如何获得路径的正确大小写? [英] How can I get the proper capitalization for a path?

问题描述

假设我有一个代表目录的类(当然是简化示例):

Let's say I have a class which represents a directory (simplified example of course):

import os
class Dir:
    def __init__(self, path):
        self.path = os.path.normcase(path)

为了使事情更容易在内部实现，我在呼叫path参数上的> os.path.normcase 保存到属性之前.这很好用，但是会小写路径:

To make things easier to implement internally, I am calling os.path.normcase on the path argument before I save it into an attribute. This works great, but it lowercases the path:

>>> import os
>>> os.path.normcase(r'C:\Python34\Lib')
'c:\\python34\\lib'
>>>

我想要一种将路径转换为正确的大写形式的C:\Python34\Lib的方法.我计划在__repr__方法中执行此操作，以便获得不错的输出，例如:

I would like a way to turn the path back into its properly capitalized form of C:\Python34\Lib. I plan to do this inside the __repr__ method so that I can get nice outputs such as:

>>> my_dir
Dir(r'C:\Python34\Lib')
>>>

当我在交互式口译员中时.标准库中有类似的东西吗?

when I am in the interactive interpreter. Is there anything like this in the standard library?

注意:我不是指用户作为path参数提供的字符串.如果用户这样做:

Note: I am not referring to the string that the user supplied as the path argument. If a user does:

my_dir = Dir('c:\PYTHON34\lib')

我仍然希望在解释器中打印Dir('C:\Python34\Lib')，因为这是正确的大写形式.基本上，我希望输出的路径与文件资源管理器中的路径相同.

I still want Dir('C:\Python34\Lib') to be printed in the interpreter because that is the proper capitalization. Basically, I want the outputed paths to be the same as they are in the file explorer.

推荐答案

更新:

对于使用新版Python的用户，新的 pathlib模块以 pathlib.Path.resolve 的形式具有此功能:

For those using the newer versions of Python, the new pathlib module possesses this functionality in the form of pathlib.Path.resolve:

>>> from pathlib import Path
>>> Path(r'c:\python34\lib').resolve()
WindowsPath('C:/Python34/Lib')
>>> str(Path(r'c:\python34\lib').resolve())
'C:\\Python34\\Lib'
>>>

因此，您可以将用户提供的路径存储为Path对象:

So, you could store the user-supplied path as a Path object:

from pathlib import Path
class Dir:
    def __init__(self, path):
        self.path = Path(path)

，然后像这样实现__repr__方法:

and then implement the __repr__ method like so:

def __repr__(self):
    return "Dir('{}')".format(self.path.resolve())

另外，由于Path对象直接支持不区分大小写的比较，因此我们不再需要os.path.normcase函数.

As an added bonus, we no longer need the os.path.normcase function since Path objects support case-insensitive comparisons directly.

pathlib的一个缺点是，它仅在Python 3.4(当前最新版本)中可用.因此，使用早期版本的用户将需要向后移植到其版本，或使用如下所示的os.path._getfinalpathname函数.

One downside to pathlib though is that it is only available in Python 3.4 (the currently newest version). So, those using earlier versions will need to either get a backport to their version or use the os.path._getfinalpathname function as demonstrated below.

在我浏览标准库时，我在名为_getfinalpathname的os.path模块中遇到了一个未记录的函数:

While I was digging through the standard library, I came across an undocumented function in the os.path module named _getfinalpathname:

>>> import os
>>> os.path._getfinalpathname(r'c:\python34\lib')
'\\\\?\\C:\\Python34\\Lib'
>>>

使用 str.lstrip ，我可以获得输出我需要:

Using str.lstrip, I can get the output I need:

>>> os.path._getfinalpathname(r'c:\python34\lib').lstrip(r'\?')
'C:\\Python34\\Lib'
>>>

此方法的唯一缺点是该函数没有文档说明，并且有些隐藏.但这现在适合我的需求(当然，如果您知道一种方法，我很想听听更好的方法:)

The only downside to this approach is that the function is undocumented and somewhat hidden. But it suits my needs for now (of course, I'd love to hear a better approach if you know of one :)

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