问题
I've seen it once or twice before, but I can't seem to find any official docs on it: Using python range
objects as indices in numpy.
import numpy as np
a = np.arange(9).reshape(3,3)
a[range(3), range(2,-1,-1)]
# array([2, 4, 6])
Let's trigger an index error just to confirm that ranges are not in the official range (pun intended) of legal indexing methods:
a['x']
# Traceback (most recent call last):
# File "<stdin>", line 1, in <module>
# IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices
Now, a slight divergence between numpy and its docs is not entirely unheard of and does not necessarily indicate that a feature is not intended (see for example here).
So, does anybody know why this works at all? And if it is an intended feature what are the exact semantics / what is it good for? And are there any ND generalizations?
回答1:
Not a proper answer, but too long for comment.
In fact, it seems to work with about any indexable object:
import numpy as np
class MyIndex:
def __init__(self, n):
self.n = n
def __getitem__(self, i):
if i < 0 or i >= self.n:
raise IndexError
return i
def __len__(self):
return self.n
a = np.array([1, 2, 3])
print(a[MyIndex(2)])
# [1 2]
I think the relevant lines in NumPy's code are below this comment in core/src/multiarray/mapping.c:
/*
* Some other type of short sequence - assume we should unpack it like a
* tuple, and then decide whether that was actually necessary.
*/
But I'm not entirely sure. For some reason, this hangs if you remove the if i < 0 or i >= self.n: raise IndexError
, even though there is a __len__
, so at some point it seems to be iterating through the given object until IndexError
is raised.
回答2:
Just to wrap this up (thanks to @WarrenWeckesser in the comments): This behavior is actually documented. One only has to realize that range
objects are python sequences in the strict sense.
So this is just a case of fancy indexing. Be warned, though, that it is very slow:
>>> a = np.arange(100000)
>>> timeit(lambda: a[range(100000)], number=1000)
12.969507368048653
>>> timeit(lambda: a[list(range(100000))], number=1000)
7.990526253008284
>>> timeit(lambda: a[np.arange(100000)], number=1000)
0.22483703796751797
来源:https://stackoverflow.com/questions/53123009/using-python-range-objects-to-index-into-numpy-arrays