Question

我想交错两个相同大小的numpy数组的行。我提出了这个解决方案。

# A and B are same-shaped arrays
A = numpy.ones((4,3))
B = numpy.zeros_like(A)
C = numpy.array(zip(A[::1], B[::1])).reshape(A.shape[0]*2, A.shape[1])
print C

输出

[[ 1.  1.  1.]
 [ 0.  0.  0.]
 [ 1.  1.  1.]
 [ 0.  0.  0.]
 [ 1.  1.  1.]
 [ 0.  0.  0.]
 [ 1.  1.  1.]
 [ 0.  0.  0.]]

是否有更干净，更快，更好，更无耻的方式？

Answer 1

这可能更清楚：

A = np.ones((4,3))
B = np.zeros_like(A)

C = np.empty((A.shape[0]+B.shape[0],A.shape[1]))

C[::2,:] = A
C[1::2,:] = B

并且它可能也快一点，我猜。

Answer 2

您可以堆叠，转置和重塑：

numpy.dstack((A, B)).transpose(0, 2, 1).reshape(A.shape[0]*2, A.shape[1])

Answer 3

我发现以下方法使用numpy.hstack()非常易读：

import numpy as np

a = np.ones((2,3))
b = np.zeros_like(a)

c = np.hstack([a, b]).reshape(4, 3)

print(c)

输出：

[[ 1.  1.  1.]
 [ 0.  0.  0.]
 [ 1.  1.  1.]
 [ 0.  0.  0.]]

很容易将其概括为相同形状的数组列表：

arrays = [a, b, c,...]

shape = (len(arrays)*a.shape[0], a.shape[1])

interleaved_array = np.hstack(arrays).reshape(shape)

它似乎比@JoshAdel对小阵列的接受答案慢一点，但在大型阵列上同样快或快：

a = np.random.random((3,100))
b = np.random.random((3,100))

%%timeit
...: C = np.empty((a.shape[0]+b.shape[0],a.shape[1]))
...: C[::2,:] = a
...: C[1::2,:] = b
...:

The slowest run took 9.29 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.3 µs per loop

%timeit c = np.hstack([a,b]).reshape(2*a.shape[0], a.shape[1])
The slowest run took 5.06 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 10.1 µs per loop

a = np.random.random((4,1000000))
b = np.random.random((4,1000000))

%%timeit
...: C = np.empty((a.shape[0]+b.shape[0],a.shape[1]))
...: C[::2,:] = a
...: C[1::2,:] = b
...: 

10 loops, best of 3: 23.2 ms per loop

%timeit c = np.hstack([a,b]).reshape(2*a.shape[0], a.shape[1])
10 loops, best of 3: 21.3 ms per loop

在Python中交错两个numpy数组的行

3 个答案: