Python3遍历目录树实现方法
本文实例讲述了Python3遍历目录树的方法。分享给大家供大家参考。具体实现方法如下:
import os, fnmatch # 检查一个目录,后者某个包含子目录的目录树,并根据某种模式迭代所有文件 # patterns如:*.html,若大小写敏感可写*.[Hh][Tt][Mm][Ll] # single_level 为True表示只检查第一层 # yield_folders 表示是否显示子目录,为False只遍历子目录中的文件, # 但不返回字母名 def all_files(root, patterns='*', single_level=False, yield_folders=False): # 将模式从字符串中取出放入列表中 patterns = patterns.split(';') for path, subdirs, files in os.walk(root): if yield_folders: files.extend(subdirs) files.sort() for name in files: for pattern in patterns: if fnmatch.fnmatch(name, pattern): yield os.path.join(path, name) break if single_level: break for file in all_files('d:\\pm', '*.s;*.c', False, False): print(file)
希望本文所述对大家的Python3程序设计有所帮助。

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