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python - 给定两个字符串列表,如何将它们转换为字典?

转载 作者:太空宇宙 更新时间:2023-11-04 07:04:52 25 4
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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']
}

最佳答案

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。

关于python - 给定两个字符串列表,如何将它们转换为字典?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56519731/

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