blocks|key|1945383|text|在您的例子中，一次计算总和是很简单的。基本上，您创建的数组中的元素分别是vm和va的适当乘积和差(使用bsxfun)，然后是各行的逐个元素乘法和求和。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1945384|pcal+=+sum(bsxfun(@times,vm,vm')+.*+(...
++++yr.*cos(bsxfun(@minus,va,va'))+%2B+...
++++yi.*sin(bsxfun(@minus,va,va'))),2);|code-block|syntax|javascript|1945385|请注意，在矢量化中，您倾向于在内存与CPU周期之间进行权衡。如果您没有足够的RAM，您可能会结束分页，这将使矢量化的解决方案变得缓慢。|1945386|entityMap^0|10|2|13|2|1F|6|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

In your case, it's straightforward to calculate the sum in one go. Basically, you create arrays where the elements are the appropriate products and differences of <code>vm</code> and <code>va</code>, respectively (using <code>bsxfun</code>), followed by element-wise multiplication and summation across the rows.

<pre><code>pcal = sum(bsxfun(@times,vm,vm') .* (...
 yr.*cos(bsxfun(@minus,va,va')) + ...
 yi.*sin(bsxfun(@minus,va,va'))),2);
</code></pre>

Note that in vecorization, you tend to trade memory vs CPU cycles. If you do not have enough RAM, you may end up paging, which would slow the vectorized solution to a crawl.

blocks|key|1945388|text|最快的选择是@Joans+answer。然而，如果你遇到内存问题，这里有一个半矢量化的选项(只有一个循环)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1945389|pcal+=+zeros(N,1);+%25+N+is+the+10000+in+your+example
for+m+=+1:+N
++++va_va+=+va(m)-va(1:N);
++++pcal(m)+=+sum(vm(m)*vm(1:N).*(yr(m,1:N).'.*cos(va_va)%2Byi(m,1:N).'.*sin(va_va)));
end|code-block|syntax|javascript|1945390|下面是该方法的基准测试，以及您的方法和@Joans方法，以及另一个使用ndgrid的方法|offset|length|style|CODE|1945391|function+sum_time
N+=+10000;
vm+=+rand(N,1);
va+=+rand(N,1);
yr+=+rand(N);
yi+=+rand(N);

loop_time+=+timeit(@()+loop(N,vm,va,yr,yi))
loop2_time+=+timeit(@()+loop2(N,vm,va,yr,yi))
bsx_time+=+timeit(@()+bsx(vm,va,yr,yi))
ndg_time+=+timeit(@()+ndg(N,vm,va,yr,yi))
end

function+pcal+=+loop(N,vm,va,yr,yi)
pcal+=+zeros(N,1);
for+m+=+1:+N
++++psum+=+0;
++++for+n+=+1:+N
++++++++psum+=+psum+%2B+vm(m)*vm(n)*(yr(m,n)*cos(va(m)-va(n))+%2B...
++++++++++++yi(m,n)*sin(va(m)-va(n)));
++++end
++++pcal(m)+=+psum;
end
end

function+pcal+=+loop2(N,vm,va,yr,yi)
pcal+=+zeros(N,1);
for+m+=+1:+N
++++va_va+=+va(m)-va(1:N);+%25+to+avoid+calculating+twice
++++pcal(m)+=+sum(vm(m)*vm(1:N).*(yr(m,1:N).'.*cos(va_va)%2Byi(m,1:N).'.*sin(va_va)));
end
end

function+pcal+=+bsx(vm,va,yr,yi)
pcal+=+sum(bsxfun(@times,vm,vm')+.*+(...
++++yr.*cos(bsxfun(@minus,va,va'))+%2B+...
++++yi.*sin(bsxfun(@minus,va,va'))),2);
end

function+pcal+=+ndg(N,vm,va,yr,yi)
[n,m]+=+ndgrid((1:N).',1:N);
yr_t+=+yr.';
yi_t+=+yi.';
va_va+=+va(m(:))-va(n(:));
vmt+=+vm(m(:)).*vm(n(:));
psum+=+vmt.*(yr_t(1:N%5E2).'.*cos(va_va)%2Byi_t(1:N%5E2).'.*sin(va_va));
pcal+=+sum(reshape(psum,N,N)).';
end|1945392|以及结果(针对N+=+10000)：|1945393|loop_time+=
+++++++7.0296
loop2_time+=
+++++++3.3722
bsx_time+=
+++++++1.2716
ndg_time+=
+++++++6.3568|1945394|因此，减少一次循环可节省约50%25的时间。|1945395|entityMap^0|0|0|Z|6|0|0|7|9|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$I|13|J|14|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

The fastest option will be @Joans answer. However, in case you run into memory problems, here is a semi-vectorized option (only one loop):

<pre><code>pcal = zeros(N,1); % N is the 10000 in your example
for m = 1: N
 va_va = va(m)-va(1:N);
 pcal(m) = sum(vm(m)*vm(1:N).*(yr(m,1:N).'.*cos(va_va)+yi(m,1:N).'.*sin(va_va)));
end
</code></pre>

And here is a benchmarking of this method along with yours and @Joans one, and another method using <code>ndgrid</code>:

<pre><code>function sum_time
N = 10000;
vm = rand(N,1);
va = rand(N,1);
yr = rand(N);
yi = rand(N);

loop_time = timeit(@() loop(N,vm,va,yr,yi))
loop2_time = timeit(@() loop2(N,vm,va,yr,yi))
bsx_time = timeit(@() bsx(vm,va,yr,yi))
ndg_time = timeit(@() ndg(N,vm,va,yr,yi))
end

function pcal = loop(N,vm,va,yr,yi)
pcal = zeros(N,1);
for m = 1: N
 psum = 0;
 for n = 1: N
 psum = psum + vm(m)*vm(n)*(yr(m,n)*cos(va(m)-va(n)) +...
 yi(m,n)*sin(va(m)-va(n)));
 end
 pcal(m) = psum;
end
end

function pcal = loop2(N,vm,va,yr,yi)
pcal = zeros(N,1);
for m = 1: N
 va_va = va(m)-va(1:N); % to avoid calculating twice
 pcal(m) = sum(vm(m)*vm(1:N).*(yr(m,1:N).'.*cos(va_va)+yi(m,1:N).'.*sin(va_va)));
end
end

function pcal = bsx(vm,va,yr,yi)
pcal = sum(bsxfun(@times,vm,vm') .* (...
 yr.*cos(bsxfun(@minus,va,va')) + ...
 yi.*sin(bsxfun(@minus,va,va'))),2);
end

function pcal = ndg(N,vm,va,yr,yi)
[n,m] = ndgrid((1:N).',1:N);
yr_t = yr.';
yi_t = yi.';
va_va = va(m(:))-va(n(:));
vmt = vm(m(:)).*vm(n(:));
psum = vmt.*(yr_t(1:N^2).'.*cos(va_va)+yi_t(1:N^2).'.*sin(va_va));
pcal = sum(reshape(psum,N,N)).';
end
</code></pre>

and the results (for <code>N = 10000</code>):

<pre><code>loop_time =
 7.0296
loop2_time =
 3.3722
bsx_time =
 1.2716
ndg_time =
 6.3568
</code></pre>

so taking one loop out saving ~50% of the time.

blocks|key|1965563|text|你可以根据这些trigonometric+identities重新表述你的方程。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1965564|sin(a-b)+=+sin+a+cos+b+-+cos+a+sin+b;
cos(a-b)+=+cos+a+cos+b+%2B+sin+a+sin+b;|code-block|syntax|javascript|1965565|因此，预先计算正弦和余弦，并在loop或bsxfun.中使用它们。这是循环版本：|1965566|yr+=+rand(10000);
yi+=+rand(10000);
va+=+rand(1,10000);
vm+=+rand(1,10000);
sin_va+=+sin(va);
cos_va+=+cos(va);
for+i+=+1:+10000
++++pcal(i)+=++sum(vm(i)*vm.*(yr(i,:).*(cos_va(i)+*+cos_va+%2B+sin_va(i)+*+sin_va)+%2B+yi(i,:).*(sin_va(i)+*+cos_va+-+cos_va(i)+*+sin_va)));
end|1965567|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/List_of_trigonometric_identities^0|7|O|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

you can reformulate your equation according to theses <a href="https://en.wikipedia.org/wiki/List_of_trigonometric_identities" rel="nofollow">trigonometric identities</a>:

<pre><code>sin(a-b) = sin a cos b - cos a sin b;
cos(a-b) = cos a cos b + sin a sin b;
</code></pre>

so precompute sine and cosine and use them in loop or bsxfun. this is loop version:

<pre><code>yr = rand(10000);
yi = rand(10000);
va = rand(1,10000);
vm = rand(1,10000);
sin_va = sin(va);
cos_va = cos(va);
for i = 1: 10000
 pcal(i) = sum(vm(i)*vm.*(yr(i,:).*(cos_va(i) * cos_va + sin_va(i) * sin_va) + yi(i,:).*(sin_va(i) * cos_va - cos_va(i) * sin_va)));
end
</code></pre>

I have two for loops for very large arrays (10k x 10k) or more. Obviously this program's part is a huge bottleneck and very time-consuming task. 

There are 4 arrays: <code>vm(10000,1)</code>, <code>va(10000,1)</code>, <code>yr(10000,10000)</code>, and <code>yi(10000,10000)</code>

<pre><code>for i = 1: 10000
 psum = 0;
 for j = 1: 10000
 psum = psum + vm(i)*vm(j)*(yr(i,j)*cos(va(i)-va(j)) + yi(i,j)*sin(va(i)-va(j)));
 end
pcal(i) = psum;
end
</code></pre>

vectorizing summation with double for loops in matlab

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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对于非常大的数组(10k X 10k)或更多，我有两个for循环。显然，这个程序的一部分是一个巨大的瓶颈和非常耗时的任务。共有4个阵列：vm(10000,1)、va(10000,1)、yr(10000,10000)和yi(10000,10000)for i = 1: 10000    psum = 0;    for ...

问matlab中双for循环的矢量化求和
EN

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问matlab中双for循环的矢量化求和EN