删除定义字符串 - 蟒蛇

Question

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

我需要从我的名单（FAA）和文件扩展名删除目录名（例如，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

你可以做到以下几点：

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 

这将使你在下面的列表：

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

如果你总是想那么文件名：

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 

，并提供：

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

Answer 3

尝试;

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

Answer 4

你可以做到以下几点：

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 5

可能：

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

Answer 6

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) 

或者不富：

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 7

你凑LD使用re模块和re.findall实现与背后(?<=)正面看你foo串并提前积极找extension锚字符串的结尾：

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

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

注：你可以检查regex表达与regex101.com