是否有从文件名中提取扩展名的功能?


当前回答

如果你喜欢正则表达式,那就是一个真正的单行。即使中间有额外的“.”也无关紧要

import re

file_ext = re.search(r"\.([^.]+)$", filename).group(1)

查看此处查看结果:单击此处

其他回答

从Python中的文件名提取扩展名

Python os模块splitext()

splitext()函数将文件路径拆分为具有两个值的元组:root和extension。

import os
# unpacking the tuple
file_name, file_extension = os.path.splitext("/Users/Username/abc.txt")
print(file_name)
print(file_extension)

使用Pathlib模块获取文件扩展名

获取文件扩展名的Pathlib模块

import pathlib
pathlib.Path("/Users/pankaj/abc.txt").suffix
#output:'.txt'

最简单的获取方法是使用mimtypes,下面是示例:

import mimetypes

mt = mimetypes.guess_type("file name")
file_extension =  mt[0]
print(file_extension)

另一种右拆分解决方案:

# to get extension only

s = 'test.ext'

if '.' in s: ext = s.rsplit('.', 1)[1]

# or, to get file name and extension

def split_filepath(s):
    """
    get filename and extension from filepath 
    filepath -> (filename, extension)
    """
    if not '.' in s: return (s, '')
    r = s.rsplit('.', 1)
    return (r[0], r[1])

如果您想提取最后一个文件扩展名,如果它有多个

class functions:
    def listdir(self, filepath):
        return os.listdir(filepath)
    
func = functions()

os.chdir("C:\\Users\Asus-pc\Downloads") #absolute path, change this to your directory
current_dir = os.getcwd()

for i in range(len(func.listdir(current_dir))): #i is set to numbers of files and directories on path directory
    if os.path.isfile((func.listdir(current_dir))[i]): #check if it is a file
        fileName = func.listdir(current_dir)[i] #put the current filename into a variable
        rev_fileName = fileName[::-1] #reverse the filename
        currentFileExtension = rev_fileName[:rev_fileName.index('.')][::-1] #extract from beginning until before .
        print(currentFileExtension) #output can be mp3,pdf,ini,exe, depends on the file on your absolute directory

输出为mp3,即使只有一个扩展名也能正常工作

即使这个问题已经得到了回答,我也会在Regex中添加解决方案。

>>> import re
>>> file_suffix = ".*(\..*)"
>>> result = re.search(file_suffix, "somefile.ext")
>>> result.group(1)
'.ext'