Question

我有以下numpy数组：

X = [[1],
     [2],
     [3],
     [4]]

Y = [[5],
     [6],
     [7],
     [8]]

Z = [[9],
     [10],
     [11],
     [12]]

我想得到以下输出：

H = [[1,5,9],
     [2,6,10],
     [3,7,11]
     [4,8,12]]

有没有办法使用numpy.reshape得到这个结果？

Answer 1

您可以使用np.column_stack -

np.column_stack((X,Y,Z))

或axis=1上的np.concatenate -

np.concatenate((X,Y,Z),axis=1)

或np.hstack -

np.hstack((X,Y,Z))

或axis=0沿着np.stack((X,Y,Z),axis=0).T，然后进行多重调换 -

reshape

重塑应用于数组，而不是堆叠或连接数组。所以，np.reshape在这里没有任何意义。

有人可能会争辩使用np.reshape((X,Y,Z),(3,4)).T为我们提供所需的输出，就像这样 -

np.asarray

但是，在它做堆叠操作的引擎盖下，AFAIK可以用In [453]: np.asarray((X,Y,Z)) Out[453]: array([[[ 1], [ 2], [ 3], [ 4]], [[ 5], [ 6], [ 7], [ 8]], [[ 9], [10], [11], [12]]])转换为数组 -

multi-dim transpose

我们只需要在其上使用3D，为我们提供预期输出的In [454]: np.asarray((X,Y,Z)).T Out[454]: array([[[ 1, 5, 9], [ 2, 6, 10], [ 3, 7, 11], [ 4, 8, 12]]])数组版本 -

{{1}}

Answer 2

这个（更快）的解决方案怎么样？

In [16]: np.array([x.squeeze(), y.squeeze(), z.squeeze()]).T
Out[16]: 
array([[ 1,  5,  9],
       [ 2,  6, 10],
       [ 3,  7, 11],
       [ 4,  8, 12]])

效率（降序）

# proposed (faster) solution
In [17]: %timeit np.array([x.squeeze(), y.squeeze(), z.squeeze()]).T
The slowest run took 7.40 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 7.36 µs per loop

# Other solutions
In [18]: %timeit np.column_stack((x, y, z))
The slowest run took 5.18 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 9.18 µs per loop

In [19]: %timeit np.hstack((x, y, z))
The slowest run took 4.49 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 16 µs per loop

In [20]: %timeit np.reshape((x,y,z),(3,4)).T
10000 loops, best of 3: 21.6 µs per loop

In [20]: %timeit np.c_[x, y, z]
10000 loops, best of 3: 55.9 µs per loop

Answer 3

不要忘记np.c_（我不认为需要np.reshape）：

np.c_[X,Y,Z]
# array([[ 1,  5,  9],
#        [ 2,  6, 10],
#        [ 3,  7, 11],
#        [ 4,  8, 12]])

重塑多个numpy数组

3 个答案: