问题描述
我想在 MATLAB 中实现逐个矩阵乘法，这可以使用 numpy.einsum 在 Python 中如下:
I want to realize component-wise matrix multiplication in MATLAB, which can be done using numpy.einsum in Python as below:
import numpy as np
M = 2
N = 4
I = 2000
J = 300
A = np.random.randn(M, M, I)
B = np.random.randn(M, M, N, J, I)
C = np.random.randn(M, J, I)
# using einsum
D = np.einsum('mki, klnji, lji -> mnji', A, B, C)
# naive for-loop
E = np.zeros(M, N, J, I)
for i in range(I):
for j in range(J):
for n in range(N):
E[:,n,j,i] = B[:,:,i] @ A[:,:,n,j,i] @ C[:,j,i]
print(np.sum(np.abs(D-E))) # expected small enough
到目前为止，我使用 i、j 和 n 的 for 循环，但我不想这样做，至少对于-n 的循环.
So far I use for-loop of i, j, and n, but I don't want to, at least for-loop of n.
推荐答案
选项 1:从 MATLAB 调用 numpy
假设您的系统已设置 根据文档，并且你已经安装了 numpy 包，你可以这样做(在 MATLAB 中):
Option 1: Calling numpy from MATLAB
Assuming your system is set up according to the documentation, and you have the numpy package installed, you could do (in MATLAB):
np = py.importlib.import_module('numpy');
M = 2;
N = 4;
I = 2000;
J = 300;
A = matpy.mat2nparray( randn(M, M, I) );
B = matpy.mat2nparray( randn(M, M, N, J, I) );
C = matpy.mat2nparray( randn(M, J, I) );
D = matpy.nparray2mat( np.einsum('mki, klnji, lji -> mnji', A, B, C) );
在哪里可以找到 matpy 这里.
这里最重要的部分是正确排列排列，因此我们需要跟踪我们的维度.我们将使用以下顺序:
Here the most important part is to get the permutations right, so we need to keep track of our dimensions. We'll be using the following order:
I(1) J(2) K(3) L(4) M(5) N(6)
现在，我将解释如何获得正确的置换顺序(我们以 A 为例): einsum 期望维度顺序为 mki，根据我们的编号是5 3 1.这告诉我们 A 的第 1 维需要是第 5，第 2 需要是 3 并且 3 需要是 1 (简而言之 1->5, 2->3, 3->1).这也意味着无源维度"会被(意味着那些没有原始尺寸成为它们的尺寸；在本例中为 2 4 6)应该是单例的.使用 ipermut 写起来真的很简单:
Now, I'll explain how I got the correct permute order (let's take the example of A): einsum expects the dimension order to be mki, which according to our numbering is 5 3 1. This tells us that the 1 dimension of A needs to be the 5, the 2 needs to be 3 and the 3 needs to be 1 (in short 1->5, 2->3, 3->1). This also means that the "sourceless dimensions" (meaning those that have no original dimensions becoming them; in this case 2 4 6) should be singleton. Using ipermute this is really simple to write:
pA = ipermute(A, [5,3,1,2,4,6]);
在上面的例子中，1->5 表示我们先写 5，其他两个维度也是如此(产生 [5,3,1]).然后我们只需在末尾添加单例 (2,4,6) 即可得到 [5,3,1,2,4,6].最后:
In the above example, 1->5 means we write 5 first, and the same goes for the other two dimensions (yielding [5,3,1]). Then we just add the singletons (2,4,6) at the end to get [5,3,1,2,4,6]. Finally:
A = randn(M, M, I);
B = randn(M, M, N, J, I);
C = randn(M, J, I);
% Reference dim order: I(1) J(2) K(3) L(4) M(5) N(6)
pA = ipermute(A, [5,3,1,2,4,6]); % 1->5, 2->3, 3->1; 2nd, 4th & 6th are singletons
pB = ipermute(B, [3,4,6,2,1,5]); % 1->3, 2->4, 3->6, 4->2, 5->1; 5th is singleton
pC = ipermute(C, [4,2,1,3,5,6]); % 1->4, 2->2, 3->1; 3rd, 5th & 6th are singletons
pD = sum( ...
permute(pA .* pB .* pC, [5,6,2,1,3,4]), ... 1->5, 2->6, 3->2, 4->1; 3rd & 4th are singletons
[5,6]);
(请参阅帖子底部有关 sum 的说明.)
(see note regarding sum at the bottom of the post.)
在 MATLAB 中执行此操作的另一种方法，@AndrasDeak 提到，如下所示:
Another way to do it in MATLAB, as mentioned by @AndrasDeak, is the following:
rD = squeeze(sum(reshape(A, [M, M, 1, 1, 1, I]) .* ...
reshape(B, [1, M, M, N, J, I]) .* ...
... % same as: reshape(B, [1, size(B)]) .* ...
... % same as: shiftdim(B,-1) .* ...
reshape(C, [1, 1, M, 1, J, I]), [2, 3]));
另请参阅:squeeze, reshape, permute, ipermute, shiftdim.
这是一个完整的示例，用于测试这些方法是否等效:
Here's a full example that shows that tests whether these methods are equivalent:
function q55913093
M = 2;
N = 4;
I = 2000;
J = 300;
mA = randn(M, M, I);
mB = randn(M, M, N, J, I);
mC = randn(M, J, I);
%% Option 1 - using numpy:
np = py.importlib.import_module('numpy');
A = matpy.mat2nparray( mA );
B = matpy.mat2nparray( mB );
C = matpy.mat2nparray( mC );
D = matpy.nparray2mat( np.einsum('mki, klnji, lji -> mnji', A, B, C) );
%% Option 2 - native MATLAB:
%%% Reference dim order: I(1) J(2) K(3) L(4) M(5) N(6)
pA = ipermute(mA, [5,3,1,2,4,6]); % 1->5, 2->3, 3->1; 2nd, 4th & 6th are singletons
pB = ipermute(mB, [3,4,6,2,1,5]); % 1->3, 2->4, 3->6, 4->2, 5->1; 5th is singleton
pC = ipermute(mC, [4,2,1,3,5,6]); % 1->4, 2->2, 3->1; 3rd, 5th & 6th are singletons
pD = sum( permute( ...
pA .* pB .* pC, [5,6,2,1,3,4]), ... % 1->5, 2->6, 3->2, 4->1; 3rd & 4th are singletons
[5,6]);
rD = squeeze(sum(reshape(mA, [M, M, 1, 1, 1, I]) .* ...
reshape(mB, [1, M, M, N, J, I]) .* ...
reshape(mC, [1, 1, M, 1, J, I]), [2, 3]));
%% Comparisons:
sum(abs(pD-D), 'all')
isequal(pD,rD)
运行上面我们得到的结果确实是等价的:
Running the above we get that the results are indeed equivalent:
>> q55913093
ans =
2.1816e-10
ans =
logical
1
请注意，这两种调用 sum 的方法是在最近的版本中引入的，因此如果您的 MATLAB 相对较旧，您可能需要替换它们:
Note that these two methods of calling sum were introduced in recent releases, so you might need to replace them if your MATLAB is relatively old:
S = sum(A,'all') % can be replaced by ` sum(A(:)) `
S = sum(A,vecdim) % can be replaced by ` sum( sum(A, dim1), dim2) `
根据评论中的要求，这是比较方法的基准:
As requested in the comments, here's a benchmark comparing the methods:
function t = q55913093_benchmark(M,N,I,J)
if nargin == 0
M = 2;
N = 4;
I = 2000;
J = 300;
end
% Define the arrays in MATLAB
mA = randn(M, M, I);
mB = randn(M, M, N, J, I);
mC = randn(M, J, I);
% Define the arrays in numpy
np = py.importlib.import_module('numpy');
pA = matpy.mat2nparray( mA );
pB = matpy.mat2nparray( mB );
pC = matpy.mat2nparray( mC );
% Test for equivalence
D = cat(5, M1(), M2(), M3());
assert( sum(abs(D(:,:,:,:,1) - D(:,:,:,:,2)), 'all') < 1E-8 );
assert( isequal (D(:,:,:,:,2), D(:,:,:,:,3)));
% Time
t = [ timeit(@M1,1), timeit(@M2,1), timeit(@M3,1)];
function out = M1()
out = matpy.nparray2mat( np.einsum('mki, klnji, lji -> mnji', pA, pB, pC) );
end
function out = M2()
out = permute( ...
sum( ...
ipermute(mA, [5,3,1,2,4,6]) .* ...
ipermute(mB, [3,4,6,2,1,5]) .* ...
ipermute(mC, [4,2,1,3,5,6]), [3,4]...
), [5,6,2,1,3,4]...
);
end
function out = M3()
out = squeeze(sum(reshape(mA, [M, M, 1, 1, 1, I]) .* ...
reshape(mB, [1, M, M, N, J, I]) .* ...
reshape(mC, [1, 1, M, 1, J, I]), [2, 3]));
end
end
在我的系统上，这会导致:
On my system this results in:
>> q55913093_benchmark
ans =
1.3964 0.1864 0.2428
这意味着 2 方法更可取(至少对于默认输入大小而言).
Which means that the 2 method is preferable (at least for the default input sizes).
这篇关于多维数组的元素矩阵乘法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！