有一个成员是字典的列表，如何把其中 key 相同的字典的 value 合并一下

[aixia]

For example

d 合并后变成 new_d

d = [
 {1: [12, 24]},
 {1: [24, 36]},
 {2: [111,222]}
 ]

new_d = [
 {1: [12, 24, 36]},
 {2: [111, 222]}
 ]

[SuperMild]

既然是列表，貌似一般就是遍历了吧，用 set 来做字典的 value 。最后再遍历一次把 set 转换为列表。

[sagaxu]

s = {k: set() for x in d for k,v in x.items()}
[s[k].update(v) for x in d for k,v in x.items()]
{k: list(v) for k,v in s.items()}

[freestyle]

https://gist.github.com/hanjm/883f2544d05a0750cb733221496ace37

[popbones]

[ {id: val} for (id, val) in [[(new_d, new_d.update({k: new_d.get(k, set()).union(set(v))})) for (k, v) in m.items()] for m in d][0][0][0].items() ]

[popbones]

new_d = {}

[TJT]

另外一种思路，前提是 d 得排序好

from itertools import groupby, chain
[{k: [x for x, _ in groupby(sorted(chain(*[v[k] for v in g])))]} for k, g in groupby(d, lambda x: list(x.keys())[0])]

[mingyun]

@sagaxu 厉害了