Python2:获取最长的公共前缀路径

Question

我现正尝试在下列路径列表中找到最后一条匹配的公共路径：

"Linux-aarch64/gcc-7.3.0/aarch64-unknown-linux-gnu/include/c++/7.3.0/aarch64-unknown-linux-gnu/bits/" 
"Linux-aarch64/gcc-7.3.0/aarch64-test-linux-gnu/lib64/"
"Linux-aarch64/gcc-7.3.0/aarch64-unknown-bin"

也就是说，输出应该是Linux-aarch64/gcc-7.3.0

import os
dirs = ["Linux-aarch64/gcc-7.3.0/aarch64-unknown-linux-gnu/include/c++/7.3.0/aarch64-unknown-linux-gnu/bits/", "Linux-aarch64/gcc-7.3.0/aarch64-test-linux-gnu/lib64/", "Linux-aarch64/gcc-7.3.0/aarch64-unknown-bin"];

prefix = os.path.commonprefix(dirs)

print("Longest common path prefix:", prefix)

print prefix

它提供了正确的输出:- Linux-aarch64/gcc-7.3.0/aarch64-

但我期待着Linux-aarch64/gcc-7.3.0。上面的API似乎是逐个字符进行比较。还有其他建议吗？

Answer 1

您可以使用公共路径代替。

>>> os.path.commonpath(dirs)
'Linux-aarch64\\gcc-7.3.0'

也可以自己写：

>>> def longest_commmon_dir(dirs):
        paths = [path.split('/') for path in dirs]
        min_length = min(len(path) for path in paths)
        for i in range(min_length):
            if not all(path[i] == paths[0][i] for path in paths):
                break
        subdir = '/'.join(paths[0][:i])
        return subdir

>>> longest_commmon_dir(dirs)
'Linux-aarch64/gcc-7.3.0'

>>> longest_commmon_dir(['a', 'b'])
''