Is there a clever way to pass the key to defaultdict's default_factory?

It hardly qualifies as clever - but subclassing is your friend:

class keydefaultdict(defaultdict):    def __missing__(self, key):        if self.default_factory is None:            raise KeyError( key )        else:            ret = self[key] = self.default_factory(key)            return retd = keydefaultdict(C)d[x] # returns C(x)

No, there is not.

The defaultdict implementation can not be configured to pass missing key to the default_factory out-of-the-box. Your only option is to implement your own defaultdict subclass, as suggested by @JochenRitzel, above.

But that isn't "clever" or nearly as clean as a standard library solution would be (if it existed). Thus the answer to your succinct, yes/no question is clearly "No".

It's too bad the standard library is missing such a frequently needed tool.

I don't think you need defaultdict here at all. Why not just use dict.setdefault method?

>>> d = {}>>> d.setdefault('p', C('p')).v'p'

That will of course would create many instances of C. In case it's an issue, I think the simpler approach will do:

>>> d = {}>>> if 'e' not in d: d['e'] = C('e')

It would be quicker than the defaultdict or any other alternative as far as I can see.

ETA regarding the speed of in test vs. using try-except clause:

>>> def g():    d = {}    if 'a' in d:        return d['a']>>> timeit.timeit(g)0.19638929363557622>>> def f():    d = {}    try:        return d['a']    except KeyError:        return>>> timeit.timeit(f)0.6167065411074759>>> def k():    d = {'a': 2}    if 'a' in d:        return d['a']>>> timeit.timeit(k)0.30074866358404506>>> def p():    d = {'a': 2}    try:        return d['a']    except KeyError:        return>>> timeit.timeit(p)0.28588609450770264