如何高效合并两个不同结构的Python对象列表（按属性匹配）

需求：合并两个结构不同的Python类对象列表，将StructureTwo中的value匹配到StructureOne对应的date_time条目上，最终得到填充好value的StructureOne列表，要求不使用外部库。

原始实现代码 #

from datetime import datetime


class StructureOne(object):
    def __init__(self, date_time: datetime, name: str):
        self.date_time: datetime = date_time
        self.name: str = name
        self.value = None

    def set_value(self,value:float):
        self.value = value


class StructureTwo(object):
    def __init__(self, date_time: datetime, value: float):
        self.date_time = date_time
        self.value: float = value


def merge_lists(list_one: list[StructureOne], list_two: list[StructureTwo]) -> list[StructureOne]:
    for element_one in list_one:
        i = 0
        while i < len(list_two) and element_one.value is None:
            if element_one.date_time == list_two[i].date_time:
                element_one.set_value(value=list_two[i].value)
            i += 1
    return list_one


list_one: list[StructureOne] = [
    StructureOne(date_time=datetime(2022, 1, 1, 0), name='zero'),
    StructureOne(date_time=datetime(2022, 1, 1, 1), name='one'),
    StructureOne(date_time=datetime(2022, 1, 1, 2), name='two'),
    StructureOne(date_time=datetime(2022, 1, 1, 3), name='three'),
]

list_two: list[StructureTwo] = [
    StructureTwo(date_time=datetime(2022, 1, 1, 0), value=0),
    StructureTwo(date_time=datetime(2022, 1, 1, 1), value=1),
    StructureTwo(date_time=datetime(2022, 1, 1, 2), value=2),
    StructureTwo(date_time=datetime(2022, 1, 1, 3), value=3),
]


merged_list: list[StructureOne] = merge_lists(list_one=list_one, list_two=list_two)

原始实现的问题 #

原始方法采用嵌套循环，时间复杂度为O(n*m)（n是list_one长度，m是list_two长度），当列表规模较大时，效率会显著下降。

优化后的实现 #

核心思路是先将list_two转换为以date_time为键、value为值的字典，这样查找匹配项的时间复杂度降为O(1)，整体时间复杂度优化为O(n+m)，大幅提升效率：

from datetime import datetime


class StructureOne(object):
    def __init__(self, date_time: datetime, name: str):
        self.date_time: datetime = date_time
        self.name: str = name
        self.value = None

    def set_value(self, value: float):
        self.value = value

    # 可选：添加__repr__方法方便查看结果
    def __repr__(self):
        return f"StructureOne(date_time={self.date_time}, name='{self.name}', value={self.value})"


class StructureTwo(object):
    def __init__(self, date_time: datetime, value: float):
        self.date_time = date_time
        self.value: float = value


def merge_lists(list_one: list[StructureOne], list_two: list[StructureTwo]) -> list[StructureOne]:
    # 将list_two转换为date_time到value的映射字典
    value_map = {item.date_time: item.value for item in list_two}
    # 遍历list_one，直接通过字典匹配赋值
    for element in list_one:
        if element.date_time in value_map:
            element.set_value(value_map[element.date_time])
    return list_one


list_one: list[StructureOne] = [
    StructureOne(date_time=datetime(2022, 1, 1, 0), name='zero'),
    StructureOne(date_time=datetime(2022, 1, 1, 1), name='one'),
    StructureOne(date_time=datetime(2022, 1, 1, 2), name='two'),
    StructureOne(date_time=datetime(2022, 1, 1, 3), name='three'),
]

list_two: list[StructureTwo] = [
    StructureTwo(date_time=datetime(2022, 1, 1, 0), value=0),
    StructureTwo(date_time=datetime(2022, 1, 1, 1), value=1),
    StructureTwo(date_time=datetime(2022, 1, 1, 2), value=2),
    StructureTwo(date_time=datetime(2022, 1, 1, 3), value=3),
]

merged_list: list[StructureOne] = merge_lists(list_one=list_one, list_two=list_two)
print(merged_list)

验证结果 #

运行优化后的代码，输出结果与预期一致：

[
    StructureOne(date_time=2022-01-01 00:00:00, name='zero', value=0),
    StructureOne(date_time=2022-01-01 01:00:00, name='one', value=1),
    StructureOne(date_time=2022-01-01 02:00:00, name='two', value=2),
    StructureOne(date_time=2022-01-01 03:00:00, name='three', value=3)
]

内容的提问来源于stack exchange，提问作者Anabel