Question

我试图编写一个函数来反转字典，但是我在找不到合适的方法而没有重写代码，使用不同的方法并在每次迭代时避免使用if / else时遇到了麻烦。什么是最蟒蛇的方式呢？

def invert_dict(dic, type=None):
    if type == 'list':
        return _invert_dict_list(dic)
    return _invert_dict(dic)


# if there's only one value per key
def _invert_dict(dic):
    inverted = defaultdict()

    for k,v in dic.items():
        for item in v:
            inverted[item]=k
    return dict(inverted)


# if there are multiple values for the same key
def _invert_dict_list(dic):
    inverted = defaultdict(list)

    for k,v in dic.items():
        for item in v:
            inverted[item].append(k)
    return dict(inverted)

Answer 1

我不会评论实际的建议，但对于基于类型的分支，有functools.singledispatch：

import functools

@functools.singledispatch
def inv_item(value, key, dest):
    < fallback implementation >

# special case based on type
@inv_item.register(list)
@inv_item.register(tuple)
def inv_sequence(value, key, dest):
    < handle sequence values >

...

def invert_dict(In):
    Out = {}
    for k, v in In.items():
        inv_item(v, k, Out)
    return Out

Answer 2

您可以使用itertools.groupby和词典理解：

import itertools 
d1 = {'val1':[4, 5, 2, 4], 13:'v2', 'val2':'v2', 'age':17}
new_d = [(a, list(b)) for a, b in itertools.groupby(sorted([(b, a) for a, b in d1.items()], key=lambda x:x[0]), key=lambda x:x[0])]
final_d = {tuple(a) if isinstance(a, list) else a:[i[-1] for i in b][0] if len([i[-1] for i in b]) == 1 else [i[-1] for i in b] for a, b in new_d}

输出：

{(4, 5, 2, 4): 'val1', 17: 'age', 'v2': ['val2', 13]}

Pythonic方式处理函数重载

2 个答案: