python合并两个列表成字典

问题

有两个列表如下：

keys = [1, 2, 3]
values = ['A', 'B', 'C']

最终结果，生成如下字典：

d = {1: 'A', 2: 'B', 3: 'C'}

解决办法

用zip函数生成一个列表（每个元素都是输入列表对应元素组成的元组），然后再用dict构造字典。

# Good

keys = [1, 2, 3]
values = ['A', 'B', 'C']

d = dict(zip(keys, values))

上面的办法要优于下面的办法：

# bad

keys = [1, 2, 3]
values = ['A', 'B', 'C']

d = {}
for i, key in enumerate(keys):
    d[key] = values[i]

讨论

zip函数的原型如下：

def zip(seq1, seq2, *more_seqs):
    """
    zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]
    
    Return a list of tuples, where each tuple contains the i-th element
    from each of the argument sequences.  The returned list is truncated
    in length to the length of the shortest argument sequence.
    """
    pass

dict构造字典有以下几种方式：

class dict(object):
    """
    dict() -> new empty dictionary
    dict(mapping) -> new dictionary initialized from a mapping object's
        (key, value) pairs
    dict(iterable) -> new dictionary initialized as if via:
        d = {}
        for k, v in iterable:
            d[k] = v
    dict(**kwargs) -> new dictionary initialized with the name=value pairs
        in the keyword argument list.  For example:  dict(one=1, two=2)
    """
    pass