How to create or fill an numpy array with another array?

问题

How to create an numpy array with shape [2, 2, 3], where the elements at axis 2 is another array, for example [1, 2, 3]?

So I would like to do something like this invalid code:

a = np.arange(1, 4)
b = np.full((3, 3), a)

Resulting in an array like:

[[[ 1.  2.  3.]
  [ 1.  2.  3.]]
 [[ 1.  2.  3.]
  [ 1.  2.  3.]]]

Could of course make the loop for filling like, but thought there may be a shortcut:

for y in range(b.shape[0]):
    for x in range(b.shape[1]):
        b[y, x, :] = a

回答1:

There are multiple ways to achieve this. One is to use np.full in np.full((2,2,3), a) as pointed out by Divakar in the comments. Alternatively, you can use np.tile for this, which allows you to construct an array by repeating an input array a given number of times. To construct your example you could do:

import numpy as np

np.tile(np.arange(1, 4), [2, 2, 1])

回答2:

If your numpy version is >= 1.10 you can use broadcast_to

a = np.arange(1,4)
a.shape = (1,1,3)
b = np.broadcast_to(a,(2,2,3))

This produces a view rather than copying so will be quicker for large arrays. EDIT this looks to be the result you're asking for with your demo.

回答3:

Based on Divakar comment, an answer can also be:

import numpy as np
np.full([2, 2, 3], np.arange(1, 4))

Yet another possibility is:

import numpy as np
b = np.empty([2, 2, 3])
b[:] = np.arange(1, 4)

回答4:

Also using np.concatenate or it's wrapper np.vstack

In [26]: a = np.arange(1,4)

In [27]: np.vstack([a[np.newaxis, :]]*4).reshape(2,2, 3)
Out[27]: 
array([[[1, 2, 3],
        [1, 2, 3]],

       [[1, 2, 3],
        [1, 2, 3]]])

In [28]: np.concatenate([a[np.newaxis, :]]*4, axis=0).reshape(2,2, 3)
Out[28]: 
array([[[1, 2, 3],
        [1, 2, 3]],

       [[1, 2, 3],
        [1, 2, 3]]])

来源：https://stackoverflow.com/questions/43537439/how-to-create-or-fill-an-numpy-array-with-another-array

标签

python

arrays

numpy

vectorization

numpy-ndarray