ints ['123 121','42 23','23 23']中的字符串列表
['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。ints ['123 121','42 23','23 23']中的字符串列表
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]
>>> 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]
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]
我想我只喜欢'chain.from_iterable'当它围绕整个表达式 – jamylak 2013-04-05 11:44:54
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]
你不需要这种方法的正则表达式 – jamylak 2013-04-05 11:48:59
>>> 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]
列表中的每个元素represends 2个整数... – 2013-04-05 11:39:02
你想要什么输出? – mgilson 2013-04-05 11:39:28