2

是否有一种方便的 Pythonic 方法可以通过搜索字符串拆分列表(即使列表包含非字符串并且具有嵌套列表)。例如,假设我想用“,”分割以下内容:

[[ 'something', ',', 'eh' ], ',', ['more'], ',', 'yet more', '|', 'even more' ]

这将变成:

[[[ 'something', ',', 'eh' ]], [['more']], ['yet more', '|', 'even more']]
4

2 回答 2

7

看看itertools.groupby

In [1]: from itertools import groupby

In [2]: lst = [[ 'something', ',', 'eh' ], ',', ['more'], ',', 'yet more', '|', 'even more' ]

In [3]: [list(group) for key, group in groupby(lst, lambda x: x!=',') if key]
Out[3]: [[['something', ',', 'eh']], [['more']], ['yet more', '|', 'even more']]

它基本上根据标准 () 将列表中的项目分成组,item != ','并且理解检查if k过滤掉那些组False - 即等于的项目','

In [4]: for key, group in groupby(lst, lambda x: x!=','):
   ...:     print key, list(group)
   ...:     
True [['something', ',', 'eh']]
False [',']
True [['more']]
False [',']
True ['yet more', '|', 'even more']
于 2013-04-19T17:31:28.900 回答
0

派对迟到了,但 FWIW,我认为结合哨兵价值itertools.takewhileiter与哨兵价值提供了一个快速的解决方案

from itertools import takewhile

z = [[ 'something', ',', 'eh' ], ',', ['more'], ',',
  'yet more', '|', 'even more' ]
z = iter(z)
def provider():
  return list(takewhile(lambda x: x != ',', z))

for i in iter(provider, []):
  print i

... 
[['something', ',', 'eh']]
[['more']]
['yet more', '|', 'even more']
于 2013-04-19T21:29:09.717 回答