python列表解压字典

Question

d = {
  'a': [1, 2, 3],
  'b': [4, 5, 6]
}

def magic_fun(d): 
  ??

magic_fun(d) -> 
[[('a', 1), ('a', 2), ('a', 3)],
 [('b', 4), ('b', 5), ('b', 6)]]

Replace ?? with a python one-liner list comprenhension. Is it doable ?

If I do this:

>>> [(key, value) for key, values in d.iteritems() for value in values]
[('a', 1), ('a', 2), ('a', 3), ('b', 4), ('b', 5), ('b', 6)]

But I loose the first level of list. Everything is flattened...

PS: It is not really useful to me, I could do with a for-loop, but I am suprised I can't find a way to do it with a list-comprehension...

Answer 1

def magic_fun(d):
    return [[(key,el) for el in value] for key,value in d.items()]
    # for key,value in d.items():
    #     element_list = []
    #     for el in value:
    #         element_list.append((key,el))
    #     yield element_list