Python groupby

[DenisSouth]

data = [{"name": "A", "group_id": None},
        {"name": "B", "group_id": None},
        {"name": "C", "group_id": None},
        {"name": "D", "group_id": 123},
        {"name": "E", "group_id": 123},
        {"name": "F", "group_id": None},
        {"name": "G", "group_id": 234},
        {"name": "H", "group_id": 123},
        {"name": "M", "group_id": 234}]

grouped_data = [
    [{"name": "A", "group_id": None}],
    [{"name": "B", "group_id": None}],
    [{"name": "C", "group_id": None}],
    [{"name": "D", "group_id": 123}, {"name": "E", "group_id": 123}, {"name": "H", "group_id": 123}],
    [{"name": "F", "group_id": None}],
    [{"name": "G", "group_id": 234}, {"name": "M", "group_id": 234}]
]

my solution is

data = sorted(data, key=lambda x: x['group_id'] or 0)
rez = []
for key, group in itertools.groupby(data, lambda x: x['group_id']):
    if key is None:
        rez = list(group)
    else:
        rez.append(list(group))

print(*rez, sep="\n")

I think it is not ideal, how can I make 'data' out of 'grouped_data' more efficient?

[James159758]

dict_a = {i:[j for j in data if j['group_id'] == i] for i in set(i['group_id'] for i in data)} print(*list(dict_a.values()),sep='\n')

[James159758]

print(*dict_a.values(), sep='\n')

[Blind4Basics]

bad time complexity...

[James159758]

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