小编rug*_*tal的帖子

In Python 3, I defined two paths using pathlib, say:

from pathlib import Path

origin = Path('middle-earth/gondor/minas-tirith/castle').resolve()
destination = Path('middle-earth/gondor/osgiliath/tower').resolve()

How can I get the relative path that leads from origin to destination? In this example, I'd like a function that returns ../../osgiliath/tower or something equivalent.

Ideally, I'd have a function relative_path that always satisfies

origin.joinpath(
    relative_path(origin, destination)
).resolve() == destination.resolve()

(well, ideally there would be an operator - such that destination == origin / (destination - origin) …

题

根据此答案，在Python 3.5或更高版本中，可以合并两个字典x并y解压缩它们：

z = {**x, **y}

是否可以解压缩各种字典表？就像是

def merge(*dicts):
    return {***dicts} # this fails, of course. What should I use here?

例如，我希望

list_of_dicts = [{'a': 1, 'b': 2}, {'c': 3}, {'d': 4}]
{***list_of_dicts} == {'a': 1, 'b': 2, 'c': 3, 'd': 4}

请注意，此问题不是关于如何合并字典列表的问题，因为上面的链接提供了对此的答案。这里的问题是：是否有可能以及如何解压缩字典列表？

编辑

正如评论指出，这个问题是非常相似的这一个。但是，解压缩字典列表与简单地合并它们是不同的。假设有一个运算符***旨在解开字典列表，并给出

def print_values(a, b, c, d):
    print('a =', a)
    print('b =', b)
    print('c =', c)
    print('d =', d)

list_of_dicts = [{'a': 1, 'b': …