Question

我想拆分＆＃34; C：\ My Work \ Tester Related \ A_B_C.txt＆＃34; 在Python中使用[C：\，My Work，Tester Related，A，B，C，txt]。

我刚开始学习Python。我确实导入了os.path模块，然后玩了交互式提示中的某些功能但无法获得所需的结果。谢谢您的帮助。我能够使用os.path.basename来获取文件，但是无法根据给定的分隔符＆＃39; _＆＃39;将文件拆分为组件。

Answer 1

s = "C:\\My Work\\Tester Related\\A_B_C.txt"

import re
print (re.split(r"\\|\.|\_",s))
['C:', 'My Work', 'Tester Related', 'A', 'B', 'C', 'txt']

Answer 2

original_path = "C:\My Work\Tester Related\A_B_C.txt"

# Must quote the backslash in this case.
split_path = original_path.split("\\")

# At this point, split_path looks like:
# ['C:', 'My Work', 'Tester Related', 'A_B_C.txt']

split_filename = split_path[-1].split("_")
# split_filename is ['A', 'B', 'C.txt']

split_ext = split_filename[:-1].split(".")
# split_ext is ['C', 'txt']

split_filename = split_filename[:-1] + split_ext
# split_filename is ['A', 'B', 'C', 'txt']

split_path = split_path[:-1] + split_filename
# ['C:', 'My Work', 'Tester Related', 'A', 'B', 'C', 'txt']

它有点hacky但它会起作用。

Answer 3

一个丑陋的两个班轮！

import itertools
txt = r"C:\\My Work\\Tester Related\\A_B_C.txt"

result = list(itertools.chain(*map(lambda x: x.split('_'), txt.split(os.path.sep))))
result = result[:-1] + list(os.path.splitext(result[-1]))

更好的方法是使用re，正如@PaedricCunningham在答案中所做的那样。

re.split(r"\\|\.|_", txt)

如何在python中将文件路径和文件名拆分为组件？

3 个答案: