numpy数组切片，为什么有时2-d数组，有时是1-d数组

Question

My question is about array slicing in numpy. What's the logic for the following behavior?

x = arange(25)
x.shape = (5, 5)
# This results in a 2-d array in which rows and columns are excerpted from x
y = x[0:2, 0:2]
# But this results in a 1-d array
y2 = x[0:2, 0]

I would have expected y2 to be a 2-d array which contains the values in rows 0 and 1, column 0.

Answer 1

You can get your expected behavior doing x[0:2, 0:1] , ie with a single item slice. But whenever a single element is selected, that dimension is collapsed. You may not like it, but if you think about it a little bit, you should realize it is the most consistent behavior: following your logic, x[0, 0] would be a 2d array of 1 row and 1 column, instead of the item stored at that position.

Answer 2

This follows standard Python conventions. Look at the results of these analogous expressions:

>>> a = [0, 1, 2, 3, 4, 5]
>>> a[4]
4
>>> a[4:5]
[4]

As you can see, one returns one item , while the other returns a list containing one item . This is always the way python works, and numpy is just following that convention, but at a higher dimension. Whenever you pass a slice rather than an individual item, a list is returned; this is true even if there are no items in the list, either because the end index is too low, or because the starting index is too high:

>>> a[4:4]
[]
>>> a[6:6]
[]

So in all situations, passing a slice means "return a sequence (along the given dimension)," while passing an integer means "return a single item (along the given dimension)."

Answer 3

When you access an array using a single element instead of a slice, it will collapse that dimension. For that reason, if you have

x = arange(25)
y = x[10]

You would expect y to be 10 and not array([10]) .

So, if you use

y2 = x[0:2, 0]
print y2.shape
(2,)

It will collapse the second dimension. If you want to keep that second dimension, you need to access that dimension using a slice.

y2 = x[0:2, 0:1]
print y2.shape
(2, 1)