What does a numpy shape starting with zero mean

问题

Okay, so I found out that you can have arrays with 0s in their shape.
For the case where you have 0 as the only dimension, this makes sense to me. It is an empty array.

np.zeros(0)

But the case where you have something like this:

np.zeros((0, 100))

Is confusing for me. Why is it defined like this?

回答1:

As far as I know it's just a redundant way to express an empty array. It doesn't seems to matter for python if you have rows of "emptiness".

Let's say we have a give array a:

import numpy as np

a = np.zeros((0,100))

If we print a all we get is the empty array itself:

print(a)

>>> []

Moreover we can actually see that despite this a maintain it's shape"

np.shape(a)

>>> (0, 100)

But if you try to access a given element by position, e.g:

print(a[0])

or

print(a[0][0])

You get an IndexError :

IndexError: index 0 is out of bounds for axis 0 with size 0

Therefore I believe that the mathematical meaning of the empty arrays, despite the shape you assign to them, is the same.

回答2:

Well, in this particular case, the two statements are equivilant:

print(np.zeros(0))
>>>[]

print(np.zeros((0,100)))
>>>[]

This is because an empty array is an empty array. Your intuitions are correct there. If you put in:

np.zeros(10)

or

np.zeros((1,10))

you'd also get the same array, namely [0,0,0,0,0,0,0,0,0,0]. The shape only matters when you are referring to a number that actually changes the shape of the array. For instance:

print(np.zeros((2,3)))
>>>[[0,0,0]
   [0,0,0]]

but:

print(np.zeros((3,2)))
>>>[[0,0]
    [0,0]
    [0,0]]

Nothing particularly opaque about it. Your common sense actually applies here. If an array is empty, none of the other dimensions you add to it matter.

来源：https://stackoverflow.com/questions/53747948/what-does-a-numpy-shape-starting-with-zero-mean