Puzzled on the ndim from Numpy

import numpy as np

a = np.zeros((5,2,3,4), dtype=np.int16)
print a.ndim

b = np.zeros((2,3,4), dtype=np.int16)
print b.ndim

Above is my code. the output is:

4
3

I have checked the page from [here] (Link)

I expected a.dim = 2 or 3, but it is 4. Why?

Could you give me some hints? Thanks

Answer

The tuple you give to zeros and other similar functions gives the ‘shape’ of the array (and is available for any array as a.shape). The number of dimensions is the number of entries in that shape.

If you were instead printing len(a.shape) and len(b.shape) you would expect the result you get. The ndim is always equal to this (because that’s how it’s defined).

The link you give shows the same behavior, except that the reshape method takes arbitary numbers of positional arguments rather than a tuple. But in the tutorial’s example:

>>> a = arange(15).reshape(3, 5)
>>> a
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14]])
>>> a.shape
(3, 5)
>>> a.ndim
2

The number of arguments given to reshape, and the number of elements in the tuple given back by a.shape, is 2. If this was instead:

>>> a = np.arange(30).reshape(3, 5, 2)
>>> a.ndim
3

Then we see the same behavior: the ndim is the number of shape entries we gave to numpy.