Python dictionary that defaults to key?

Question

Is there a way to get a defaultdict to return the key by default? Or some data structure with equivalent behavior? I.e., after initializing dictionary d,

>>> d['a'] = 1
>>> d['a']
1
>>> d['b']
'b'
>>> d['c']
'c'

I've only seen default dictionaries take functions that don't take parameters, so I'm not sure if there's a solution other than creating a new kind of dictionary .

This is a somewhat too specialized task. I suggest you to consider subclassing the 'dict' object/type. — heltonbiker, May 29 '12 at 00:27
I'm intrigued by this. Can you give an idea of what you need it for? — Ben Butler-Cole, May 29 '12 at 09:34

score 25 · Accepted Answer · answered May 29 '12 at 00:28

25

I'd override the __missing__ method of dict:

>>> class MyDefaultDict(dict):
...     def __missing__(self, key):
...         self[key] = key
...         return key
...
>>> d = MyDefaultDict()
>>> d['joe']
'joe'
>>> d
{'joe': 'joe'}

answered May 29 '12 at 00:28

pillmuncher

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Is there a specific reason to store the key in addition to returning it? Unless I *need* to know which keys were queried, I see no reason to do this. – Joachim Sauer May 29 '12 at 10:25
@JoachimSauer: Normally one expects that if `d['joe']` doesn't raise a `KeyError`, then `'joe' in d` evaluates to `True`. If I didn't store the key that wouldn't be the case. – pillmuncher May 29 '12 at 12:26
1

But this solution makes `'joe' in d` evaluate to `False` *unless* you used `d['joe']` *before*. So the pure "read-only" access changes the result of *another* read-only access. I'm not sure that's any better. – Joachim Sauer May 29 '12 at 12:31
2

@JoachimSauer: You're right, but storing the key is consistent with how `collections.defaultdict` works. – pillmuncher May 29 '12 at 12:35

score 7 · Answer 2 · edited May 23 '17 at 12:21

7

Edit: Oops, I just realized that code in my file originally came from another stackoverflow answer! https://stackoverflow.com/a/2912455/456876, go upvote that one.

This is what I use - it's a defaultdict variant that passes the key as an argument to the default-value factory function that's passed as an argument to init, instead of no arguments:

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

This is the use you want:

>>> d = keybased_defaultdict(lambda x: x)
>>> d[1]
1
>>> d['a']
'a'

Other possibilities:

>>> d = keybased_defaultdict(lambda x: len(x))
>>> d['a']
1
>>> d['abc']
3

edited May 23 '17 at 12:21

Community

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answered May 29 '12 at 00:36

weronika

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I was actually somewhat surprised to [re]discover that `defaultdict` does not already work like this... – May 29 '12 at 00:41
@pst I suppose it theoretically makes some sense that's not the default behavior - if you're storing the same information in the key and the value, you may be using the wrong data type. That said, I do use this in my programs at work. – weronika May 29 '12 at 00:46

score 4 · Answer 3 · answered May 29 '12 at 00:59

If you don't want to subclass dict, you can try using

d.get('a', 'a')
d.get('b', 'b')
d.get('c', 'c')

Which I think is clearer and less magical for this purpose

If you are a DRY fanatic and only have single char keys, you can do this :)

d.get(*'a'*2)
d.get(*'b'*2)
d.get(*'c'*2)

score 3 · Answer 4 · answered May 29 '12 at 00:27

3

You'll likely have to write your own class that inherits from (or is similar to) defaultdict and override the __getitem__ method.

answered May 29 '12 at 00:27

inspectorG4dget

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Python dictionary that defaults to key?

4 Answers4