将numpy.array中的每个元素与numpy.array中的每个元素相乘

Given two numpy.arrays a and b,

鉴于两个numpy.arrays a和b,

c = numpy.outer(a, b)

returns an two-dimensional array where c[i, j] == a[i] * b[j]. Now, imagine a having k dimensions.

返回二维数组,其中c [i,j] == a [i] * b [j]。现在,想象一下有k个维度。

Which operation returns an array c of dimension k+1 where c[..., j] == a * b[j]?

哪个操作返回维度为k + 1的数组c,其中c [...,j] == a * b [j]?

Additionally, let b have l dimensions.

另外,让b具有l维度。

Which operation returns an array c of dimension k+1 where c[..., i1, i2, i3] == a * b[i1, i2, i3]?

哪个操作返回维度为k + 1的数组c,其中c [...,i1,i2,i3] == a * b [i1,i2,i3]?

5 个解决方案

#1

The outer method of NumPy ufuncs treats multidimensional input the way you want, so you could do

NumPy ufuncs的外部方法以您想要的方式处理多维输入,因此您可以这样做

numpy.multiply.outer(a, b)

rather than using numpy.outer.

而不是使用numpy.outer。

All solutions suggested here are equally fast; for small arrays, multiply.outer has a slight edge

这里建议的所有解决方案同样快速;对于小数组,multiply.outer有一个轻微的边缘

Code for generating the image:

生成图像的代码:

import numpy
import perfplot


def multiply_outer(data):
    a, b = data
    return numpy.multiply.outer(a, b)


def outer_reshape(data):
    a, b = data
    return numpy.outer(a, b).reshape((a.shape + b.shape))


def tensor_dot(data):
    a, b = data
    return numpy.tensordot(a, b, 0)


perfplot.show(
        setup=lambda n: (numpy.random.rand(n, n), numpy.random.rand(n, n)),
        kernels=[multiply_outer, outer_reshape, tensor_dot],
        n_range=[2**k for k in range(7)],
        logx=True,
        logy=True,
        )

#2

One approach would be using np.outer and then reshape -

一种方法是使用np.outer然后重塑 -

np.outer(a,b).reshape((a.shape + b.shape))

#3

I think np.tensordot also works

我认为np.tensordot也有效

c = np.tensordot(a, b, 0)

inds = np.reshape(np.indices(b.shape), (b.ndim, -1))
for ind in inds.T:
    ind = tuple(ind)
    assert np.allclose(a * b[ind], c[(...,) + ind])
else:
    print('no error')
# no error

#4

I think you're looking for kroneker product

我想你正在寻找kroneker产品

for example

> np.kron(np.eye(2), np.ones((2,2)))

array([[ 1.,  1.,  0.,  0.],
       [ 1.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  1.],
       [ 0.,  0.,  1.,  1.]])

#5

np.einsum is what you are looking for.

np.einsum正是你要找的。

c[..., j] == a * b[j]

c [...,j] == a * b [j]

should be

c = np.einsum('...i,j -> ...ij', a, b)

c = np.einsum('... i,j - > ... ij',a,b)

and c[..., i1, i2, i3] == a * b[i1, i2, i3] should be

和c [...,i1,i2,i3] == a * b [i1,i2,i3]应该是

c = np.einsum('i,...jkl -> ...ijkl', a, b)

c = np.einsum('i,... jkl - > ... ijkl',a,b)

#1