替换文件路径和扩展名元素的最快方法？

Question

给定文件路径：

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

如何快速替换为：

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

同时更改“newdirectory”的目录和“.jpg”的“.MOV”

Answer 1

好吧，这可以用不同的方法来做，但我会这样做

先换分机。这可以很容易地通过os.path.splitext实现，比如

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

这使路径成为

"/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。你知道吗

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

最后使用join，用您最喜欢的目录替换列表中的第二项，并用“/”作为分隔符

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

所以最后我们有

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

总而言之，可以归结为三行

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

我会利用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

结果：

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