Using a Python Dictionary as a Key (Non-nested)

Question

Python doesn't allow dictionaries to be used as keys in other dictionaries. Is there a workaround for using non-nested dictionaries as keys?

The general problem with more complicated non-hashable objects and my specific use case has been moved here. My original description of my use case was incorrect.

Solution 1

If you have a really immutable dictionary (although it isn't clear to me why you don't just use a list of pairs: e.g. [('content-type', 'text/plain'), ('host', 'example.com')]), then you may convert your dict into:

1. A tuple of pairs. You've already done that in your question. A tuple is required instead of list because the results rely on the ordering and the immutability of the elements.

   >>> tuple(sorted(a.items()))
   

2. A frozen set. It is a more suitable approach from the mathematical point of view, as it requires only the equality relation on the elements of your immutable dict, while the first approach requires the ordering relation besides equality.

   >>> frozenset(a.items())
   

Solution 2

If I needed to use dictionaries as keys, I would flatten the dictionary into a tuple of tuples.

You might find this SO question useful: What is the best way to implement nested dictionaries?

And here is an example of a flatten module that will flatten dictionaries: http://yawpycrypto.sourceforge.net/html/public/Flatten.Flatten-module.html

I don't fully understand your use case and I suspect that you are trying to prematurely optimize something that doesn't need optimization.

Solution 3

To turn a someDictionary into a key, do this

key = tuple(sorted(someDictionary .items())

You can easily reverse this with dict( key )

Solution 4

One way to do this would be to subclass the dict and provide a hash method. ie:

class HashableDict(dict):
    def __hash__(self):
        return hash(tuple(sorted(self.iteritems())))

>>> d = HashableDict(a=1, b=2)
>>> d2 = { d : "foo"}
>>> d2[HashableDict(a=1, b=2)]
"foo"

However, bear in mind the reasons why dicts (or any mutable types) don't do this: mutating the object after it has been added to a hashtable will change the hash, which means the dict will now have it in the wrong bucket, and so incorrect results will be returned.

If you go this route, either be very sure that dicts will never change after they have been put in the other dictionary, or actively prevent them (eg. check that the hash never changes after the first call to __hash__, and throw an exception if not.)

Solution 5

Hmm, isn't your use case just memoizing function calls? Using a decorator, you will have easy support for arbitrary functions. And yes, they often pickle the arguments, and using circular reasoning, this works for non-standard types as long as they can be pickled.

See e.g. this memoization sample

View more solutions

Comments

[mmmmmm]

I think repl(a) would be better here as str might not be unique

[Casebash]

This solves nesting, but does it do if the values are non-standard types? Does it pickle the values as well?

[Andrey Vlasovskikh]

repr of the different objects might not be different. The "difference" among Python objects is usually coded as __eq__, not __repr__.

[Brian]

repr and str are actually the same for dicts anyway. However, you could run into trouble this way - it's possible to get dicts with different internal state so that, while they contain the same items, they list their keys in a different order, and would thus produce a different key. You'll also run into trouble if you store objects without the property that repr(x)==repr(y) <=> x==y in the dict (eg. most user created classes).

[Andrey Vlasovskikh]

IMO, serialization is an overhead here. And @Casebash made a good point by mentioning the problem with non-standard types.

[user1066101]

+1: tuple( someDictionary.items() ) works really, really well for making a dictionary into an immutable key.

[user1066101]

-1: Use tuple( someDictionary.items() ) instead of repr. it gives you a structure that can be trivially transformed back into a dictionary without resorting to eval.

[Brian]

Make that tuple(sorted(somedictionary.items()) - the order of keys is not guaranteed, which means equal dicts might produce different reprs by listing the items in a different order.

[Ryu]

Overriding all mutating methods to raise an error will catch mistakes earlier at the cost of more code.

[Denis Otkidach]

Sorting in important. To see why you have to find 2 different key values with equal hash (it's probably hard to find for strings, but can be easily achieved with user defined objects), then construct 2 equal dictionaries by inserting them in different order. You'll get equal dictionaries with different order in .items().

[Denis Otkidach]

This won't work when you get hash collision. See my comment to other answer to see how to demonstrate the problem.

[Andrey Vlasovskikh]

+1, though I find my solution with frozenset more "correct", see my answer.

[user1066101]

+1: Good point on ordering. A dictionary can always be transformed to a frozenset because the keys must be unique, assuring each tuple will be preserved in the set. Elegant.

[Casebash]

This is a nice solution which is more general than than the specific problem I posed, but doesn't this doesn't handle dictionaries within dictionaries.

[Casebash]

The difficulty with this is that each dictionary has to be sorted

[Casebash]

Again, this is a solution to the more specific problem, but the more general problem is open

[Casebash]

Now moved this to a new question

[AnukuL]

Note that, Frozen set solution will not work, if the dict contains list or any other mutable object as values. For ex: a={'key1': 'val1', 'key2': ['val2', 'val3']}

[BallpointBen]

tuple(sorted()) will work as long as the dict's keys are comparable. frozenset requires hashable dict values.