Sorry to grouse, but isn't this maybe being a bit too clever?
Using your example,
p1 = partial(operator.add)
is creating a callable, p1, i.e. a sort of function. Yes I know technically
it's not a function, but it behaves very much like one.
Now, if I write
def f1(x,y): return x+y
def f2(x,y): return x+y
I don't expect f1==f2 to be True, even though f1 and f2 behave in exactly
the same way,
and indeed it is not.
If I wanted it to be true, I should have written
def f1(x): return x+y
f2=f1
I find this behaviour natural and expected, both in my example and yours
(although maybe I've just got used to Python's behaviour).
Similarly, if you wanted p1==p2, why not write
p1 = partial(operator.add)
p2 = p1
Maybe I could be persuaded otherwise by a convincing use case, but I rather
doubt it.
Rob Cliffe
----- Original Message -----
From: "VanL" <van.lindb...@gmail.com>
To: <python-dev@python.org>
Sent: Friday, May 07, 2010 3:37 PM
Subject: [Python-Dev] Possible patch for functools partial - Interested?
Howdy all -
I have an app where I am using functools.partial to bundle up jobs to
do, where a job is defined as a callable + args. In one case, I wanted
to keep track of whether I had previously seen a job, so I started
putting them into a set... only to find out that partials never test
equal to each other:
import operator
from functools import partial
p1 = partial(operator.add)
p2 = partial(operator.add)
p1 == p2
False
seen = set();seen.add(p1)
p2 in seen
False
I created a subclass of functools.partial that provides appropriate
__eq__ and __hash__ methods, so that this works as expected. I called
the subclass a Job:
j1 = Job(operator.add)
j2 = Job(operator.add)
j1 == j2
True
seen = set();seen.add(j1)
j2 in seen
True
j1 is j2
False
While I was at it, I also added a nice repr. Would this group be
interested in a patch, or is this not interesting?
Thanks,
Van
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