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How to implement an efficient bidirectional hash table?

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?

like image 903
Juanjo Conti Avatar asked Jul 23 '10 13:07

Juanjo Conti


4 Answers

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.
    1. The inverse directory bd.inverse[value] is always a list of key such that bd[key] == value.
    1. 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']}
like image 191
Basj Avatar answered Oct 16 '22 13:10

Basj


You can use the same dict itself by adding key,value pair in reverse order.

d={'a':1,'b':2}
revd=dict([reversed(i) for i in d.items()])
d.update(revd)
like image 33
Emil Avatar answered Oct 16 '22 12:10

Emil


A poor man's bidirectional hash table would be to use just two dictionaries (these are highly tuned datastructures already).

There is also a bidict package on the index:

  • https://pypi.python.org/pypi/bidict

The source for bidict can be found on github:

  • https://github.com/jab/bidict
like image 40
miku Avatar answered Oct 16 '22 11:10

miku


The below snippet of code implements an invertible (bijective) map:

class BijectionError(Exception):
    """Must set a unique value in a BijectiveMap."""

    def __init__(self, value):
        self.value = value
        msg = 'The value "{}" is already in the mapping.'
        super().__init__(msg.format(value))


class BijectiveMap(dict):
    """Invertible map."""

    def __init__(self, inverse=None):
        if inverse is None:
            inverse = self.__class__(inverse=self)
        self.inverse = inverse

    def __setitem__(self, key, value):
        if value in self.inverse:
            raise BijectionError(value)

        self.inverse._set_item(value, key)
        self._set_item(key, value)

    def __delitem__(self, key):
        self.inverse._del_item(self[key])
        self._del_item(key)

    def _del_item(self, key):
        super().__delitem__(key)

    def _set_item(self, key, value):
        super().__setitem__(key, value)

The advantage of this implementation is that the inverse attribute of a BijectiveMap is again a BijectiveMap. Therefore you can do things like:

>>> foo = BijectiveMap()
>>> foo['steve'] = 42
>>> foo.inverse
{42: 'steve'}
>>> foo.inverse.inverse
{'steve': 42}
>>> foo.inverse.inverse is foo
True
like image 7
jme Avatar answered Oct 16 '22 12:10

jme