问题
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
So far I use for-loop of i, j, and n, but I don't want to, at least for-loop of n.
回答1:
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) );
Where matpy can be found here.
Option 2: Native MATLAB
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)
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 1st dimension of A needs to be the 5th, the 2nd needs to be 3rd and the 3rd needs to be 1st (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]);
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]);
(see note regarding sum at the bottom of the post.)
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]));
See also: squeeze, reshape, permute, ipermute, shiftdim.
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)
Running the above we get that the results are indeed equivalent:
>> q55913093
ans =
2.1816e-10
ans =
logical
1
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) `
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
On my system this results in:
>> q55913093_benchmark
ans =
1.3964 0.1864 0.2428
Which means that the 2nd method is preferable (at least for the default input sizes).
来源：https://stackoverflow.com/questions/55913093/element-wise-matrix-multiplication-for-multi-dimensional-array