如何在python中从上到下遍历每个路径？

Question

给我一串这个

s='MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout|Custom'

在Python2.7中获得这样的列表最简单的方法是什么？在

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例如：

import re
for t in s.split('|'):
    print re.search(r'.*%s'%t,s).group()

但如果路径中的某些文件夹名称不唯一，则会中断，例如

s='MayaWindow|toolBar2|MainShelfLayout|formLayout14|formLayout14|Custom'

import re
for t in s.split('|'):
    print re.search(r'.*%s'%t,s).group()

MayaWindow
MayaWindow|toolBar2
MayaWindow|toolBar2|MainShelfLayout
MayaWindow|toolBar2|MainShelfLayout|formLayout14|formLayout14
MayaWindow|toolBar2|MainShelfLayout|formLayout14|formLayout14
MayaWindow|toolBar2|MainShelfLayout|formLayout14|formLayout14|Custom

Answer 1

在python3.2中，itertools.accumulate允许我们避免使用索引

from itertools import accumulate
s = 'MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout|Custom'

new_list = list(accumulate(s.split('|'), lambda x,y: '|'.join((x,y))))

print(new_list)

生产

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另一方面，在python2中，您可以沿着这条线做一些事情

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Answer 2

一句话：

>>> [s.rsplit("|", i)[0] for i in range(s.count("|"), -1, -1)]
['MayaWindow',
 'MayaWindow|toolBar2',
 'MayaWindow|toolBar2|MainShelfLayout',
 'MayaWindow|toolBar2|MainShelfLayout|formLayout14',
 'MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout',
 'MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout|Custom']

Answer 3

在感兴趣的分隔符上拆分它，然后在每个切片大小之间循环：

s = 'MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout|Custom'
l = s.split('|')
new_list = [l[:idx] for idx in range(1, len(l)+1)]

结果：

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或者，如果您想要字符串而不是lists，join()它们回到分隔符上：

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结果：

>>> for item in new_list:
...     print(item)
...
MayaWindow
MayaWindow|toolBar2
MayaWindow|toolBar2|MainShelfLayout
MayaWindow|toolBar2|MainShelfLayout|formLayout14
MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout
MayaWindow|toolBar2|MainShelfLayout|formLayout14|ShelfLayout|Custom