[英]Interleave rows of two numpy arrays in Python
我想交错两个相同大小的 numpy 数组的行。 我想出了这个解决方案。
# 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)
产出
[[ 1. 1. 1.]
[ 0. 0. 0.]
[ 1. 1. 1.]
[ 0. 0. 0.]
[ 1. 1. 1.]
[ 0. 0. 0.]
[ 1. 1. 1.]
[ 0. 0. 0.]]
是否有更清洁、更快、更好、仅限 numpy 的方式?
这样做可能更清楚一些:
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
我猜它也可能会快一点。
我发现以下使用numpy.hstack()
的方法非常可读:
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 在小型阵列上接受的答案慢一点,但在大型阵列上同样快或更快:
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
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])
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.