I would like to create a 3D array in Python (2.7) to use like this:
distance[i][j][k]
And the sizes of the array should be the size of a variable I have. (n*n*n)
I tried using:
distance = [[[]*n]*n]
but that didn't seem to work.
Any ideas? Thanks a lot!
EDIT: I can only use the deafult libraries, and the method of multiplying (ie [[0]*n]*n) wont work because they are linked to the same pointer and I need all of the values to be individual
EDIT2: Already solved by answer below.
You should use a list comprehension:
>>> import pprint
>>> n = 3
>>> distance = [[[0 for k in xrange(n)] for j in xrange(n)] for i in xrange(n)]
>>> pprint.pprint(distance)
[[[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]]]
>>> distance[0][1]
[0, 0, 0]
>>> distance[0][1][2]
0
You could have produced a data structure with a statement that looked like the one you tried, but it would have had side effects since the inner lists are copy-by-reference:
>>> distance=[[[0]*n]*n]*n
>>> pprint.pprint(distance)
[[[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]]]
>>> distance[0][0][0] = 1
>>> pprint.pprint(distance)
[[[1, 0, 0], [1, 0, 0], [1, 0, 0]],
[[1, 0, 0], [1, 0, 0], [1, 0, 0]],
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]]
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)
If you insist on everything initializing as empty, you need an extra set of brackets on the inside ([[]] instead of [], since this is "a list containing 1 empty list to be duplicated" as opposed to "a list containing nothing to duplicate"):
distance=[[[[]]*n]*n]*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
