另一种更实用的方法是:

my_map = { 'a': 1, 'b':2 }
dict(map(reversed, my_map.items()))

假设字典中的值是唯一的:

Python 3:

dict((v, k) for k, v in my_map.items())

Python 2:

dict((v, k) for k, v in my_map.iteritems())

字典值为集合的一种情况。如:

some_dict = {"1":{"a","b","c"},
        "2":{"d","e","f"},
        "3":{"g","h","i"}}

逆函数是:

some_dict = {vi: k  for k, v in some_dict.items() for vi in v}

输出如下:

{'c': '1',
 'b': '1',
 'a': '1',
 'f': '2',
 'd': '2',
 'e': '2',
 'g': '3',
 'h': '3',
 'i': '3'}

Python 3 +:

inv_map = {v: k for k, v in my_map.items()}

Python 2:

inv_map = {v: k for k, v in my_map.iteritems()}

我认为最好的方法是定义一个类。下面是一个“对称字典”的实现:

class SymDict:
    def __init__(self):
        self.aToB = {}
        self.bToA = {}

    def assocAB(self, a, b):
        # Stores and returns a tuple (a,b) of overwritten bindings
        currB = None
        if a in self.aToB: currB = self.bToA[a]
        currA = None
        if b in self.bToA: currA = self.aToB[b]

        self.aToB[a] = b
        self.bToA[b] = a
        return (currA, currB)

    def lookupA(self, a):
        if a in self.aToB:
            return self.aToB[a]
        return None

    def lookupB(self, b):
        if b in self.bToA:
            return self.bToA[b]
        return None

如果需要，删除和迭代方法很容易实现。

这种实现比颠倒整个字典(这似乎是本页上最流行的解决方案)更有效。更不用说，您可以随心所欲地从SymDict中添加或删除值，并且您的反向字典将始终保持有效——如果您只是将整个字典反向一次，这是不正确的。

函数对于list类型的值是对称的;执行reverse_dict(reverse_dict(dictionary))时，元组被转换为列表

def reverse_dict(dictionary):
    reverse_dict = {}
    for key, value in dictionary.iteritems():
        if not isinstance(value, (list, tuple)):
            value = [value]
        for val in value:
            reverse_dict[val] = reverse_dict.get(val, [])
            reverse_dict[val].append(key)
    for key, value in reverse_dict.iteritems():
        if len(value) == 1:
            reverse_dict[key] = value[0]
    return reverse_dict