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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:

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#-*- 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')

Copy after login

希望本文所述对大家的Python程序设计有所帮助。

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