Python 目录操作

知道两个文件的绝对目录，怎样计算出两个文件的相对目录，例如：知道
　　
a = '/usr/share/pyshared/test/a.py'
b = '/usr/lib/dist/test/a.py'

可以直接应用OS模块中的os.path.reppath得到结果：

　　>>> print(os.path.relpath(a,b))
　　........sharepyshared esta.py

  但是计算结果是怎样实现的呢？

def _abspath_split(path):
    abs = abspath(normpath(path))
    prefix, rest = splitunc(abs)
    is_unc = bool(prefix)
    if not is_unc:
        prefix, rest = splitdrive(abs)
    return is_unc, prefix, [x for x in rest.split(sep) if x]

def relpath(path, start=curdir):
    """Return a relative version of a path"""

    if not path:
        raise ValueError("no path specified")

    start_is_unc, start_prefix, start_list = _abspath_split(start)
    path_is_unc, path_prefix, path_list = _abspath_split(path)

    if path_is_unc ^ start_is_unc:
        raise ValueError("Cannot mix UNC and non-UNC paths (%s and %s)"
                                                            % (path, start))
    if path_prefix.lower() != start_prefix.lower():
        if path_is_unc:
            raise ValueError("path is on UNC root %s, start on UNC root %s"
                                                % (path_prefix, start_prefix))
        else:
            raise ValueError("path is on drive %s, start on drive %s"
                                                % (path_prefix, start_prefix))
    # Work out how much of the filepath is shared by start and path.
    i = 0
    for e1, e2 in zip(start_list, path_list):
        if e1.lower() != e2.lower():
            break
        i += 1

    rel_list = [pardir] * (len(start_list)-i) + path_list[i:]
    if not rel_list:
        return curdir
    return join(*rel_list)

上述代码是relpath在Python中的内部实现，实现比较复杂，作为简单的可替代形式：

def ownrelpath(a,b):
    relpath = '';
    la = a.split('/');
    lb = b.split('/');
    L = len(lb)-1;
    tt = '';
    for i in range(len(la)-1):
        if la[i] == lb[i] and tt =='':
            continue;
        else:
            tt = tt+la[i]+"\";
            relpath =  "..\"*(L-1);
    print(relpath+tt);