Python从JSON列表中选择最佳4

我有一个排序的JSON列表，如下所示：

[{ "id": "1", "score": "5" },
{ "id": "1", "score": "4" },
{ "id": "2", "score": "9" },
{ "id": "2", "score": "8" },
{ "id": "3", "score": "99" }, 
{ "id": "3", "score": "98" }]

---

它按id排序，并根据id对分数进行排序。现在，我想选择每个id最好的4个分数，并将它们存储到一个新列表中。id的分数可能超过4分，也可能不超过4分。排序时间应该是O（n），知道吗？

根据

id

和

score

对列表进行排序，其中

0（n）

采用

id

属性，该属性也采用

0（n）

---

因为它已经按分数排序了，只需对它进行迭代，并为每个id获得最好的四个，就完成了，时间复杂度为

O（n）

以下是方法：

import itertools

new_lst = []
for _, g in itertools.groupby(lst, key=lambda x: x['id']):
    new_lst.extend(sorted(g, key=lambda x: x['score'], reverse=True)[:4])

不是真正的测试：

>>> lst = [{ "id": "1", "score": "5" },
{ "id": "1", "score": "4" },
{ "id": "2", "score": "9" },
{ "id": "2", "score": "8" },
{ "id": "3", "score": "99" }, 
{ "id": "3", "score": "98" }]
>>> new_lst = []
>>> for _, g in itertools.groupby(lst, key=lambda x: x['id']):
    new_lst.extend(sorted(g, key=lambda x: x['score'], reverse=True)[:4])

>>> new_lst
[{'id': '1', 'score': '5'}, {'id': '1', 'score': '4'}, {'id': '2', 'score': '9'}, {'id': '2', 'score': '8'}, {'id': '3', 'score': '99'}, {'id': '3', 'score': '98'}]

>>> lst = [{ "id": "1", "score": "5" },
{ "id": "1", "score": "4" },
{ "id": "2", "score": "9" },
{ "id": "2", "score": "8" },
{ "id": "3", "score": "99" }, 
{ "id": "3", "score": "98" }]
>>> new_lst = []
>>> for _, g in itertools.groupby(lst, key=lambda x: x['id']):
    new_lst.extend(sorted(g, key=lambda x: x['score'], reverse=True)[:4])

>>> new_lst
[{'id': '1', 'score': '5'}, {'id': '1', 'score': '4'}, {'id': '2', 'score': '9'}, {'id': '2', 'score': '8'}, {'id': '3', 'score': '99'}, {'id': '3', 'score': '98'}]