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'}

试试python 2.7/3.x

inv_map={};
for i in my_map:
    inv_map[my_map[i]]=i    
print inv_map

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'}

即使在原始字典中有非唯一的值，这种方法也有效。

def dict_invert(d):
    '''
    d: dict
    Returns an inverted dictionary 
    '''
    # Your code here
    inv_d = {}
    for k, v in d.items():
        if v not in inv_d.keys():
            inv_d[v] = [k]
        else:
            inv_d[v].append(k)
        inv_d[v].sort()
        print(f"{inv_d[v]} are the values")
        
    return inv_d

如果my_map中的值不是唯一的:

Python 3:

inv_map = {}
for k, v in my_map.items():
    inv_map[v] = inv_map.get(v, []) + [k]

Python 2:

inv_map = {}
for k, v in my_map.iteritems():
    inv_map[v] = inv_map.get(v, []) + [k]

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()}