正确方式

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

(你试图做些什么,请参看NxNxN。)

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

但这不正确,见“Adaman评论为什么”。

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

多功能阵列使用<代码>numpy.ndarray。

也可使用<<条码>。 如下所示:

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)

解决你的问题有很多办法。

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

例如:

>>> 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 如果你热衷于验证精益的正确价值,即只有自然数字,那么你就可以在要求履行这一职能之前写出一个精彩的职能。

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

由于这些职能相互矛盾,产出如下:

>>> 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)

我只想指出:

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

可缩短时间,以便

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

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