Tim Peters <t...@python.org> added the comment:
This won't be changed. The dict type doesn't support efficient random choice
(neither do sets, by the way), and it's been repeatedly decided that it would
do a disservice to users to hide that. As you know, you can materialize the
keys in a list (or tuple) first if you _want_ to pay that cost. Otherwise you
should use a different data structure.
Note that there's really no differnce between Pythons 2 and 3 here. If you
_happen_ to have a dict that uses little integers as keys, then it can _appear_
to work, when a random integer picked from range(len(the_dict)) happens to be
one of the keys. But then you get back the associated dict value, not the key.
For example, here under Python 2.7.11:
>>> import random
>>> random.choice({0: "a", 1: "b"})
'b'
>>> random.choice({0: "a", 1: "b"})
'b'
>>> random.choice({0: "a", 1: "b"})
'a'
But if the keys don't happen to be little integers, it always fails:
>>> random.choice({"a": 1, "b": 2})
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python27\lib\random.py", line 275, in choice
return seq[int(self.random() * len(seq))] # raises IndexError if seq is
empty
KeyError: 1
----------
nosy: +tim.peters
resolution: -> wont fix
stage: -> resolved
status: open -> closed
_______________________________________
Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue33098>
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