如何避免Python的os.path.commonprefix的谬误? [英] How to circumvent the fallacy of Python's os.path.commonprefix?

问题描述

我的问题是找到给定文件集的通用 path 前缀.

My problem is to find the common path prefix of a given set of files.

从字面上看，我期望"os.path.commonprefix"能够做到这一点.不幸的是，commonprefix位于path中的事实颇具误导性，因为它实际上会搜索字符串前缀.

Literally I was expecting that "os.path.commonprefix" would do just that. Unfortunately, the fact that commonprefix is located in path is rather misleading, since it actually will search for string prefixes.

我的问题是，如何才能真正解决路径问题?该问题在此答案(评级很高)中被简短提及，但仅作为旁注和建议的解决方案(附加斜线输入commonprefix的输入)，恕我直言有问题，因为它将失败，例如:

The question to me is, how can this actually be solved for paths? The issue was briefly mentioned in this (fairly high rated) answer but only as a side-note and the proposed solution (appending slashes to the input of commonprefix) imho has issues, since it will fail for instance for:

os.path.commonprefix(['/usr/var1/log/', '/usr/var2/log/'])
# returns /usr/var but it should be /usr

为防止其他人陷入同一陷阱，可能有必要在一个单独的问题中进行讨论:是否有一个不依赖于文件系统讨厌检查的简单/可移植解决方案来解决此问题(即，访问commonprefix的结果，并检查它是否是目录，如果不是，则返回结果的os.path.dirname?

To prevent others from falling into the same trap, it might be worthwhile to discuss this issue in a separate question: Is there a simple / portable solution for this problem that does not rely on nasty checks on the file system (i.e., access the result of commonprefix and check whether it is a directory and if not returns a os.path.dirname of the result)?

推荐答案

前一段时间，我遇到了这个问题，其中os.path.commonprefix是字符串前缀，而不是预期的路径前缀.所以我写了以下内容:

Awhile ago I ran into this where os.path.commonprefix is a string prefix and not a path prefix as would be expected. So I wrote the following:

def commonprefix(l):
    # this unlike the os.path.commonprefix version
    # always returns path prefixes as it compares
    # path component wise
    cp = []
    ls = [p.split('/') for p in l]
    ml = min( len(p) for p in ls )

    for i in range(ml):

        s = set( p[i] for p in ls )         
        if len(s) != 1:
            break

        cp.append(s.pop())

    return '/'.join(cp)

通过将'/'替换为os.path.sep，可以使其具有更高的可移植性.

it could be made more portable by replacing '/' with os.path.sep.

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