Multidimensional array in Python

Question

I have a little Java problem I want to translate to Python. Therefor I need a multidimensional array. In Java it looks like:

double dArray[][][] = new double[x.l         


        

Answer 1

If you restrict yourself to the Python standard library, then a list of lists is the closest construct:

arr = [[1,2],[3,4]]

gives a 2d-like array. The rows can be accessed as arr[i] for i in {0,..,len(arr}, but column access is difficult.

If you are willing to add a library dependency, the NumPy package is what you really want. You can create a fixed-length array from a list of lists using:

import numpy
arr = numpy.array([[1,2],[3,4]])

Column access is the same as for the list-of-lists, but column access is easy: arr[:,i] for i in {0,..,arr.shape[1]} (the number of columns).

In fact NumPy arrays can be n-dimensional.

Empty arrays can be created with

numpy.empty(shape)

where shape is a tuple of size in each dimension; shape=(1,3,2) gives a 3-d array with size 1 in the first dimension, size 3 in the second dimension and 2 in the 3rd dimension.

If you want to store objects in a NumPy array, you can do that as well:

 arr = numpy.empty((1,), dtype=numpy.object)
 arr[0] = 'abc'

For more info on the NumPy project, check out the NumPy homepage.

Answer 2

Another option is to use a dictionary:

>>> from collections import defaultdict
>>> array = defaultdict(int) # replace int with the default-factory you want
>>> array[(0,0)]
0
>>> array[(99,99)]
0

You'll need to keep track of the upper & lower bounds as well.

Answer 3

Take a look at numpy

here's a code snippet for you

import numpy as npy

d = npy.zeros((len(x)+1, len(y)+1, len(x)+len(y)+3))
d[0][0][0] = 0 # although this is unnecessary since zeros initialises to zero
d[i][j][k] = npy.inf

I don't think you need to be implementing a scientific application to justify the use of numpy. It is faster and more flexible and you can store pretty much anything. Given that I think it is probably better to try and justify not using it. There are legitimate reasons, but it adds a great deal and costs very little so it deserves consideration.

P.S. Are your array lengths right? It looks like a pretty peculiar shaped matrix...

Answer 4

I've just stepped into a similar need and coded this:

def nDimensionsMatrix(dims, elem_count, ptr=[]):
    if (dims > 1):
        for i in range(elem_count[dims-1]):
            empty = []
            ptr.append(empty)
            nDimensionsMatrix(dims-1, elem_count, empty)
        return ptr
    elif dims == 1:
        ptr.extend([0 for i in range(elem_count[dims])])
        return ptr

matrix = nDimensionsMatrix(3, (2,2,2))

I'm not looking at speed, only funcionality ;)

I want to create a matrix with N dimensions and initialize with 0 (a elem_count number of elements in each dimension).

Hope its helps someone

Answer 5

Probably not relevant for you but if you are doing serious matrix work see numpy

Answer 6

For numeric data, Numpy Arrays:

>>> matrix1 = array(([0,1],[1,3]))
>>> print matrix1
[[0 1]
[1 3]]

For general data (e.g. strings), you can use a list of lists, list of tuples, ...

matrix2 = [['a','b'], ['x','y']]