与自己交错一个numpy数组

Question

我有一个N x 2维numpy数组.我想做一个(2*N)x 2,其中每列重复.我很好奇是否有比我下面写的更有效的方法来完成这项任务.

>>> a = np.array([[1,2,3,4],
                  [2,4,6,8]])
>>> b = np.array(zip(a.T,a.T))
>>> b.shape = (2*len(a[0]), 2)
>>> b.T
array([[1, 1, 2, 2, 3, 3, 4, 4],
       [2, 2, 4, 4, 6, 6, 8, 8]])

上面的代码是缓慢的标准,很可能是因为zip.有没有numpy我可以替换的功能zip？还是一个更好的方法来完成这个？

Answer 1

你可以使用repeat:

import numpy as np

def slow(a):
    b = np.array(zip(a.T,a.T))
    b.shape = (2*len(a[0]), 2)
    return b.T

def fast(a):
    return a.repeat(2).reshape(2, 2*len(a[0]))

def faster(a):
    # compliments of WW
    return a.repeat(2, axis=1)

给

In [42]: a = np.array([[1,2,3,4],[2,4,6,8]])

In [43]: timeit slow(a)
10000 loops, best of 3: 59.4 us per loop

In [44]: timeit fast(a)
100000 loops, best of 3: 4.94 us per loop

In [45]: a = np.arange(100).reshape(2, 50)

In [46]: timeit slow(a)
1000 loops, best of 3: 489 us per loop

In [47]: timeit fast(a)
100000 loops, best of 3: 6.7 us per loop

[更新]:

In [101]: timeit faster(a)
100000 loops, best of 3: 4.4 us per loop