numpy:将两个数组组合成一个矩阵,然后用lambda映射

对于您的特定示例,可以使用 np.meshgrid 和 reduce :

import numpy as np


def mesh(values):
    return np.array(np.meshgrid(*values)).T

X = [1,2,3]
Y = [4,5,6,7]

Z = mesh([X, Y])

result = np.multiply.reduce(Z, axis=2)
print(result)

产量

[[ 4  5  6  7]
 [ 8 10 12 14]
 [12 15 18 21]]

对于自定义函数,可以使用 np.frompyfunc :

def_mult = np.frompyfunc(lambda x, y: x * y, 2, 1)
result = def_mult.reduce(Z, axis=2)
print(result)

产量

[[4 5 6 7]
 [8 10 12 14]
 [12 15 18 21]]

您可以使用列表理解:

X = [1,2,3]
Y = [4,5,6,7]

使用 itertools.product 在获取两个列表的笛卡尔积的列表理解中,保留指定的嵌套列表结构:

Z = [list(product([x],Y)) for x in X]
#[[(1, 4), (1, 5), (1, 6), (1, 7)],
# [(2, 4), (2, 5), (2, 6), (2, 7)],
# [(3, 4), (3, 5), (3, 6), (3, 7)]]

并使用嵌套列表压缩应用保留结构的函数:

[[x*y for x,y in z] for z in Z]
#[[4, 5, 6, 7], [8, 10, 12, 14], [12, 15, 18, 21]]

你可以这样做:

import numpy as np

X = [1, 2, 3]
Y = [4, 5, 6, 7]
Z = np.tensordot(X, Y, axes=0)
print(Z)
# [[ 4  5  6  7]
#  [ 8 10 12 14]
#  [12 15 18 21]]

对于其他操作,您可以执行以下操作:

import numpy as np

X = [1, 2, 3]
Y = [4, 5, 6, 7]
X2d = np.asarray(X)[:, np.newaxis]
Y2d = np.asarray(Y)[np.newaxis, :]

print(X2d * Y2d)
# [[ 4  5  6  7]
#  [ 8 10 12 14]
#  [12 15 18 21]]
print(X2d + Y2d)
# [[ 5  6  7  8]
#  [ 6  7  8  9]
#  [ 7  8  9 10]]
print(X2d ** Y2d)
# [[   1    1    1    1]
#  [  16   32   64  128]
#  [  81  243  729 2187]]

编辑:或者实际上只是使用:

import numpy as np

X = [1, 2, 3]
Y = [4, 5, 6, 7]

print(np.multiply.outer(X, Y))
# [[ 4  5  6  7]
#  [ 8 10 12 14]
#  [12 15 18 21]]
print(np.add.outer(X, Y))
# [[ 5  6  7  8]
#  [ 6  7  8  9]
#  [ 7  8  9 10]]
print(np.power.outer(X, Y))
# [[   1    1    1    1]
#  [  16   32   64  128]
#  [  81  243  729 2187]]