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. (nnn)
I tried using:
distance = [[[]*n]*n]
but that didn't seem to work.
I can only use the default libraries, and the method of multiplying (i.e.,[[0]*n]*n
) wont work because they are linked to the same pointer and I need all of the values to be individual
This question is related to
python
arrays
python-2.7
multidimensional-array
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)
"""
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.
numpy.array
s 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.
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
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).
There are many ways to address your problem.
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.
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
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]]
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
Source: Stackoverflow.com