Python 3: range objects cannot be sliced

[Alan G Isaac]

Is the behavior below expected? Documented?
(The error msg is misleading.)
Thanks,
Alan Isaac

>>> x = range(20)
>>> s = slice(None,None,2)
>>> x

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: sequence index must be integer, not 'slice'

 

[Fuzzyman]

Is the behavior below expected? Documented?
(The error msg is misleading.)
Thanks,
Alan Isaac

 >>> x = range(20)
 >>> s = slice(None,None,2)
 >>> x
Traceback (most recent call last):
   File "<stdin>", line 1, in <module>
TypeError: sequence index must be integer, not 'slice'

Well, it has the same behaviour as the iterator returned by xrange in
Python 2.X - so expected I guess. The error message is also the same
in Python 2.X.

Michael Foord

 

[Paul Rubin]

x = range(20)
s = slice(None,None,2)
x

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: sequence index must be integer, not 'slice'

range is an iterator now. Try itertools.islice.

 

[Alan G Isaac]

range is an iterator now. Try itertools.islice.

Well yes, it behaves like xrange did.
But (also like xrange) it supports indexing. (!)
So why not slicing?
I expected this (to keep it functionally
more similar to the old range).

Alan Isaac

 

[Alan G Isaac]

It is documented:
http://docs.python.org/3.0/library/stdtypes.html#sequence-types-str-bytes-bytearray-list-tuple-range