views:

585

answers:

7

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.

A: 

I don't see why you'd ever want to do this, but if you really really do need to, you could try pickling the dictionary:

mydict = {"a":1, "b":{"c":10}}
import pickle
key = pickle.dumps(mydict)

d[key] = value
Daniel Roseman
This solves nesting, but does it do if the values are non-standard types? Does it pickle the values as well?
Casebash
IMO, serialization is an overhead here. And @Casebash made a good point by mentioning the problem with non-standard types.
Andrey Vlasovskikh
A: 

I don't know whether I understand your question correctly, but i'll give it a try

    d[repr(a)]=value

You can interate over the dictionary like this

for el1 in d:
        for el2 in eval(el1):
                print el2,eval(el1)[el2]
Viswanadh
I think repl(a) would be better here as str might not be unique
Mark
Mark, I already changed it..
Viswanadh
`repr` of the different objects might not be different. The "difference" among Python objects is usually coded as `__eq__`, not `__repr__`.
Andrey Vlasovskikh
`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).
Brian
-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`.
S.Lott
This won't work when you get hash collision. See my comment to other answer to see how to demonstrate the problem.
Denis Otkidach
+11  A: 

If you have a really immutable dictionary (although it isn't clear to me why you don't use just 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 an 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())
    
Andrey Vlasovskikh
+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.
S.Lott
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
Now moved this to a new question
Casebash
+3  A: 

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.)

Brian
Overriding all mutating methods to raise an error will catch mistakes earlier at the cost of more code.
Ants Aasma
Again, this is a solution to the more specific problem, but the more general problem is open
Casebash
+7  A: 

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: http://stackoverflow.com/questions/635483/what-is-the-best-way-to-implement-nested-dictionaries-in-python

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.

Michael Dillon
+1: `tuple( someDictionary.items() )` works really, really well for making a dictionary into an immutable key.
S.Lott
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.
Brian
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
The difficulty with this is that each dictionary has to be sorted
Casebash
+3  A: 

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

Frank Bechmann
+2  A: 

To turn a someDictionary into a key, do this

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

You can easily reverse this with dict( key )

S.Lott
+1, though I find my solution with `frozenset` more "correct", see my answer.
Andrey Vlasovskikh