python - 整数字符串列表 ['123 121','42 23','23 23']

Question

['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153'] 通过上面的列表，我想将元素转换为整数，我知道仅使用是[int(x) for x in mylist]行不通的。所以我的问题是如何将我拥有的列表变成整数列表。

Answer 1

>>> L = ['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153']
>>> [int(y) for x in L for y in x.split()]
[136, 145, 136, 149, 137, 145, 138, 145, 139, 145, 142, 149, 142, 153, 145, 153]

Answer 2

先拆分文本，然后转换为 int：

[map(int, elem.split()) for elem in originallist]

对于 Python 3，其中map()返回的是生成器，而不是列表，您可以嵌套列表推导：

[[int(n) for n in elem.split()] for elem in originallist]

这在 Python 2 下同样有效。

快速演示：

>>> originallist = ['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153']
>>> [[int(n) for n in elem.split()] for elem in originallist]
[[136, 145], [136, 149], [137, 145], [138, 145], [139, 145], [142, 149], [142, 153], [145, 153]]

elem.split()您可以通过将循环移动到外部列表理解来删除嵌套：

[int(n) for elem in originallist for n in elem.split()]

这使：

[136, 145, 136, 149, 137, 145, 138, 145, 139, 145, 142, 149, 142, 153, 145, 153]

Answer 3

由于我倾向于竭尽全力避免嵌套列表推导（我永远记不起顺序），因此我会执行以下操作：

from itertools import chain
x = ['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153']
gen = chain.from_iterable(elem.split() for elem in x)
integers = [int(elem) for elem in gen]

Answer 4

你可以这样试试

>>> import re
>>> l=['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153']    
>>> map(int, re.findall(r'\d+',' '.join(l)))
[136, 145, 136, 149, 137, 145, 138, 145, 139, 145, 142, 149, 142, 153, 145, 153]

Answer 5

>>> L = ['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153']
>>> map(int, ' '.join(L).split())
[136, 145, 136, 149, 137, 145, 138, 145, 139, 145, 142, 149, 142, 153, 145, 153]