如何在不使用聚合函数的情况下按列id和manager_id分组

# 假设数据集存储在一个名为data的列表中，每个元素是一个字典，包含id和manager_id两列的值
data = [
    {"id": 1, "manager_id": 100, "name": "John"},
    {"id": 2, "manager_id": 100, "name": "Jane"},
    {"id": 3, "manager_id": 200, "name": "Mike"},
    {"id": 4, "manager_id": 200, "name": "Sarah"},
    {"id": 5, "manager_id": 100, "name": "Tom"},
    {"id": 6, "manager_id": 200, "name": "Emily"}
]

# 按id和manager_id排序
sorted_data = sorted(data, key=lambda x: (x["id"], x["manager_id"]))

# 创建一个空字典用于存储分组结果
groups = {}

# 遍历排序后的数据集
for row in sorted_data:
    key = (row["id"], row["manager_id"])
    
    # 检查字典中是否存在以当前id和manager_id为键的分组
    if key not in groups:
        groups[key] = []
    
    # 将当前行数据添加到分组中
    groups[key].append(row)

# 打印分组结果
for key, rows in groups.items():
    print("Group:", key)
    for row in rows:
        print(row)
    print()