Can I do an ordered, default dict in Python?

问题:

I would like to combine OrderedDict() and defaultdict() from collections in one object, which shall be an ordered, default dict. Is this possible?

回答1:

The following (using a modified version of this recipe) works for me:

from collections import OrderedDict, Callable

class DefaultOrderedDict(OrderedDict):
    # Source: http://stackoverflow.com/a/6190500/562769
    def __init__(self, default_factory=None, *a, **kw):
        if (default_factory is not None and
           not isinstance(default_factory, Callable)):
            raise TypeError(\'first argument must be callable\')
        OrderedDict.__init__(self, *a, **kw)
        self.default_factory = default_factory

    def __getitem__(self, key):
        try:
            return OrderedDict.__getitem__(self, key)
        except KeyError:
            return self.__missing__(key)

    def __missing__(self, key):
        if self.default_factory is None:
            raise KeyError(key)
        self[key] = value = self.default_factory()
        return value

    def __reduce__(self):
        if self.default_factory is None:
            args = tuple()
        else:
            args = self.default_factory,
        return type(self), args, None, None, self.items()

    def copy(self):
        return self.__copy__()

    def __copy__(self):
        return type(self)(self.default_factory, self)

    def __deepcopy__(self, memo):
        import copy
        return type(self)(self.default_factory,
                          copy.deepcopy(self.items()))

    def __repr__(self):
        return \'OrderedDefaultDict(%s, %s)\' % (self.default_factory,
                                               OrderedDict.__repr__(self))

回答2:

Here is another possibility, inspired by Raymond Hettinger\'s super() Considered Super, tested on Python 2.7.X and 3.4.X:

from collections import OrderedDict, defaultdict

class OrderedDefaultDict(OrderedDict, defaultdict):
    def __init__(self, default_factory=None, *args, **kwargs):
        #in python3 you can omit the args to super
        super(OrderedDefaultDict, self).__init__(*args, **kwargs)
        self.default_factory = default_factory

If you check out the class\'s MRO (aka, help(OrderedDefaultDict)), you\'ll see this:

class OrderedDefaultDict(collections.OrderedDict, collections.defaultdict)
 |  Method resolution order:
 |      OrderedDefaultDict
 |      collections.OrderedDict
 |      collections.defaultdict
 |      __builtin__.dict
 |      __builtin__.object

meaning that when an instance of OrderedDefaultDict is initialized, it defers to the OrderedDict\'s init, but this one in turn will call the defaultdict\'s methods before calling __builtin__.dict, which is precisely what we want.

回答3:

Here\'s another solution to think about if your use case is simple like mine and you don\'t necessarily want to add the complexity of a DefaultOrderedDict class implementation to your code.

from collections import OrderedDict

keys = [\'a\', \'b\', \'c\']
items = [(key, None) for key in keys]
od = OrderedDict(items)

(None is my desired default value.)

Note that this solution won\'t work if one of your requirements is to dynamically insert new keys with the default value. A tradeoff of simplicity.

Update 3/13/17 - I learned of a convenience function for this use case. Same as above but you can omit the line items = ... and just:

od = OrderedDict.fromkeys(keys)

Output:

OrderedDict([(\'a\', None), (\'b\', None), (\'c\', None)])

And if your keys are single characters, you can just pass one string:

OrderedDict.fromkeys(\'abc\')

This has the same output as the two examples above.

You can also pass a default value as the second arg to OrderedDict.fromkeys(...).

回答4:

If you want a simple solution that doesn\'t require a class, you can just use OrderedDict.setdefault(key, default=None) or OrderedDict.get(key, default=None). If you only get / set from a few places, say in a loop, you can easily just setdefault.

totals = collections.OrderedDict()

for i, x in some_generator():
    totals[i] = totals.get(i, 0) + x

It is even easier for lists with setdefault:

agglomerate = collections.OrderedDict()

for i, x in some_generator():
    agglomerate.setdefault(i, []).append(x)

But if you use it more than a few times, it is probably better to set up a class, like in the other answers.

回答5:

A simpler version of @zeekay \'s answer is:

from collections import OrderedDict

class OrderedDefaultListDict(OrderedDict): #name according to default
    def __missing__(self, key):
        self[key] = value = [] #change to whatever default you want
        return value

回答6:

Another simple approach would be to use dictionary get method

>>> from collections import OrderedDict
>>> d = OrderedDict()
>>> d[\'key\'] = d.get(\'key\', 0) + 1
>>> d[\'key\'] = d.get(\'key\', 0) + 1
>>> d
OrderedDict([(\'key\', 2)])
>>> 

回答7:

A simple and elegant solution building on @NickBread. Has a slightly different API to set the factory, but good defaults are always nice to have.

class OrderedDefaultDict(OrderedDict):
    factory = list

    def __missing__(self, key):
        self[key] = value = self.factory()
        return value

回答8:

Inspired by other answers on this thread, you can use something like,

from collections import OrderedDict

class OrderedDefaultDict(OrderedDict):
    def __missing__(self, key):
        value = OrderedDefaultDict()
        self[key] = value
        return value

I would like to know if there\'re any downsides of initializing another object of the same class in the missing method.

回答9:

i tested the default dict and discovered it\'s also sorted! maybe it was just a coincidence but anyway you can use the sorted function:

sorted(s.items())

i think it\'s simpler