python - How to implement an efficient bidirectional hash table?

Question

Python dict is a very useful data-structure:

d = {'a': 1, 'b': 2}

d['a'] # get 1

Sometimes you'd also like to index by values.

d[1] # get 'a'

Which is the most efficient way to implement this data-structure? Any official recommend way to do it?

Answer 1

Here is a class for a bidirectional dict, inspired by Finding key from value in Python dictionary and modified to allow the following 2) and 3).

Note that :

• 1) The inverse directory bd.inverse auto-updates itself when the standard dict bd is modified.
• 2) The inverse directory bd.inverse[value] is always a list of key such that bd[key] == value.
• 3) Unlike the bidict module from https://pypi.python.org/pypi/bidict, here we can have 2 keys having same value, this is very important.

Code:

class bidict(dict):
    def __init__(self, *args, **kwargs):
        super(bidict, self).__init__(*args, **kwargs)
        self.inverse = {}
        for key, value in self.items():
            self.inverse.setdefault(value,[]).append(key) 

    def __setitem__(self, key, value):
        if key in self:
            self.inverse[self[key]].remove(key) 
        super(bidict, self).__setitem__(key, value)
        self.inverse.setdefault(value,[]).append(key)        

    def __delitem__(self, key):
        self.inverse.setdefault(self[key],[]).remove(key)
        if self[key] in self.inverse and not self.inverse[self[key]]: 
            del self.inverse[self[key]]
        super(bidict, self).__delitem__(key)

Usage example:

bd = bidict({'a': 1, 'b': 2})  
print(bd)                     # {'a': 1, 'b': 2}                 
print(bd.inverse)             # {1: ['a'], 2: ['b']}
bd['c'] = 1                   # Now two keys have the same value (= 1)
print(bd)                     # {'a': 1, 'c': 1, 'b': 2}
print(bd.inverse)             # {1: ['a', 'c'], 2: ['b']}
del bd['c']
print(bd)                     # {'a': 1, 'b': 2}
print(bd.inverse)             # {1: ['a'], 2: ['b']}
del bd['a']
print(bd)                     # {'b': 2}
print(bd.inverse)             # {2: ['b']}
bd['b'] = 3
print(bd)                     # {'b': 3}
print(bd.inverse)             # {2: [], 3: ['b']}