A data-structure for 1:1 mappings in python?

Question

I have a problem which requires a reversable 1:1 mapping of keys to values.

That means sometimes I want to find the value given a key, but at other times I want to

Answer 1

The other alternative is to make a new class which unites two dictionaries, one for each > kind of lookup. That would most likely use up twice as much memory as a single dict.

Not really, since they would just be holding two references to the same data. In my mind, this is not a bad solution.

Have you considered an in-memory database lookup? I am not sure how it will compare in speed, but lookups in relational databases can be very fast.

Answer 2

It so happens that I find myself asking this question all the time (yesterday in particular). I agree with the approach of making two dictionaries. Do some benchmarking to see how much memory it's taking. I've never needed to make it mutable, but here's how I abstract it, if it's of any use:

class BiDict(list):
    def __init__(self,*pairs):
        super(list,self).__init__(pairs)
        self._first_access = {}
        self._second_access = {}
        for pair in pairs:
            self._first_access[pair[0]] = pair[1]
            self._second_access[pair[1]] = pair[0]
            self.append(pair)

    def _get_by_first(self,key):
        return self._first_access[key]

    def _get_by_second(self,key):
        return self._second_access[key]

    # You'll have to do some overrides to make it mutable
    # Methods such as append, __add__, __del__, __iadd__
    # to name a few will have to maintain ._*_access

class Constants(BiDict):
    # An implementation expecting an integer and a string
    get_by_name = BiDict._get_by_second
    get_by_number = BiDict._get_by_first

t = Constants(
        ( 1, 'foo'),
        ( 5, 'bar'),
        ( 8, 'baz'),
    )

>>> print t.get_by_number(5)
bar
>>> print t.get_by_name('baz')
8
>>> print t
[(1, 'foo'), (5, 'bar'), (8, 'baz')]

Answer 3

The other alternative is to make a new class which unites two dictionaries, one for each kind of lookup. That would most likely be fast but would use up twice as much memory as a single dict.

Not really. Have you measured that? Since both dictionaries would use references to the same objects as keys and values, then the memory spent would be just the dictionary structure. That's a lot less than twice and is a fixed ammount regardless of your data size.

What I mean is that the actual data wouldn't be copied. So you'd spend little extra memory.

Example:

a = "some really really big text spending a lot of memory"

number_to_text = {1: a}
text_to_number = {a: 1}

Only a single copy of the "really big" string exists, so you end up spending just a little more memory. That's generally affordable.

I can't imagine a solution where you'd have the key lookup speed when looking by value, if you don't spend at least enough memory to store a reverse lookup hash table (which is exactly what's being done in your "unite two dicts" solution).

Answer 4

Here is my own solution to this problem: http://github.com/spenthil/pymathmap/blob/master/pymathmap.py

The goal is to make it as transparent to the user as possible. The only introduced significant attribute is partner.

OneToOneDict subclasses from dict - I know that isn't generally recommended, but I think I have the common use cases covered. The backend is pretty simple, it (dict1) keeps a weakref to a 'partner' OneToOneDict (dict2) which is its inverse. When dict1 is modified dict2 is updated accordingly as well and vice versa.

From the docstring:

>>> dict1 = OneToOneDict()
>>> dict2 = OneToOneDict()
>>> dict1.partner = dict2
>>> assert(dict1 is dict2.partner)
>>> assert(dict2 is dict1.partner)
>>> dict1['one'] = '1'
>>> dict2['2'] = '1'
>>> dict1['one'] = 'wow'
>>> assert(dict1 == dict((v,k) for k,v in dict2.items()))
>>> dict1['one'] = '1'
>>> assert(dict1 == dict((v,k) for k,v in dict2.items()))
>>> dict1.update({'three': '3', 'four': '4'})
>>> assert(dict1 == dict((v,k) for k,v in dict2.items()))
>>> dict3 = OneToOneDict({'4':'four'})
>>> assert(dict3.partner is None)
>>> assert(dict3 == {'4':'four'})
>>> dict1.partner = dict3
>>> assert(dict1.partner is not dict2)
>>> assert(dict2.partner is None)
>>> assert(dict1.partner is dict3)
>>> assert(dict3.partner is dict1)
>>> dict1.setdefault('five', '5')
>>> dict1['five']
'5'
>>> dict1.setdefault('five', '0')
>>> dict1['five']
'5'

When I get some free time, I intend to make a version that doesn't store things twice. No clue when that'll be though :)

Answer 5

Assuming that you have a key with which you look up a more complex mutable object, just make the key a property of that object. It does seem you might be better off thinking about the data model a bit.

Answer 6

class TwoWay:
    def __init__(self):
       self.d = {}
    def add(self, k, v):
       self.d[k] = v
       self.d[v] = k
    def remove(self, k):
       self.d.pop(self.d.pop(k))
    def get(self, k):
       return self.d[k]