问题
Several times (even several in a row) I've been bitten by the defaultdict bug: forgetting that something is actually a defaultdict and treating it like a regular dictionary.
d = defaultdict(list)
...
try:
v = d["key"]
except KeyError:
print "Sorry, no dice!"
For those who have been bitten too, the problem is evident: when d has no key 'key', the v = d["key"] magically creates an empty list and assigns it to both d["key"] and v instead of raising an exception. Which can be quite a pain to track down if d comes from some module whose details one doesn't remember very well.
I'm looking for a way to take the sting out of this bug. For me, the best solution would be to somehow disable a defaultdict's magic before returning it to the client.
回答1:
You can prevent creation of default values by assigning d.default_factory = None. However, I don't quite like the idea of object suddenly changing behavior. I'd prefer copying values to the new dict unless it imposes severe performance penalty.
回答2:
You may still convert it to an normal dict.
d = collections.defaultdict(list)
d = dict(d)
回答3:
use different idiom:
if 'key' not in d:
print "Sorry, no dice!"
回答4:
That is exactly the behavior you want from a defaultdict and not a bug. If you dont't want it, dont use a defaultdict.
If you keep forgetting what type variables have, then name them appropriately - for example suffix your defaultdict names with "_ddict".
回答5:
Using rkhayrov's idea of resetting self.default_factory, here is a toggleable subclass of defaultdict:
class ToggleableDefaultdict(collections.defaultdict):
def __init__(self,default_factory):
self._default_factory=default_factory
super(ToggleableDefaultdict,self).__init__(default_factory)
def off(self):
self.default_factory=None
def on(self):
self.default_factory=self._default_factory
For example:
d=ToggleableDefaultdict(list)
d['key'].append(1)
print(d)
# defaultdict(<type 'list'>, {'key': [1]})
d.off()
d['newkey'].append(2)
# KeyError: 'newkey'
d.on()
d['newkey'].append(2)
# defaultdict(<type 'list'>, {'newkey': [2], 'key': [1]})
来源:https://stackoverflow.com/questions/3031817/how-do-i-make-a-defaultdict-safe-for-unexpecting-clients