Find the dimensions of a multidimensional Python array

Question

In Python, is it possible to write a function that returns the dimensions of a multidimensional array (given the assumption that the array's dimensions are not jagged)?

For example, the dimensions of [[2,3], [4,2], [3,2]] would be [3, 2], while the dimensions of [[[3,2], [4,5]],[[3,4],[2,3]]] would be [2,2,2].

Does Python have any built-in functions that will return all of the dimensions of a multidimensional array, or will I need to implement this function myself?

Atc

isn't the second example [2, 2, 2] ? – Suppression

Out of curiosity, what would it do on [[1,2],[3,4,5]]? Or [[1,2],[3,4],[5,[6,7]]]? – Scopas

@Scopas I'm working with the assumption that the input array will not be jagged. I'd want to check the regularity of the array's dimensions first, and then throw an error if the array is jagged. – Atc

Unclear why this is marked as a duplicate: the "duplicate" is about numpy arrays and this is about Python lists. The answers there do not help the OP. – Ponder

Answer 1

No, there's nothing built-in because with such "arrays"¹ it can be jagged and the concept of "dimensions" or "shape" doesn't make any sense at all. So, you'll have to write your own. If you can make an assumption of uniformity along all dimensions, you can proceed as follows:

dim1 = len(a)
dim2 = len(a[0])
dim3 = len(a[0][0])
.
.
.

It'd be pretty easy to make this recursive to handle all dimensions. This should do it:

def dim(a):
    if not type(a) == list:
        return []
    return [len(a)] + dim(a[0])

But if you need something like this, you might want to consider looking at NumPy arrays which have numpy.ndarray.shape which would give you what you're looking for.

from numpy import array
l = [[2, 3], [4, 2], [3, 2]]
a = array(l)
print a.shape

Output

(3, 2)

¹ In scare quotes because you're not really looking at array, you're looking at a list, or a list of lists, or a list of list of lists....

Exigible

" the concept of "dimensions" or "shape" doesn't make any sense at all" Not sure why you would say this? It makes a lot of sense and your answer makes this clear. Numpy handles this kind of thing gracefully, though obviously it isn't the exact same. Seems shape should be added as an attribute frankly (the jagged case could just return an error). – Ponder

Python should have a function for this. – Testify

Answer 2

You can do it with numpy:

import numpy
l = [[2,3], [4,2], [3,2]]
m = numpy.array(l)
print m.shape

But the shape of your second example is [2,2,2], not [1,4,5], unless I've misunderstood your question...

Humic

Yes, I meant [2,2,2] instead of [1,4,5]. – Atc

print m.shape is syntactically valid in Python 2, but not in Python 3. – Atc

I use numpy so much now, these questions about lists just confuse me. Time to review lists I guess? – Ponder

Use print(m.shape) for Python 3 – Overrule

Answer 3

That is not a multi-dimensional array. It is a list. It happens to contain other lists. There's nothing to say that your list could not be:

[[2,3], [4,2], [3,2,4,5,6]]

In which case, what value would you expect such a function to return?

There is no general function that does what you ask, not least because Python itself does not define a matrix/array class. You certainly can write your own function which operates on iterable objects like lists and tuples if you are prepared to make assumptions, or write assertions, as to the uniformity of the list. Use len(a) for the first dimension, len(a[0]) for the second, and so on. Recursion will be your friend here.

If you used a numpy array for your matrix, which to be honest would make a lot of sense, then your function would exist (it is the shape property of the ndarray class) and be meaningful.

Hypoderm

Is there any difference between arrays and lists in Python? I thought they were synonymous. – Atc

OF course it would've been more constructive to assume that the OP's data does in fact have a regular shape and provide something else than a non-answer. Or ask for clarification instead of assuming. – Northern

@AndersonGreen Python does not have arrays. It has lists and tuples. But no arrays. – Hypoderm

@Northern I did not assume anything about OP's data. I was trying to explain why the designers of Python could not write such a general purpose function because they could not make such assumptions. – Hypoderm

@DavidHeffernan: well, the stdlib has a type called array (from array import array), although I admit it's not used that much. – Adeno

@AndersonGreen More to the point, Python does not have (or does not commonly use) fixed-size blocks of memory you can access using a possibly multidimensional index. Lists certainly use resizeable arrays "behind the scenes". – Northern

@DavidHeffernan I missed the bit in the OP's question where they ask about built-in functions specifically, since they start with "Is it possible to write...". You're right in that it doesn't make sense to provide such a function built-in since it would need to consider the case where the data is irregular and such genericity is too much bother for too little benefit. (Over using a library that deals with this.) – Northern

Answer 4

Assuming the input array is not jagged :

def arr_dimen(a):
  return [len(a)]+arr_dimen(a[0]) if(type(a) == list) else []

Veiled

@Northern renamed the function while posting. missed to rename in the call. Thanks for pointing this. – Veiled

Answer 5

Iterative approach

a = [[[3,2], [4,5]],[[3,4],[2,3]]]
shape = []
b = a
while type(b) == list:
    shape.append(len(b))
    b = b[0]
print(shape)

Saturniid

Answer 6

import numpy as np
# define array 3 row and 5 colums
a = np.arange(15).reshape(3, 5)
print ("the array elements:\n",a);
# print all dim 
print("all dim :\n",a.shape)
# the print the number of row only
print("the number of rows :\n",a.shape[0])
# the print the number of colume only
print("the number of cols :\n",a.shape[1])

Oxy