Create 3D array using Python

Question

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

Solution 1

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]]]

Solution 2

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.

Solution 3

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).

Solution 4

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]]

Solution 5

"""
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.

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Comments

[Amadan]

Not good. It will contain references to same array. Try this: a = [[0] * 3] * 3; a[0][0] = 1; print a

[Bruno Kim]

Don't do that, or all of them will point to the same reference! Just try distance[1][2][0].append(1)

[Laís Minchillo]

Oh, I thought this was one of the deafult libraries. I can't use anything other than that.

[mata]

no, unfortunately it's an external library. but usually extremely well suited if you need to process (large) arrays of numeric data. Specially if speed is an issue.

[Laís Minchillo]

Yes, my problem with that is if I change one of them, it will change all of them too. I need them to be separate elements.

[csteel]

For Python3, you can change xrange to range.

[gocen]

I define an array like this: new_array= numpy.zeros((6340,200,200)) but it tooks a space of 1.9GB at memory. Is it normal?

[Filip Franik]

@gocen Yup sounds about right. Default value returned by numpy zeroes is numpy.float64 and 6340*200*200*64bits = 2.0288 gigabytes. You can provide an extra argument dtype to zeros to change the type it returns and save some RAM this way.

[Mihaela Grigore]

I used the 'copy-by-reference' method without knowing this is what actually happened on the back-end. Therefore, I ended up having the 'side-effects' (all fields in my variable ended up having the same value) and did not know why. Thank you for the explanation.