以numpy反向索引索引0

Question

I have created a matrix a, and I'm trying to index it in reverse order. These are my outputs:

>>> a

array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15],
       [16, 17, 18, 19],
       [20, 21, 22, 23],
       [24, 25, 26, 27],
       [28, 29, 30, 31],
       [32, 33, 34, 35],
       [36, 37, 38, 39],
       [40, 41, 42, 43],
       [44, 45, 46, 47]])

>>> print a[2:0:-1]
[[ 8  9 10 11]
 [ 4  5  6  7]]

As you can see I'm not getting the first row. Now, when I try this, I get an empty arrow output, as it treats it as 2 .. n - 1.

>>> print a[2:-1:-1]
[]

I'm wondering is there anyway to achieve a reverse iteration including the 0th index using indexing notation in numpy?

I'd like to do this in a general sense, so that as I iterate some array like so, I can make use of this:

>>> b = []
>>> for i in range(a.shape[0] - 3):
...     b.append(np.hstack(a[i+3:i:-1]))

Extra clarifications to show the issue:

Now, if I modify this to:

>>> b = []
>>> for i in range(a.shape[0] - 3):
...     b.append(np.hstack(a[i+2:i-1:-1]))

The above will simply output an error. Because, in the first iteration it will be a[2:-1:-1] which will be an empty array.

The first form simply generates the wrong output because it goes it will be interpreted as [3,0), and the second form will be interpreted as [2,n-1) which is completely wrong.

I'm using python 2.7, but it maybe relevant to python 3 as well.

Answer 1

Note that slices never include the item at the stop index [start, stop) , so the zeroth item is excluded from your slice.

To slice all the way to the start, don't specify the stop index (index is passed as None ):

>>> a[2::-1]
array([[ 8,  9, 10, 11],
       [ 4,  5,  6,  7],
       [ 0,  1,  2,  3]])

Specifying stop as -1 returns an empty slice since -1 connotes the last item; one can't (in a useful sense) go from index 2 to the last item striding backwards.

For your use case, you can play around with expressions in the slice like short-circuiting:

...
a[2:(i or None):-1] # 0 is falsy, so at i=0, we have None

Or use a ternary conditional to explicitly specify how you want the edge case to be handled:

...
a[2:(None if i==0 else i):-1]

Answer 2

Another way to do that is to use numpy.ix function which is very useful in slicing.

Try this:

print a[np.ix_(np.arange(2,-1,-1)), :]

np.arange returns array [2, 1, 0] - sequence of rows you need. You can pass any sequence you want inside np.ix to get rows or columns you need. See examples here .