如何将多值真值表转换为条件判断/表达式并生成Python函数

问题背景 #

有如下包含所有输入组合的多值真值表，输入字段（Location、Weather、Temperature、Time of Day）均有固定取值范围，需要实现一个高效无冗余的Python函数，输入指定条件后返回对应的Activity：

实现方案 #

方案1：字典映射（最简便高效） #

利用Python字典的O(1)查询特性，将输入的组合元组作为键，对应的Activity作为值存入字典。这种方式实现简单，查询速度快，完全无冗余，适合输入组合固定的场景。

# 预定义所有输入组合与Activity的映射
activity_map = {
    ("Indoors", "Sunny", "Hot", "Morning"): "Reading",
    ("Indoors", "Sunny", "Hot", "Evening"): "Watching TV",
    ("Indoors", "Sunny", "Cool", "Morning"): "Reading",
    ("Indoors", "Sunny", "Cool", "Evening"): "Watching TV",
    ("Indoors", "Rainy", "Hot", "Morning"): "Reading",
    ("Indoors", "Rainy", "Hot", "Evening"): "Watching TV",
    ("Indoors", "Rainy", "Cool", "Morning"): "Reading",
    ("Indoors", "Rainy", "Cool", "Evening"): "Watching TV",
    ("Outdoors", "Sunny", "Hot", "Morning"): "Gardening",
    ("Outdoors", "Sunny", "Hot", "Evening"): "Barbecue",
    ("Outdoors", "Sunny", "Cool", "Morning"): "Playing Sports",
    ("Outdoors", "Sunny", "Cool", "Evening"): "Barbecue",
    ("Outdoors", "Rainy", "Hot", "Morning"): "Shopping",
    ("Outdoors", "Rainy", "Hot", "Evening"): "Barbecue",
    ("Outdoors", "Rainy", "Cool", "Morning"): "Shopping",
    ("Outdoors", "Rainy", "Cool", "Evening"): "Barbecue",
    ("None", "Sunny", "Hot", "Morning"): "Reading",
    ("None", "Sunny", "Hot", "Evening"): "Barbecue",
    ("None", "Sunny", "Cool", "Morning"): "Reading",
    ("None", "Sunny", "Cool", "Evening"): "Shopping",
    ("None", "Rainy", "Hot", "Morning"): "Reading",
    ("None", "Rainy", "Hot", "Evening"): "Barbecue",
    ("None", "Rainy", "Cool", "Morning"): "Shopping",
    ("None", "Rainy", "Cool", "Evening"): "Shopping",
}

def get_activity(location, weather, temperature, time_of_day):
    # 输入合法性校验，避免非法参数
    valid_locations = {"Indoors", "Outdoors", "None"}
    valid_weathers = {"Sunny", "Rainy"}
    valid_temperatures = {"Hot", "Cool"}
    valid_times = {"Morning", "Evening"}
    
    if location not in valid_locations:
        raise ValueError(f"无效的Location：{location}，可选值为{valid_locations}")
    if weather not in valid_weathers:
        raise ValueError(f"无效的Weather：{weather}，可选值为{valid_weathers}")
    if temperature not in valid_temperatures:
        raise ValueError(f"无效的Temperature：{temperature}，可选值为{valid_temperatures}")
    if time_of_day not in valid_times:
        raise ValueError(f"无效的Time of Day：{time_of_day}，可选值为{valid_times}")
    
    return activity_map[(location, weather, temperature, time_of_day)]

方案2：规则提炼（精简代码） #

先分析真值表提炼重复逻辑，再基于规则编写函数，代码更简洁易维护，适合需要后续修改规则的场景。从表中可提炼出以下规则：

• Location为Indoors时，Activity仅由Time of Day决定：Morning是Reading，Evening是Watching TV，与其他输入无关
• Location为Outdoors时，Evening的Activity固定为Barbecue；Morning时，Sunny+Hot对应Gardening，Sunny+Cool对应Playing Sports，Rainy不管温度都是Shopping
• Location为None时，Morning的Activity由Weather决定：Sunny对应Reading，Rainy对应Shopping；Evening的Activity由Temperature决定：Hot对应Barbecue，Cool对应Shopping

def get_activity(location, weather, temperature, time_of_day):
    # 输入合法性校验
    valid_locations = {"Indoors", "Outdoors", "None"}
    valid_weathers = {"Sunny", "Rainy"}
    valid_temperatures = {"Hot", "Cool"}
    valid_times = {"Morning", "Evening"}
    
    if location not in valid_locations:
        raise ValueError(f"无效的Location：{location}，可选值为{valid_locations}")
    if weather not in valid_weathers:
        raise ValueError(f"无效的Weather：{weather}，可选值为{valid_weathers}")
    if temperature not in valid_temperatures:
        raise ValueError(f"无效的Temperature：{temperature}，可选值为{valid_temperatures}")
    if time_of_day not in valid_times:
        raise ValueError(f"无效的Time of Day：{time_of_day}，可选值为{valid_times}")
    
    if location == "Indoors":
        return "Reading" if time_of_day == "Morning" else "Watching TV"
    elif location == "Outdoors":
        if time_of_day == "Evening":
            return "Barbecue"
        else:
            if weather == "Sunny":
                return "Gardening" if temperature == "Hot" else "Playing Sports"
            else:
                return "Shopping"
    elif location == "None":
        if time_of_day == "Morning":
            return "Reading" if weather == "Sunny" else "Shopping"
        else:
            return "Barbecue" if temperature == "Hot" else "Shopping"

方案对比 #

• 字典映射：实现成本最低，查询速度最快，适合输入组合固定、无需频繁修改规则的场景。如果输入组合数量大，可以通过脚本读取CSV/表格文件自动生成字典，避免手动编写。
• 规则提炼：代码更精简，可读性和可维护性更强，适合能提炼出明确逻辑的场景，后续修改规则只需调整对应分支即可。

内容的提问来源于stack exchange，提问作者Saif