我有一个3xNxM numpy数组a,我想迭代最后两个轴:a [:,x,y].优雅的方法是:
import numpy as np
a = np.arange(60).reshape((3,4,5))
M = np. array([[1,0,0],
[0,0,0],
[0,0,-1]])
for x in arange(a.shape[1]):
for y in arange(a.shape[2]):
a[:,x,y] = M.dot(a[:,x,y])
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这可以用nditer完成吗?这样做的目的是对每个条目执行矩阵乘法,例如[:,x,y] = M [:,:,x,y] .dot(a [:,x,y]).另一种MATLAB风格的方法是将a(3,N*M)和M重塑为(3,3*N*M)并采用点积,但这往往会占用大量内存.
虽然对形状进行愚弄可能会使你想要完成的事情变得更加清晰,但是在不考虑太多问题的情况下处理这类问题的最简单方法是np.einsum:
In [5]: np.einsum('ij, jkl', M, a)
Out[5]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[ 10, 11, 12, 13, 14],
[ 15, 16, 17, 18, 19]],
[[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0]],
[[-40, -41, -42, -43, -44],
[-45, -46, -47, -48, -49],
[-50, -51, -52, -53, -54],
[-55, -56, -57, -58, -59]]])
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此外,它通常还带有性能奖励:
In [17]: a = np.random.randint(256, size=(3, 1000, 2000))
In [18]: %timeit np.dot(M, a.swapaxes(0,1))
10 loops, best of 3: 116 ms per loop
In [19]: %timeit np.einsum('ij, jkl', M, a)
10 loops, best of 3: 60.7 ms per loop
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编辑 einsum是非常强大的伏都教.你也可以在下面的评论中做OP所要求的内容,如下所示:
>>> a = np.arange(60).reshape((3,4,5))
>>> M = np.array([[1,0,0], [0,0,0], [0,0,-1]])
>>> M = M.reshape((3,3,1,1)).repeat(4,axis=2).repeat(5,axis=3)
>>> np.einsum('ijkl,jkl->ikl', M, b)
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[ 10, 11, 12, 13, 14],
[ 15, 16, 17, 18, 19]],
[[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0]],
[[-40, -41, -42, -43, -44],
[-45, -46, -47, -48, -49],
[-50, -51, -52, -53, -54],
[-55, -56, -57, -58, -59]]])
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