Initialize empty list with same shape as array

Question

Say I have an array of the shape:

import numpy as np
a = np.zeros(shape=(3, 4, 2))

which looks like:

print a

[[[ 0.  0.]
  [ 0.  0.]
  [ 0.  0.]
  [ 0.  0.]]

 [[ 0.  0.]
  [ 0.  0.]
  [ 0.  0.]
  [ 0.  0.]]

 [[ 0.  0.]
  [ 0.  0.]
  [ 0.  0.]
  [ 0.  0.]]]

How do I create an empty list with the same shape, where every 0. element is replaced by an empty sub-list?

In the specific case shown above, it would look like:

[[[[], []]
  [[], []]
  [[], []]
  [[], []]],
 [[[], []]
  [[], []]
  [[], []]
  [[], []]],
 [[[], []]
  [[], []]
  [[], []]
  [[], []]]]

but I need a way that works in general for arrays of any shape. Is there a built in function to do this?

Answer 1

np.empty(shape=(3, 4, 2, 0)) might be what you are looking for. Or more generally, np.empty(shape=your_shape+(0,)) where your_shape is a tuple like (3, 4, 2).

Now to get the desired list of lists, you can call the tolist method:

np.empty(shape=your_shape+(0,)).tolist()

Otherwhise, you could do a wrapper function that returns nested list comprehensions:

a = [[[[] for j in range(2)] for i in range(4)] for k in range(3)]

and if you want a numpy array:

a = np.array(a)

Such a function could be:

import copy
def empty(shape):
    if len(shape) == 1:
        return [[] for i in range(shape[0])]
    items = shape[0]
    newshape = shape[1:]
    sublist = empty(newshape)
    return [copy.deepcopy(sublist) for i in range(items)]

and you would call it like this:

a = empty([3,4,2])