Question

在假设下,我想做一个“普通”矩阵,像这样重复:

C=A*B

地点

A是“2D-kind”矩阵,但每个矩阵元件都形成(1 5)

以及

B是“1D-kind”病媒,但每个病媒元素已形成(20,5)

结果

C应为“1D”病媒,但每一病媒元素再次形成(20,5个)。

I tried to produce the elements C1 以及 C2 of C manually:

>>> A.shape
(2, 2, 1, 5)
>>> B.shape
(2, 20, 5)
>>> C0 = A[0,0]*B[0]+A[0,1]*B[1]
>>> C0.shape
(20, 5)
>>> C1 = A[1,0]*B[0]+A[1,1]*B[1]
>>> C1.shape
(20, 5)
>>>

Broadcasting (1,5) from A with (20,5) of B works as expected.

然而,我无法发现,这怎么可以像一个矩阵表那样写成:

C = np.matmul(A, B)

当然,这并不奏效,因为假设能够知道我想要总结的指数。但我认为,必须存在一些简单的“静态”解决办法。

Answer 1

您可使用einsum:

C = np.einsum( ij...,j...->i... , A, B)

This is a contraction of the second dimension of A with the first dimension of B. The remaining dimensions are carried along for the ride.

An example in an ipython session:

In [114]: rng = np.random.default_rng(8779780912870439733)

In [115]: A = rng.integers(0, 5, size=(2, 2, 1, 5))

In [116]: B = rng.integers(0, 5, size=(2, 20, 5))

In [117]: C = np.einsum( ij...,j...->i... , A, B)

In [118]: C.shape
Out[118]: (2, 20, 5)

In [119]: C0 = A[0,0]*B[0]+A[0,1]*B[1]

In [120]: C1 = A[1,0]*B[0]+A[1,1]*B[1]

In [121]: np.all(C[0] == C0)
Out[121]: True

In [122]: np.all(C[1] == C1)
Out[122]: True

Answer 2

You can directly multiply A and B and sum the second dimension:

C = (A*B).sum(1)

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