Matlab 三重加权和
我试图对某个加权和进行矢量化,但不知道如何实现。我在下面创建了一个简单的最小工作示例。我猜解决方案涉及bsxfun或Reforme和kronecker产品,但我仍然没有设法让它工作Matlab 三重加权和,matlab,vectorization,weighted-average,Matlab,Vectorization,Weighted Average,我试图对某个加权和进行矢量化,但不知道如何实现。我在下面创建了一个简单的最小工作示例。我猜解决方案涉及bsxfun或Reforme和kronecker产品,但我仍然没有设法让它工作 rng(1); N = 200; T1 = 5; T2 = 7; T3 = 10; A = rand(N,T1,T2,T3); w1 = rand(T1,1); w2 = rand(T2,1); w3 = rand(T3,1); B = zeros(N,1); for i = 1:N for j1=1:T1
rng(1);
N = 200;
T1 = 5;
T2 = 7;
T3 = 10;
A = rand(N,T1,T2,T3);
w1 = rand(T1,1);
w2 = rand(T2,1);
w3 = rand(T3,1);
B = zeros(N,1);
for i = 1:N
for j1=1:T1
for j2=1:T2
for j3=1:T3
B(i) = B(i) + w1(j1) * w2(j2) * w3(j3) * A(i,j1,j2,j3);
end
end
end
end
A = B;
对于二维情况,有一个聪明的答案。您可以使用额外的乘法修改上一个答案的
w1*w2'
网格,然后再乘以w3
。然后,您可以再次使用矩阵乘法与a
的“展平”版本进行乘法
W = reshape(w1 * w2.', [], 1) * w3.';
B = reshape(A, size(A, 1), []) * W(:);
您可以将权重的创建封装到它自己的函数中,并将其概括为N
weights。由于这使用递归,N
被限制为当前的递归限制(默认为500)
并将其用于:
W = createWeights(w1, w2, w3);
B = reshape(A, size(A, 1), []) * W(:);
更新
使用@CKT的一部分非常好的建议来使用kron
,我们可以稍微修改createWeights
function W = createWeights(W, varargin)
if numel(varargin) > 0
W = createWeights(kron(varargin{1}, W), varargin{2:end});
end
end
这就是背后的逻辑:
ww1 = repmat (permute (w1, [4, 1, 2, 3]), [N, 1, T2, T3]);
ww2 = repmat (permute (w2, [3, 4, 1, 2]), [N, T1, 1, T3]);
ww3 = repmat (permute (w3, [2, 3, 4, 1]), [N, T1, T2, 1 ]);
B = ww1 .* ww2 .* ww3 .* A;
B = sum (B(:,:), 2)
通过首先在适当的维度中创建w1
、w2
和w3
,可以避免permute
。您也可以使用bsxfun
而不是repmat
,以获得更高的性能,我只是在这里展示逻辑,而且repmat
更容易理解
编辑:任意输入维度的通用版本:
Dims = {N, T1, T2, T3}; % add T4, T5, T6, etc as appropriate
Params = cell (1, length (Dims));
Params{1} = rand (Dims{:});
for n = 2 : length (Dims)
DimSubscripts = ones (1, length (Dims)); DimSubscripts(n) = Dims{n};
RepSubscripts = [Dims{:}]; RepSubscripts(n) = 1;
Params{n} = repmat (rand (DimSubscripts), RepSubscripts);
end
B = times (Params{:});
B = sum (B(:,:), 2)
同样,除非你做了一些函数来构造Kronecker积向量,否则你不能把它推广到N-D,但是呢
A = reshape(A, N, []) * kron(w3, kron(w2, w1));
如果我们要走有功能的路线,并且倾向于表现在优雅/简洁上,那么考虑一下:
function B = weightReduce(A, varargin)
B = A;
for i = length(varargin):-1:1
N = length(varargin{i});
B = reshape(B, [], N) * varargin{i};
end
end
这是我看到的性能比较:
tic;
for i = 1:10000
W = createWeights(w1,w2,w3);
B = reshape(A, size(A,1), [])*W(:);
end
toc
Elapsed time is 0.920821 seconds.
tic;
for i = 1:10000
B2 = weightReduce(A, w1, w2, w3);
end
toc
Elapsed time is 0.484470 seconds.
你需要概括一下吗?因为如果是这样的话,我会把你的N,T1,T2,T3换成一个数组,实际上我只是想要三维的情况。但是概括可能对其他人有用:)下面的概括:)我想你可能对基准感兴趣。对于以倍频程执行的1000次迭代,循环版本需要20分钟,我的版本使用repmat需要1.66s,不使用repmat需要0.92s,Suever的版本需要0.77s.)@太棒了!非常有用!
tic;
for i = 1:10000
W = createWeights(w1,w2,w3);
B = reshape(A, size(A,1), [])*W(:);
end
toc
Elapsed time is 0.920821 seconds.
tic;
for i = 1:10000
B2 = weightReduce(A, w1, w2, w3);
end
toc
Elapsed time is 0.484470 seconds.