是否有从文件名中提取扩展名的功能?


当前回答

使用splitext时,具有双扩展名的文件存在问题(例如file.tar.gz、file.tar.bz2等)

>>> fileName, fileExtension = os.path.splitext('/path/to/somefile.tar.gz')
>>> fileExtension 
'.gz'

但应该是:.tar.gz

可能的解决方案如下

其他回答

上面的任何解决方案都有效,但在linux上,我发现扩展字符串末尾有一个换行符,这将阻止匹配成功。将strip()方法添加到末尾。例如:

import os.path
extension = os.path.splitext(filename)[1][1:].strip() 
import os.path
extension = os.path.splitext(filename)[1][1:]

只获取扩展名的文本,而不获取点。

a = ".bashrc"
b = "text.txt"
extension_a = a.split(".")
extension_b = b.split(".")
print(extension_a[-1])  # bashrc
print(extension_b[-1])  # txt

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

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])