给定两个字符串列表，如何将它们转换为字典？

Question

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']

如何从它们创建字典：

{
'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']
}

Answer 1

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)}

时间

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，因此您可能应该在目标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 修改答案以使用zip 而不是使用索引从content 中提取值。

有趣的是，对@Michael_MacAskill 答案的字典理解并没有比使用普通 for 循环的结果更好。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，并为他们实施了一些性能改进。

大多数 Pythonic

如果速度差异并不重要，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。

Answer 2

dictionary = {}                                                         

for i in range(len(content)): 
   dictionary[labels[i]] = content[i] #setting each element in labels list as key and each corresponding index of content list's content as value

Answer 3

也许这可以通过字典理解更有效地完成，但这是一种快速而肮脏的方法：

d = {}                                                                                                                

for i, label in enumerate(labels): 
    d[label] = content[i]