python - 合并文件路径列表

Question

我有一个目录路径列表，我试图将其减少到可能的最小列表。例如:

bionic/main/installer-amd64/20101020ubuntu543/images/hd-media
bionic/main/installer-amd64/20101020ubuntu543/images/cdrom/xen
bionic/main/installer-amd64/20101020ubuntu543/images/cdrom
bionic/main/installer-amd64/20101020ubuntu543/images
bionic/main/installer-amd64/20101020ubuntu543

我可以摆脱 bionic/main/installer-amd64/20101020ubuntu543因为它可以折叠成 bionic/main/installer-amd64/20101020ubuntu543/images我也可以摆脱它，因为它可以折叠成 bionic/main/installer-amd64/20101020ubuntu543/images/cdrom然后折叠成 bionic/main/installer-amd64/20101020ubuntu543/images/cdrom/xen我知道有一种方法可以做到这一点，但我一生都无法弄清楚。

Answer 1

如果我正确理解您的问题，您可以将目录放入树结构中，然后从该树中获取叶节点，这些目录不是列表中另一个目录的父目录。

dirs = ["bionic/main/installer-amd64/20101020ubuntu543/images/hd-media",
        "bionic/main/installer-amd64/20101020ubuntu543/images/cdrom/xen",
        "bionic/main/installer-amd64/20101020ubuntu543/images/cdrom",
        "bionic/main/installer-amd64/20101020ubuntu543/images",
        "bionic/main/installer-amd64/20101020ubuntu543"]

tree = {}
for d in dirs:
    t = tree
    for x in d.split("/"):
        t = t.setdefault(x, {})
# {'bionic': {'main': {'installer-amd64': {'20101020ubuntu543': {'images': {'hd-media': {}, 'cdrom': {'xen': {}}}}}}}}

def leafs(tree):
    for x in tree:
        if tree[x]:
            for l in leafs(tree[x]):
                yield x + "/" + l
        else:
            yield x

for l in leafs(tree):
    print(l)
# bionic/main/installer-amd64/20101020ubuntu543/images/hd-media
# bionic/main/installer-amd64/20101020ubuntu543/images/cdrom/xen