Raymond Hettinger added the comment:
[Serhiy]
I'm inclined to reject this proposition. It serves very special use case.
[Silent Ghost]
Correct me if I'm wrong but this seem as a very unlikely use case
[Alec Nikolas Reiter]
Implicitly assigning the factory is a pretty good compromise, like I
Vedran Čačić added the comment:
That's usual behavior for any class method. You can call them on an instance,
but they don't have the access to it, only to its class. So transferring of
factory would in fact not be possible. Of course,it's possible to make fromkeys
an instance method, but
Gavin Panella added the comment:
It's inconsistent that defaultdict([]) should be rejected:
>>> defaultdict([])
Traceback (most recent call last):
File "", line 1, in
TypeError: first argument must be callable or None
but defaultdict.fromkeys([]) is okay:
>>>
Serhiy Storchaka added the comment:
It may be written simpler:
d = defaultdict(factory)
for i in constants:
d[i]
or
d = defaultdict(factory, {i: factory() for i in constants})
I'm inclined to reject this proposition. It serves very special use case.
--
nosy:
SilentGhost added the comment:
Correct me if I'm wrong but this seem as a very unlikely use case. Why would
you need to ensure content of the defaultdict (i.e., why would you ever use its
fromkeys method)? And, perhaps, having to implicitly assign default factory is
not such a bad tradeoff,
Raymond Hettinger added the comment:
-0 I could see some use cases for this, dict.fromkeys(contestants,
factory=list), but it adds complexity to a core container API and it starts to
overlap the use cases for collections.defaultdict().
Perhaps set comprehensions should remain the
New submission from Alec Nikolas Reiter:
Not something I've noticed until today, but defaultdict.fromkeys only accepts
an iterable of keys and an initial value. To set the default_factory for the
newly created collection, you need to prod at the attribute itself.
This isn't a huge issue for
SilentGhost added the comment:
Raymond, but Alec talks about
defaultdict.fromkeys(constants, factory=list)
as opposed to current solution
d = defaultdict.fromkeys(constants)
d.default_factory = list
for i in d:
d[i] = []
I wouldn't think that the dict.fromkeys should be
Alec Nikolas Reiter added the comment:
@SilentGhost I was playing with a voting algorithm (Instant Runoff, rather than
traditional majority). Ultimately, I ended up using Counter in place of a
defaultdict, but that's how I ended up noticing it.
Not so much ensuring the content so much as
Changes by Serhiy Storchaka storch...@gmail.com:
--
nosy: +rhettinger
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http://bugs.python.org/issue23372
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