Numpy array ndmin behavior for empty array

Question

import numpy as np
np.array([], ndmin=2)

The resulting array has shape (1, 0). But why not (0, 0), if the input array has length 0? Is there a better way to create an array of shape (0, 0) than explicitely call np.zeros((0, 0))?

There's nothing preventing that array from having a shape of (0, 0), verify for yourself by assigning that shape to it, or using reshape. The behavior of the constructor is to append 1s until the number of dimensions has been satisfied. — user3483203
– user3483203, Commented Nov 1, 2019 at 15:33

Ruzihm · Accepted Answer · 2019-11-01 15:33:02Z

2

From the documentation for numpy.array:

ndmin : int, optional -- Specifies the minimum number of dimensions that the resulting array should have. Ones will be pre-pended to the shape as needed to meet this requirement

You're giving it a shape (0,) array-like, so it prepends a 1 to make the shape size 2.

np.zeros((0, 0)) is fine.

answered Nov 1, 2019 at 15:33

Ruzihm

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3 Comments

steam_engine Over a year ago

"Ones will be pre-pended to the shape as needed to meet this requirement" - why is it even needed?

hpaulj Over a year ago

It doesn't treat the [] case special. np.array( np.ones(n), ndmin=2) becomes (1,n) shape regardless whether n is 0 or 10. Consistency like that is nice.

Ruzihm Over a year ago

@steam_engine [] is a shape (0,) array-like, and you tell it that your requirement is a shape (x,y) array-like, so it appends a 1 before the 0 to give you an array with shape (1,0).

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