python之给定两个字符串列表，如何将它们转换为字典

我有以下字符串列表:

content = [['a list with a lot of strings and chars 1'], ['a list with a lot of strings and chars 2'], ['a list with a lot of strings and chars 3'], ['a list with a lot of strings and chars 4']] 
 
labels = ['label_1','label_2','label_3','label_4'] 

我如何根据它们创建字典:

{ 
'label_1': ['a list with a lot of strings and chars 1'] 
'label_2': ['a list with a lot of strings and chars 2'] 
'label_3': ['a list with a lot of strings and chars 3'] 
'label_4': ['a list with a lot of strings and chars 4'] 
} 

请您参考如下方法:

dictionary = dict(zip(labels, content)) 

各种版本:

def f1(labels, content): 
    return dict(zip(labels, content)) 
 
def f2(labels, content): 
    d = {} 
 
    for i, label in enumerate(labels): 
        d[label] = content[i] 
    return d 
 
def f3(labels, content): 
    d = {} 
    for l, c in zip(labels, content): 
        d[l] = c 
    return d 
 
def f4(labels, content): 
    return {l : c for (l, c) in zip(labels, content)} 
 
def f5(labels, content): 
    dictionary = {} 
 
    for i in range(len(content)): 
       dictionary[labels[i]] = content[i] 
    return dictionary 
 
def f6(labels, content): 
    return {l : content[i] for (i, l) in enumerate(labels)} 

时机

这些是使用 Python 3.6.7 测试的。请注意，不同版本的 Python 可能具有不同的性能，因此您可能应该在预期平台上重新运行基准测试。

In [20]: %timeit f1(labels, content) 
637 ns ± 4.17 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 
 
In [21]: %timeit f2(labels, content) 
474 ns ± 4.44 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 
 
In [22]: %timeit f3(labels, content) 
447 ns ± 2.76 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 
 
In [23]: %timeit f4(labels, content) 
517 ns ± 4.44 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 
 
In [24]: %timeit f5(labels, content) 
529 ns ± 8.04 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 
 
In [4]: %timeit f6(labels, content) 
602 ns ± 0.64 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 

最快

最快的是 f3，@Michael_MacAskill 对答案的修改，使用 zip 而不是使用索引从 content 中提取值.

有趣的是，对@Michael_MacAskill 的答案的字典理解并不比使用普通 for 循环的更好。也许该语言的实现者意识到人们大部分时间仍然坚持使用 for 循环，并为他们实现了一些性能改进。

最Pythonic

如果速度差异不是关键，大多数有经验的 Python 程序员可能会选择 dict(zip(labels, content)) 选项，因为它是语言中的常见习语。