在Python中插入两个numpy数组的行

I wanted to interleave the rows of two numpy arrays of the same size. I came up with this solution.

我想将两个相同大小的numpy数组的行交叉。我想到了这个解决方案。

# 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

Outputs

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.]]

Is there a cleaner, faster, better, numpy-only way?

是否有一个更干净、更快、更好、更简单的方式?

3 个解决方案

#1

It is maybe a bit clearer to do:

也许这样做更清楚一些:

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

and it's probably a bit faster as well, I'm guessing.

我猜它可能也会快一点。

#2

You can stack, transpose, and reshape:

你可以叠加，转置，和重塑:

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

#3

I find the following approach using numpy.hstack() quite readable:

我发现使用numpi .hstack()的以下方法非常具有可读性:

import numpy as np

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

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

print(c)

Output:

输出:

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

It is easy to generalize this to a list of arrays of the same shape:

很容易将其归纳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)

It seems to be a bit slower than the accepted answer of @JoshAdel on small arrays but equally fast or faster on large arrays:

它似乎比@JoshAdel在小数组上的回答慢一点，但是在大数组上同样快或快:

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

#1