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Python3遍历目录树实现方法

Jun 10, 2016 pm 03:11 PM
python3 Traverse

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) 

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希望本文所述对大家的Python3程序设计有所帮助。

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