Python numpy array negative indexing

Question

I'm a bit confused about the indexing of numpy. Assume the following example:

>>> import numpy as np
>>> x = np.arange(10)
>>> x.shape = (2,5)
>>> x
array([[0, 1, 2, 3, 4],
       [5, 6, 7, 8, 9]])
>>> x[0:-1]
array([[0, 1, 2, 3, 4]])
>>> x[1:-1]
array([], shape=(0, 5), dtype=int64)
>>> x[1:]
array([[5, 6, 7, 8, 9]])

What I'm confused about, I can get the first row as 2D-array using x[0:-1]. But what does the -1 actually means in terms of indices? I would have thought, that calling x[1:-1] would then give me the second row, but instead if returns me an empty array, and to get what I want I need to use x[1:]?

I'm a little bit confused. Thanks for the help

Answer 1

You have this statement:

In [31]: x[0:-1]

This way of indexing means that "start at 1st row and go till the last row (excluded)". That's why we get the first row as a result.

Out[31]: array([[0, 1, 2, 3, 4]])

But, when you do:

 In [31]: x[1:-1]   
 Out[31]: array([], shape=(0, 5), dtype=int64)

It's asking NumPy to "start at second row and not include the last row". Since here the second row is also the last row, it is excluded and we get an empty array as a result.

---

More information: There's nothing specific about using negative indexing such as -1 here. For instance, the following ways of indexing would also return empty arrays.

# asking to "start at first row and end at first row"
In [42]: x[0:0]  
Out[42]: array([], shape=(0, 5), dtype=int64)

# asking to "start at second row and end at second row"
In [43]: x[1:1]  
Out[43]: array([], shape=(0, 5), dtype=int64)

When it comes to indexing in Python/NumPy, it's always "left inclusive and right exclusive".

Here's something in plain Python (i.e. indexing a list)

In [52]: lst = [1, 2] 

In [53]: lst[1:-1]    
Out[53]: []   # an empty list

Please note the construct of indexing which is: [start:stop:step]

If we start and stop at the same index, then we get nowhere and an empty data structure (array/list/tuple etc.) is returned as a result.