Implementing an Efficient Bidirectional Hash Table

A bidirectional hash table allows for both key-to-value and value-to-key lookups. While Python's built-in dict data structure excels at key-to-value lookups, it doesn't offer efficient value-to-key retrievals.

An effective method for implementing a bidirectional hash table is to utilize a class that extends the standard dict. This class, named bidict, maintains an inverse directory that automatically updates with any modifications to the regular dict.

Code Implementation:

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)

Key Features:

• The inverse directory (bd.inverse) is a dictionary that maps values to a list of keys with that value.
• The inverse directory automatically updates when the bidict is modified.
• Unlike some bidict implementations, this class allows multiple keys to have the same value.

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']}