ints ['123 121'，'42 23'，'23 23']中的字符串列表

Question

['136 145', '136 149', '137 145', '138 145', '139 145', '142 149', '142 153', '145 153'] 与上面的列表一样，我想将元素转换为int，我知道仅仅使用[int(x) for x in mylist]将不起作用。所以我的问题是，你如何将我的列表变成一个整数列表。

Answer 1

拆分文本第一，然后转换成int：

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

对于Python 3，凡map()返回一个发电机，而不是一个列表，你可以嵌套列表理解：

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

这将在Python 2下同样工作。

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]] 

您可以通过移动elem.split()环路外列表理解删除筑巢，到最后：

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

这给：

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

Answer 2

>>> 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 3

正如我倾向于为了避免嵌套的列表解析（我永远不会记住顺序），我会做很多事情，比如：

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

你可以尝试这样的，

>>> 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]