Python开发餐食计划程序时如何避免用大量if语句校验多参数

方案1：逐项校验快速失败（最简单易读） #

无需嵌套if，对每一项限制单独判断，只要有一项不满足就直接跳过当前餐食：

for meal in meals:
    is_qualified = True
    # 过敏限制：用户选"是"则过滤含过敏原的餐食
    if restrictions[0] and meal.getAllergy():
        is_qualified = False
    # 健康限制：用户选"是"则过滤非健康餐食
    if restrictions[1] and not meal.getHealthy():
        is_qualified = False
    # 无麸质限制：用户选"是"则过滤含麸质的餐食
    if restrictions[2] and not meal.getGluten():
        is_qualified = False
    # 素食限制：用户选"是"则过滤非素食餐食
    if restrictions[3] and not meal.getVegitarian():
        is_qualified = False
    # 烹饪时间限制：用户没选无限制则过滤超时餐食
    if restrictions[4] != 1000 and meal.getCookTime() > restrictions[4]:
        is_qualified = False
    # 所有限制都满足才加入结果列表
    if is_qualified:
        mealsR.append(meal)

这个写法逻辑直观，没有任何嵌套，新增限制只要照着加一行if判断即可。

方案2：规则抽象（易扩展易维护） #

如果后续会频繁新增限制条件，可以把规则和校验逻辑抽象成列表，新增限制仅需添加规则项，无需修改校验逻辑：

def getRestrictedMeals(meals):
    mealsR = []
    # 定义规则列表：每一项为(用户限制值, 餐食校验函数)
    rules = []
    # 收集过敏限制
    print("Would you like only foods you aren't allergic to?\n1. Yes\n2. No")
    rules.append(
        (getRestHelp(int(input("Please choose an option: "))),
        lambda meal: not meal.getAllergy())
    )
    # 收集健康限制
    print("\nWould you like only healthy foods?\n1. Yes\n2. No")
    rules.append(
        (getRestHelp(int(input("Please choose an option: "))),
        lambda meal: meal.getHealthy())
    )
    # 收集无麸质限制
    print("\nWould you like only Gluten Free foods?\n1. Yes\n2. No")
    rules.append(
        (getRestHelp(int(input("Please choose an option: "))),
        lambda meal: meal.getGluten())
    )
    # 收集素食限制
    print("\nWould you like only Vegitarian foods?\n1. Yes\n2. No")
    rules.append(
        (getRestHelp(int(input("Please choose an option: "))),
        lambda meal: meal.getVegitarian())
    )
    # 收集烹饪时间限制
    print("\nWhat is the longest cook time you want?\nPlease enter 1000 for any cook time.")
    time_limit = int(input())
    rules.append(
        (time_limit,
        lambda meal: meal.getCookTime() <= time_limit if time_limit != 1000 else True)
    )

    # 统一校验所有餐食
    for meal in meals:
        valid = True
        for restrict_val, check in rules:
            # 用户未开启该限制则跳过校验
            if not restrict_val:
                continue
            # 校验不通过直接跳出
            if not check(meal):
                valid = False
                break
        if valid:
            mealsR.append(meal)
    return mealsR

额外优化建议 #

你当前代码中提前收集所有餐食属性到facts列表的逻辑可以删掉，直接在校验时取餐食对应属性即可，减少不必要的内存占用。

内容的提问来源于stack exchange，提问作者Alayna Juneau