我有一个字符串变量,它代表一个dos路径,例如:
var = “d:\stuff\morestuff\Furtherdown\THEFILE.txt”
我想把这个字符串分成:
[ “d”, “stuff”, “morestuff”, “Furtherdown”, “THEFILE.txt” ]
我尝试过使用split()和replace(),但它们要么只处理第一个反斜杠,要么将十六进制数字插入字符串。
我需要以某种方式将这个字符串变量转换为原始字符串,以便我可以解析它。
最好的方法是什么?
我还应该添加,var的内容,即我试图解析的路径,实际上是一个命令行查询的返回值。这不是我自己生成的路径数据。它存储在一个文件中,命令行工具不会转义反斜杠。
我曾多次被那些编写自己的路径篡改函数并出错的人所困扰。空格、斜杠、反斜杠、冒号——造成混淆的可能性不是无穷尽的,但无论如何都很容易犯错误。所以我很坚持使用操作系统。路径,并在此基础上推荐它。
(However, the path to virtue is not the one most easily taken, and many people when finding this are tempted to take a slippery path straight to damnation. They won't realise until one day everything falls to pieces, and they -- or, more likely, somebody else -- has to work out why everything has gone wrong, and it turns out somebody made a filename that mixes slashes and backslashes -- and some person suggests that the answer is "not to do that". Don't be any of these people. Except for the one who mixed up slashes and backslashes -- you could be them if you like.)
你可以像这样获得驱动器和路径+文件:
drive, path_and_file = os.path.splitdrive(path)
获取路径和文件:
path, file = os.path.split(path_and_file)
获取单个文件夹名称并不是特别方便,但这是一种诚实的中等不舒服,这增加了后来发现一些实际工作良好的东西的乐趣:
folders = []
while 1:
path, folder = os.path.split(path)
if folder != "":
folders.append(folder)
elif path != "":
folders.append(path)
break
folders.reverse()
(如果路径原本是绝对路径,则会在文件夹的开头弹出“\”。如果你不想这样做,你可能会丢失一些代码。)
你可以简单地使用最python化的方法(IMHO):
import os
your_path = r"d:\stuff\morestuff\furtherdown\THEFILE.txt"
path_list = your_path.split(os.sep)
print path_list
这将给你:
['d:', 'stuff', 'morestuff', 'furtherdown', 'THEFILE.txt']
这里的线索是使用操作系统。Sep而不是“\\”或“/”,因为这使得它与系统无关。
要从驱动器号中删除冒号(尽管我看不出你为什么要这样做),你可以这样写:
path_list[0] = path_list[0][0]
让我们假设你有一个文件filedata.txt的内容:
d:\stuff\morestuff\furtherdown\THEFILE.txt
d:\otherstuff\something\otherfile.txt
您可以读取和分割文件路径:
>>> for i in open("filedata.txt").readlines():
... print i.strip().split("\\")
...
['d:', 'stuff', 'morestuff', 'furtherdown', 'THEFILE.txt']
['d:', 'otherstuff', 'something', 'otherfile.txt']