从list - python中删除已定义的字符串

Question

foo = '/input/directory/'
faa = ['/input/directory/file1.txt', '/input/directory/file2.txt']

我需要从我的列表（faa）和文件扩展名中删除目录名称（即foo），只留下

bar = ['file1', 'file2']

Answer 1

使用map

bar = map(lambda x: '.'.join(x.replace(foo, '').split('.')[:-1]), faa)

Answer 2

尝试;

>>> [x.replace(foo, '').split('.')[0] for x in faa]
['file1', 'file2']

Answer 3

也许：

bar = [elem.replace(foo, '').rsplit('.', 1)[0] for elem in faa]

Answer 4

foo = '/input/directory/'
faa = ['/input/directory/file1.txt', '/input/directory/file2.txt']

import os.path

bar = [os.path.splitext(path.replace(foo, ''))[0]
       for path in faa]
print(bar)

或者没有foo：

faa = ['/input/directory/file1.txt', '/input/directory/file2.txt']

from os.path import basename, splitext
bar = [splitext(basename(path))[0]
       for path in faa]
print(bar)

Answer 5

您可以执行以下操作：

import os

foo = '/input/directory/'
faa = ['/input/directory/file1.txt', '/input/directory/file2.txt', '/not/matching/foo/file3.txt']   

faa = [os.path.splitext(os.path.split(f)[1])[0] if f.startswith(foo) else f for f in faa]   
print faa

这会给你以下列表：

['file1', 'file2', '/not/matching/foo/file3.txt']

如果你总是想要文件名，那么：

import os

foo = '/input/directory/'
faa = ['/input/directory/file1.txt', '/input/directory/file2.txt', '/not/matching/foo/file3.txt']   

faa = [os.path.splitext(os.path.split(f)[1])[0] for f in faa]   
print faa

赠送：

['file1', 'file2', 'file3']

Answer 6

您可以执行以下操作：

import re
foo = '/input/directory/'
faa = ['/input/directory/file1.txt', '/input/directory/file2.txt']

# Use a regular expression to match '<foo>someFileName<.fileExtension>'
faa = [re.sub(r'^%s(.*)\.\w+$' % foo, '\g<1>', elem) for elem in faa]

# faa => ['file1', 'file2']

Answer 7

你可以使用re模块和re.findall来实现那个，你的foo字符串后面有正面的外观(?<=) ，而对于带有锚到字符串末尾的extension ，你可以正向前看：

res = [re.findall('(?<={}).+(?=[.].*$)'.format(foo), elem)[0] for elem in faa]

print(res)
['file1', 'file2']

注意 ：您可以使用regex101.com检查该regex表达式