多维数组的逐元素矩阵乘法

时间:2019-04-30 02:14:37

标签: matlab multidimensional-array sum elementwise-operations numpy-einsum

我想在MATLAB中实现按组件矩阵乘法,可以在Python中使用numpy.einsum来完成,如下所示:

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ijn的for循环,但我至少不想使用n的for循环。

1 个答案:

答案 0 :(得分:8)

选项1:从MATLAB调用numpy

假设您的系统设置为according to the documentation,并且已安装了numpy软件包,则可以执行以下操作(在MATLAB中):

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) );

可以在here找到matpy的地方。

选项2:本机MATLAB

最重要的部分是正确排列,因此我们需要跟踪尺寸。我们将使用以下顺序:

I(1) J(2) K(3) L(4) M(5) N(6)

现在,我将说明如何获得正确的置换顺序(以A为例):einsum期望尺寸顺序为mki,根据我们编号为5 3 1。这告诉我们A的第1 st 维必须是第5 ,第2 nd 必须是3 < sup> rd 和3 rd 必须是1 st (简称1->5, 2->3, 3->1)。这也意味着“无源尺寸”(意味着没有原始尺寸的尺寸;在这种情况下为2 4 6)应为单件。使用ipermute确实很容易编写:

pA = ipermute(A, [5,3,1,2,4,6]);

在上面的示例中,1->5意味着我们首先写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:

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的注释。)

在MATLAB中as mentioned by @AndrasDeak的另一种实现方法如下:

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]));

另请参见:squeezereshapepermuteipermuteshiftdim


下面是一个完整的示例,该示例演示测试这些方法是否等效:

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)

运行上述操作,我们得出的结果确实是等效的:

>> q55913093
ans =
   2.1816e-10 
ans =
  logical
   1

请注意,这两种调用sum的方法是在最近的发行版中引入的,因此如果您的MATLAB相对较旧,则可能需要替换它们:

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:

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,结果为:

>> q55913093_benchmark
ans =
    1.3964    0.1864    0.2428

这意味着首选2 nd 方法(至少对于默认输入大小而言)。