Create 3D array using Python

numpy.arrays are designed just for this case:

 numpy.zeros((i,j,k))

will give you an array of dimensions ijk, filled with zeroes.

depending what you need it for, numpy may be the right library for your needs.

The right way would be

[[[0 for _ in range(n)] for _ in range(n)] for _ in range(n)]

(What you're trying to do should be written like (for NxNxN)

[[[0]*n]*n]*n

but that is not correct, see @Adaman comment why).

d3 = [[[0 for col in range(4)]for row in range(4)] for x in range(6)]

d3[1][2][1]  = 144

d3[4][3][0]  = 3.12

for x in range(len(d3)):
    print d3[x]



[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 144, 0, 0], [0, 0, 0, 0]]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [3.12, 0, 0, 0]]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

"""
Create 3D array for given dimensions - (x, y, z)

@author: Naimish Agarwal
"""


def three_d_array(value, *dim):
    """
    Create 3D-array
    :param dim: a tuple of dimensions - (x, y, z)
    :param value: value with which 3D-array is to be filled
    :return: 3D-array
    """

    return [[[value for _ in xrange(dim[2])] for _ in xrange(dim[1])] for _ in xrange(dim[0])]

if __name__ == "__main__":
    array = three_d_array(False, *(2, 3, 1))
    x = len(array)
    y = len(array[0])
    z = len(array[0][0])
    print x, y, z

    array[0][0][0] = True
    array[1][1][0] = True

    print array

Prefer to use numpy.ndarray for multi-dimensional arrays.

You can also use a nested for loop like shown below

n = 3
arr = []
for x in range(n):
    arr.append([])
    for y in range(n):
        arr[x].append([])
        for z in range(n):
            arr[x][y].append(0)
print(arr)

I just want notice that

distance = [[[0 for k in range(n)] for j in range(n)] for i in range(n)]

can be shortened to

distance = [[[0] * n for j in range(n)] for i in range(n)]

There are many ways to address your problem.

1. First one as accepted answer by @robert. Here is the generalised solution for it:

def multi_dimensional_list(value, *args):
  #args dimensions as many you like. EG: [*args = 4,3,2 => x=4, y=3, z=2]
  #value can only be of immutable type. So, don't pass a list here. Acceptable value = 0, -1, 'X', etc.
  if len(args) > 1:
    return [ multi_dimensional_list(value, *args[1:]) for col in range(args[0])]
  elif len(args) == 1: #base case of recursion
    return [ value for col in range(args[0])]
  else: #edge case when no values of dimensions is specified.
    return None

Eg:

>>> multi_dimensional_list(-1, 3, 4)  #2D list
[[-1, -1, -1, -1], [-1, -1, -1, -1], [-1, -1, -1, -1]]
>>> multi_dimensional_list(-1, 4, 3, 2)  #3D list
[[[-1, -1], [-1, -1], [-1, -1]], [[-1, -1], [-1, -1], [-1, -1]], [[-1, -1], [-1, -1], [-1, -1]], [[-1, -1], [-1, -1], [-1, -1]]]
>>> multi_dimensional_list(-1, 2, 3, 2, 2 )  #4D list
[[[[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]]], [[[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]]]]

P.S If you are keen to do validation for correct values for args i.e. only natural numbers, then you can write a wrapper function before calling this function.

2. Secondly, any multidimensional dimensional array can be written as single dimension array. This means you don't need a multidimensional array. Here are the function for indexes conversion:

def convert_single_to_multi(value, max_dim):
  dim_count = len(max_dim)
  values = [0]*dim_count
  for i in range(dim_count-1, -1, -1): #reverse iteration
    values[i] = value%max_dim[i]
    value /= max_dim[i]
  return values


def convert_multi_to_single(values, max_dim):
  dim_count = len(max_dim)
  value = 0
  length_of_dimension = 1
  for i in range(dim_count-1, -1, -1): #reverse iteration
    value += values[i]*length_of_dimension
    length_of_dimension *= max_dim[i]
  return value

Since, these functions are inverse of each other, here is the output:

>>> convert_single_to_multi(convert_multi_to_single([1,4,6,7],[23,45,32,14]),[23,45,32,14])
[1, 4, 6, 7]
>>> convert_multi_to_single(convert_single_to_multi(21343,[23,45,32,14]),[23,45,32,14])
21343

3. If you are concerned about performance issues then you can use some libraries like pandas, numpy, etc.

n1=np.arange(90).reshape((3,3,-1))
print(n1)
print(n1.shape)

def n_arr(n, default=0, size=1):
    if n is 0:
        return default

    return [n_arr(n-1, default, size) for _ in range(size)]

arr = n_arr(3, 42, 3)
assert arr[2][2][2], 42