsxfun(@times,x,permute(log(m),[3 2 1])),2))
tp2 = squeeze(sum(bsxfun(@times,1-x,permute(log(1-m),[3 2 1])),2))
tp3 = tp1 + tp2
f = bsxfun(@plus,tp3,log(p))
f = bsxfun(@minus,f,max(f,[],2))

代码 2：

p = sum(z)./N;
m = bsxfun(@rdivide,squeeze(sum(bsxfun(@times,x,permute(z,[1 3 2]))))',sum(z)');

代码 2： [方法 2] -

p = sum(z)./N;
sum2_1 = sum(z);
m = zeros(K,D);
for k=1:K  
  m(k,:) = sum(bsxfun(@times,x,z(:,k)))./sum2_1(k);
end

代码 3：（假设您从某处获得 maxfL）

L_new = sum(maxfL' + log(sum(fL,2)))