numpy array.shape behaviour

Question

For following:

d = np.array([[0,1,4,3,2],[10,18,4,7,5]])
print(d.shape)

Output is:

(2, 5)

It is expected.

But, for this(difference in number of elements in individual rows):

d = np.array([[0,1,4,3,2],[10,18,4,7]])
print(d.shape)

Output is:

(2,)

How to explain this behaviour?

willeM_ Van Onsem · Accepted Answer · 2019-09-29 16:12:54Z

2

Short answer: It parses it as an array of two objects: two lists.

Numpy is used to process "rectangular" data. In case you pass it non-rectangular data, the np.array(..) function will fallback on considering it a list of objects.

Indeed, take a look at the dtype of the array here:

>>> d
array([list([0, 1, 4, 3, 2]), list([10, 18, 4, 7])], dtype=object)

It is an one-dimensional array that contains two items two lists. These lists are simply objects.

answered Sep 29, 2019 at 16:12

willeM_ Van Onsem

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2 Comments

Mandroid Over a year ago

Thanks for the input. Btw, any good numpy resource for python beginners?

hpaulj Over a year ago

There's a lot of numpy documentation. But the problem with your example is that it's a kind of error, trying to make a 2d array from lists that differ in size. np.array normally makes a multidimensional numeric array. Making the 1d object dtype array is a fall-back alternative. It get's a lot of attention in SO questions, not so much in the formal documentation or tutorials.

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