按索引列表拆分字符串
按索引列表拆分字符串
提问于 2012-06-01 21:43:23
我想通过索引列表拆分字符串，其中拆分的段从一个索引开始，在下一个索引之前结束。
示例：
s = 'long string that I want to split up'
indices = [0,5,12,17]
parts = [s[index:] for index in indices]
for part in parts:
print part这将返回：
我要拆分的
长字符串
我想拆分的字符串
我想要分开
我想分头行动
我正在尝试获取的：
long
字符串
那
我想拆分
回答 3
Stack Overflow用户
回答已采纳
发布于 2012-06-01 21:45:37
s = 'long string that I want to split up'
indices = [0,5,12,17]
parts = [s[i:j] for i,j in zip(indices, indices[1:]+[None])]返回
['long ', 'string ', 'that ', 'I want to split up']您可以使用以下命令进行打印：
print '\n'.join(parts)另一种可能性(不复制indices)是：
s = 'long string that I want to split up'
indices = [0,5,12,17]
indices.append(None)
parts = [s[indices[i]:indices[i+1]] for i in xrange(len(indices)-1)]Stack Overflow用户
发布于 2012-06-01 21:52:37
以下是大量使用itertools module的简短解决方案。tee函数用于对索引进行成对迭代。有关更多帮助，请参阅模块中的配方部分。
>>> from itertools import tee, izip_longest
>>> s = 'long string that I want to split up'
>>> indices = [0,5,12,17]
>>> start, end = tee(indices)
>>> next(end)
0
>>> [s[i:j] for i,j in izip_longest(start, end)]
['long ', 'string ', 'that ', 'I want to split up']编辑：这是一个不复制索引列表的版本，所以它应该更快。
Stack Overflow用户
发布于 2019-08-04 06:18:24
如果您不想对索引列表进行任何修改，您可以编写一个生成器：
>>> def split_by_idx(S, list_of_indices):
... left, right = 0, list_of_indices[0]
... yield S[left:right]
... left = right
... for right in list_of_indices[1:]:
... yield S[left:right]
... left = right
... yield S[left:]
...
>>>
>>>
>>> s = 'long string that I want to split up'
>>> indices = [5,12,17]
>>> [i for i in split_by_idx(s, indices)]
['long ', 'string ', 'that ', 'I want to split up']页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/10851445
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