给定两个字符串列表,如何将它们转换为dict?

ano*_*non 1 python list-comprehension data-structures python-3.x

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']
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如何从他们创建字典:

{
'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']
}
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lig*_*ist 5

dictionary = dict(zip(labels, content))
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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)}
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定时

这些已使用Python 3.6.7进行了测试。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的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)
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最快的

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

有趣的是,对@Michael_MacAskill的答案的字典理解并不比使用普通for循环的字典表现更好。也许这种语言的实现者意识到人们在大多数时间仍然坚持使用for循环,并为他们实施了一些性能改进措施。

大多数Pythonic

dict(zip(labels, content))如果速度差异不是很关键,那么大多数有经验的Python程序员可能会选择该选项,因为这是该语言中的一个常见习惯。