如何优雅地将多个字典里中相同键的值整合成列表

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示例字典：

d = [
    {"time": "09:00", "dt":{"key1": "a", "key2": "d", "key3": "g"}},
    {"time": "10:00", "dt":{"key1": "b", "key2": "e", "key3": "h"}}, 
    {"time":"11:00", "dt":{"key1": "c", "key2":"f", "key3":"i"}}
]

如果要取出所有字典中“time”， “kye1”，”key2”和”key3”并组成一个列表该怎样做呢？一般来说都是初始化空列表然后遍历所有字典，获取对应的值再填充。这样的方法简单粗暴，现在介绍一种优雅地方式，虽然可能比较花哨。

>>> time, key1, key2, key3 = zip(*map(lambda x: (x['time'], x['dt']['key1'], x['dt']['key2'], x['dt']['key3']), d))
>>> time
('09:00', '10:00', '11:00')
>>> key1
('a', 'b', 'c')
>>> key2
('d', 'e', 'f')
>>> key3
('g', 'h', 'i')