How to implement an ordered, default dict?
Solution 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))
Solution 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.
Solution 3:
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.
Solution 4:
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(...)
.
Solution 5:
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)])
>>>