按索引列表拆分字符串

Question

我想用索引列表分割一个字符串，在这个列表中，分割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

返回：

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

我想得到：

long
string
that
I want to split up

Answer 1

如果不想对索引列表进行任何修改，可以编写生成器：

>>> 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']

Answer 2

这里有一个简短的解决方案，其中大量使用了itertools module。函数tee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：此版本不复制索引列表，因此应该更快。

Answer 3

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])]

回报

['long ', 'string ', 'that ', 'I want to split up']

您可以使用：

print '\n'.join(parts)

另一种可能（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)]