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Python删除空文件和空文件夹的方法

Jun 10, 2016 pm 03:09 PM
python delete

本文实例讲述了Python删除空文件和空文件夹的方法。分享给大家供大家参考。具体实现方法如下:

#-*- coding:cp936 -*-
"""
os.walk()
函数声明:walk(top,topdown=True,onerror=None)
1>参数top表示需要遍历的目录树的路径
2>参数topdown的默认值是"True",表示首先返回目录树下的文件,然后在遍历目录树的子目录.Topdown的值为"False"时,
则表示先遍历目录树的子目录,返回子目录下的文件,最后返回根目录下的文件
3>参数onerror的默认值是"None",表示忽略文件遍历时产生的错误.如果不为空,则提供一个自定义函数提示错误信息后继续
遍历或抛出异常中止遍历
4>该函数返回一个元组,该元组有3个元素,这3个元素分别表示每次遍历的路径名,目录列表和文件列表
在2.5版中新增加了一个叫partition的函数,它可以做什么呢?举一个小例子:
>>> 'http://www.donews.net/limodou'.partition('://')
('http', '://', 'www.donews.net/limodou')
>>> 'file:/a.html'.partition('://')
('file:/a.html', ", ")
从第一个例子可以看出,它用来根据指定的分隔符将字符串进行分割,如果字符串包含指定的分隔符,则返回一个3元的tuple,
第一个为分隔符左边的子串,第二个为分隔符本身,第三个为分隔符右边的子串。第二个例子说明,如果找不到指定的分隔符,
则返回仍然是一个3元的tuple,第一个为整个字符串,第二和第三个为空串
"""
import os, sys
def remove_empty_dir(path):
 print path
 while(path[-1] == "\\"):
 path = path[:-1]
 print path
 a = {}
 for root, dirs, files in os.walk(path, False):
 if len(files) == 0:
  a[root] = 0
 else:
  for file in files: #对文件列表进行扫描
   try:
   fn = os.path.join(root, file) #将路径名和文件名拼接起来
   size = os.path.getsize(fn) #获取文件名大小
   if size != 0:
   b = root
   while(b != path):
    a[b] = 1
    b = b.rpartition("\\")[0] #保存上一级目录名
   a[path] = 1
   else: 
   try:
    os.remove(fn) #删除文件名为空的文件
    a[root] = 0
   except (WindowsError):
    b = root
    while(b != path):
    a[b] = 1
    b = b.rpartition("\\")[0]
    a[path] = 1
   except WindowsError:
   b = root
   while(b != path):
    a[b] = 1
    b = b.rpartition("\\")[0]
   a[path] = 1
   if a[root]:
   break;
 empty_dirs = []
 for i, j in a.iteritems():
 if j == 0:
  print i
  empty_dirs.insert(0, i)
 del a
 empty_dirs.sort(reverse=True) 
 for i in empty_dirs:
 try:
  os.rmdir(i)  #删除目录
  print "%s 删掉了!!" % (i)
 except (WindowsError):
  print "%s 删不掉!!" % (i)
if __name__ == "__main__":
 remove_empty_dir('f:\\b')

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希望本文所述对大家的Python程序设计有所帮助。

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