blocks|key|2880131|text|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:]%2B[None])]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2880132|返回|unstyled|2880133|['long+',+'string+',+'that+',+'I+want+to+split+up']|2880134|您可以使用以下命令进行打印：|2880135|print+'\n'.join(parts)|2880136|另一种可能性(不复制indices)是：|offset|length|style|CODE|2880137|s+=+'long+string+that+I+want+to+split+up'
indices+=+[0,5,12,17]
indices.append(None)
parts+=+[s[indices[i]:indices[i%2B1]]+for+i+in+xrange(len(indices)-1)]|2880138|entityMap^0|0|0|0|0|0|A|7|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|11|8|@$O|12|P|13|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$B|C]]|$1|U|3|-4|5|F|7|15|8|@]|9|@]|A|$]]]|V|$]]

<pre><code>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])]
</code></pre>

returns 

<pre><code>['long ', 'string ', 'that ', 'I want to split up']
</code></pre>

which you can print using:

<pre><code>print '\n'.join(parts)
</code></pre>

Another possibility (without copying <code>indices</code>) would be:

<pre><code>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)]
</code></pre>

blocks|key|226515|text|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。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|226516|>>>+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']|code-block|syntax|javascript|226517|编辑：这是一个不复制索引列表的版本，所以它应该更快。|BOLD|226518|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html^0|V|3|7|G|0|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|M]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

Here is a short solution with heavy usage of the <a href="http://docs.python.org/library/itertools.html" rel="nofollow noreferrer">itertools module</a>. The <code>tee</code> function is used to iterate pairwise over the indices. See the Recipe section in the module for more help.

<pre><code>&gt;&gt;&gt; from itertools import tee, izip_longest
&gt;&gt;&gt; s = 'long string that I want to split up'
&gt;&gt;&gt; indices = [0,5,12,17]
&gt;&gt;&gt; start, end = tee(indices)
&gt;&gt;&gt; next(end)
0
&gt;&gt;&gt; [s[i:j] for i,j in izip_longest(start, end)]
['long ', 'string ', 'that ', 'I want to split up']
</code></pre>

Edit: This is a version that does not copy the indices list, so it should be faster.

blocks|key|2880203|text|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：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2880204|>>>+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']|code-block|syntax|javascript|2880205|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can write a generator if you don't want to make any modifications to the list of indices:

<pre><code>&gt;&gt;&gt; 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:]
... 
&gt;&gt;&gt; 
&gt;&gt;&gt; 
&gt;&gt;&gt; s = 'long string that I want to split up'
&gt;&gt;&gt; indices = [5,12,17]
&gt;&gt;&gt; [i for i in split_by_idx(s, indices)]
['long ', 'string ', 'that ', 'I want to split up']
</code></pre>

I want to split a string by a list of indices, where the split segments begin with one index and end before the next one. 

Example:

<pre><code>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
</code></pre>

This will return: 

<blockquote>
 long string that I want to split up 
 string that I want to split up 
 that I want to split up 
 I want to split up
</blockquote>

I'm trying to get:

<blockquote>
 long 
 string 
 that 
 I want to split up
</blockquote>

Splitting a string by list of indices

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 part...

问按索引列表拆分字符串
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