生成唯一的numpy数组的列表

Question

I'm trying to make a list of numpy ndarrays, similar to the following:

>>> import numpy as np
>>> a = np.array([1,2,3])
>>> b = 3*[np.copy(a)]
>>> print b
[array([1, 2, 3]), array([1, 2, 3]), array([1, 2, 3])]

But each element of this list is an alias of the original array np.copy(a) , so changing one element of any ndarray changes all of the other corresponding elements, ie:

>>> b[0][0] = 0
>>> print b
[array([0, 2, 3]), array([0, 2, 3]), array([0, 2, 3])]

How can I make each of these arrays independent of each other, so that the above result would be:

[array([0, 2, 3]), array([1, 2, 3]), array([1, 2, 3])]

Answer 1

Doing 3*[np.copy(a)] actually does one copy of a and creates 3 references to this copy, so that you can not change only one because they are the same object. Doing this:

b = [np.copy(a) for i in range(3)]

will create 3 independent copies.

But it seems you should work with b as a 2D array, which you can achieve doing:

b = np.vstack((a for i in range(3)))

Answer 2

The reason why what you were trying didn't work is that

>>> b = 3*[np.copy(a)]

is essentially equivalent to

>>> c = np.copy(a)
>>> b = 3*[c]

In Python, c is not the array, c is, in this case, a reference to an array. 3*[c] just copies that reference three times. You could do,

>>> b = [np.copy(a) for i in xrange(3)]

as sgpc mentions , or you could even do,

>>> b = [np.array([1,2,3]) for i in xrange(3)]