Question

可能重复：
In Python, fastest way to build a list of files in a directory with a certain extension

我目前正在使用os.walk以递归方式扫描标识.MOV文件的目录

def fileList():
    matches = []
    for root, dirnames, filenames in os.walk(source):
        for filename in fnmatch.filter(filenames, '*.mov'):
            matches.append(os.path.join(root, filename))
    return matches

如何扩展它以识别多个文件，例如.mov，.MOV，.avi，.mpg？

感谢。

Answer 1

尝试这样的事情：

def fileList(source):
    matches = []
    for root, dirnames, filenames in os.walk(source):
        for filename in filenames:
            if filename.endswith(('.mov', '.MOV', '.avi', '.mpg')):
                matches.append(os.path.join(root, filename))
    return matches

Answer 2

使用os.path.splitext的替代方案：

for root, dirnames, filenames in os.walk(source):
    filenames = [ f for f in filenames if os.path.splitext(f)[1] in ('.mov', '.MOV', '.avi', '.mpg') ]
    for filename in filenames:
        matches.append(os.path.join(root, filename))
return matches

Answer 3

pattern = re.compile('.*\.(mov|MOV|avi|mpg)$')

def fileList(source):
   matches = []
   for root, dirnames, filenames in os.walk(source):
       for filename in filter(lambda name:pattern.match(name),filenames):
           matches.append(os.path.join(root, filename))
   return matches

使用Python os.walk来标识文件列表

3 个答案: