Question

给定文件的路径：

file = "/directory/date/2011/2009-01-11 This is a file's path/file.jpg"

如何快速将其替换为：

new_file = "/newdirectory/date/2011/2009-01-11 This is a file's path/file.MOV"

为“.MOV”更改“newdirectory”和“.jpg”的目录

Answer 1

这可以用不同的方式完成，但这就是我要做的事情

首先更改扩展名。这可以通过像{/ 3>这样的os.path.splitext轻松完成

path = "/directory/date/2011/2009-01-11 This is a file's path/file.jpg"
new_file=os.path.splitext(path)[0]+".MOV"

这将路径设为

"/directory/date/2011/2009-01-11 This is a file's path/file.MOV"

现在要将目录更改为newdirectory，我们可以使用str.split, with maxsplit选项。

new_file=new_file.split('/',2)

最后使用join，用你喜欢的目录替换列表中的第二项，用'/'作为分隔符

new_file = '/'.join([new_file[0],"newdirectory",new_file[2]])

所以我们终于有了

"/newdirectory/date/2011/2009-01-11 This is a file's path/file.MOV"

总而言之，它归结为三行

new_file=os.path.splitext(path)[0]+".MOV"
new_file=new_file.split('/',2)
new_file = '/'.join([new_file[0],"newdirectory",new_file[2]])

Answer 2

我会使用os.sep

import os

path = "/directory/date/2011/2009-01-11 This is a file's path/file.jpg"

path = os.path.splitext(path)[0] + '.mov'

path = path.split(os.sep, 2)
path[1] = 'newdirectory'
path = os.sep.join(path)

print path

<强>结果：

/newdirectory/date/2011/2009-01-11 This is a file's path/file.mov

替换文件路径和扩展名元素的最快方法？

2 个答案: