numpy 用一维数组的值填充 ndarray

Question

I want to fill Ndarray x with values from array b along dimension i without using a for loop. This snippet of code is what I'm currently using but it's not that fast. Is there a better way?

for i in range(len(b)):
    x[...,i,:,:] = b[i]

Edit 1: It's almost what I'm looking for but for higher dimensions it doesn't seem to work. x has a dimension of 8 and it's important that the shape of the Ndarray remains the same. Any more ideas?

import numpy as np

x = np.ones((2,3,4))

b = np.arange(3)

for i in range(len(b)):
    x[:,i,:] = b[i]

x
Out[5]: 
array([[[0., 0., 0., 0.],
        [1., 1., 1., 1.],
        [2., 2., 2., 2.]],

       [[0., 0., 0., 0.],
        [1., 1., 1., 1.],
        [2., 2., 2., 2.]]])

y = np.tile(b,(4,1,2)).T

y
Out[7]: 
array([[[0, 0, 0, 0]],

       [[1, 1, 1, 1]],

       [[2, 2, 2, 2]],

       [[0, 0, 0, 0]],

       [[1, 1, 1, 1]],

       [[2, 2, 2, 2]]])

Edit 2: This seems to do the job

z[...] = b.reshape(1,-1,1)

z
Out[20]: 
array([[[0., 0., 0., 0.],
        [1., 1., 1., 1.],
        [2., 2., 2., 2.]],

       [[0., 0., 0., 0.],
        [1., 1., 1., 1.],
        [2., 2., 2., 2.]]])

Answer 1

Depending on the shape of your destination array you can do something like this

>>> import numpy as np
>>> x = np.ones((4,8))
>>> x
array([[1., 1., 1., 1., 1., 1., 1., 1.],
       [1., 1., 1., 1., 1., 1., 1., 1.],
       [1., 1., 1., 1., 1., 1., 1., 1.],
       [1., 1., 1., 1., 1., 1., 1., 1.]])
>>> b = np.arange(4)
>>> b
array([0, 1, 2, 3])

>>> x[:,1] = b
>>> x
array([[1., 0., 1., 1., 1., 1., 1., 1.],
       [1., 1., 1., 1., 1., 1., 1., 1.],
       [1., 2., 1., 1., 1., 1., 1., 1.],
       [1., 3., 1., 1., 1., 1., 1., 1.]])

In this example we assigned b to column 1 of the 2D array x

If instead you are trying to repeat b a certain number of times you can use np.tile

>>> x = np.tile(b, (8,1)).T
>>> x
array([[0, 0, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 1, 1, 1, 1, 1],
       [2, 2, 2, 2, 2, 2, 2, 2],
       [3, 3, 3, 3, 3, 3, 3, 3]])

Answer 2

There is a faster way. You can reshape b to add new dimensions and get the advantages of numpy broadcasting rules:

x[...,:,:,:] = b.reshape(-1,1,1)

Here I am assuming that b is a vector.

Another equivalent way to create new dimensions is as the following code indicates:

x[...,:,:,:] = b[:, np.newaxis, np.newaxis]