Question

我有一个函数可以返回价格等于或低于指定值的餐馆。这是我目前的代码：

Restaurant = namedtuple('Restaurant', 'name cuisine phone menu')
Dish = namedtuple('Dish', 'name price calories')

r1 = Restaurant('Thai Dishes', 'Thai', '334-4433', [Dish('Mee Krob', 12.50, 500),
                                                Dish('Larb Gai', 11.00, 450)])
r2 = Restaurant('Taillevent', 'French', '01-44-95-15-01', 
            [Dish('Homard Bleu', 45.00, 750),
             Dish('Tournedos Rossini', 65.00, 950),
             Dish("Selle d'Agneau", 60.00, 850)])


collection =[r1,r2]


def Collection_is_cheap(C, price):
    result = []
    if not C:
       return ''
    else:
        for rest in C:
            for dish in rest.menu:
                if dish.price <= price:
                    result.append(rest)
    return result

但是当我尝试运行它时：

print(Collection_is_cheap(collection, 28))

我得到的是一长串正确的餐馆，但重复了一遍。

[Restaurant(name='Thai Dishes', cuisine='Thai', phone='334-4433', menu=[Dish(name='Mee Krob', price=12.5, calories=500), Dish(name='Larb Gai', price=11.0, calories=450)]), Restaurant(name='Thai Dishes', cuisine='Thai', phone='334-4433', menu=[Dish(name='Mee Krob', price=12.5, calories=500), Dish(name='Larb Gai', price=11.0, calories=450)]), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=[Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)]), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=[Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)]), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=[Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)]), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=[Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)])]

对于正确的输出，它应该只打印两个餐馆一次。如何更正此问题，以便该函数仅返回：

[Restaurant(name='Thai Dishes', cuisine='Thai', phone='334-4433', menu=[Dish(name='Mee Krob', price=12.5, calories=500), Dish(name='Larb Gai', price=11.0, calories=450)]), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=[Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)]

Answer 1

当你有比赛时，只需停止在餐厅菜单上循环;使用break：

def Collection_is_cheap(C, price):
    result = []
    for rest in C:
        for dish in rest.menu:
            if dish.price <= price:
                result.append(rest)
                break  # stop the rest.menu loop, go to the next
    return result

请注意，我删除了if not C: return ''部分;最好不要从函数中返回不同类型的对象。

演示：

>>> def Collection_is_cheap(C, price):
...     result = []
...     for rest in C:
...         for dish in rest.menu:
...             if dish.price <= price:
...                 result.append(rest)
...                 break  # stop the rest.menu loop, go to the next
...     return result
... 
>>> print(Collection_is_cheap(collection, 28))
[Restaurant(name='Thai Dishes', cuisine='Thai', phone='334-4433', menu=(Dish(name='Mee Krob', price=12.5, calories=500), Dish(name='Larb Gai', price=11.0, calories=450))), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=(Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)))]

另一种方法是使用 set 而不是列表;集合只能容纳唯一对象，因此不止一次添加餐馆不会产生任何影响：

def Collection_is_cheap(C, price):
    result = set()
    for rest in C:
        for dish in rest.menu:
            if dish.price <= price:
                result.add(rest)
    return list(result)

为了实现这个目的，你还需要让菜单使用元组，而不是列表：

r1 = Restaurant('Thai Dishes', 'Thai', '334-4433', (
    Dish('Mee Krob', 12.50, 500),
    Dish('Larb Gai', 11.00, 450)))

r2 = Restaurant('Taillevent', 'French', '01-44-95-15-01', ( 
    Dish('Homard Bleu', 45.00, 750),
    Dish('Tournedos Rossini', 65.00, 950),
    Dish("Selle d'Agneau", 60.00, 850)))

r3 = Restaurant('Pascal', 'French', '940-752-0107', (
    Dish('Escargots', 12.95, 250),
    Dish('Poached salmon', 18.50, 550),
    Dish("Rack of lamb", 24.00, 850),
    Dish("Marjolaine cake", 8.50, 950)))

因此它们完全不可变，需要使用集合。

使用set仅收集唯一餐馆的另一个影响是，返回的餐馆的顺序可能会发生变化，因为集合是无序的。相反，它们以依赖于实现的顺序保存对象，这有利于有效地测试已存在的对象。

演示;对于这个简单的例子，只有两个便宜的餐馆，订单发生以匹配第一个版本返回的：

>>> def Collection_is_cheap(C, price):
...     result = set()
...     for rest in C:
...         for dish in rest.menu:
...             if dish.price <= price:
...                 result.add(rest)
...     return list(result)
... 
>>> print(Collection_is_cheap(collection, 28))
[Restaurant(name='Thai Dishes', cuisine='Thai', phone='334-4433', menu=(Dish(name='Mee Krob', price=12.5, calories=500), Dish(name='Larb Gai', price=11.0, calories=450))), Restaurant(name='Pascal', cuisine='French', phone='940-752-0107', menu=(Dish(name='Escargots', price=12.95, calories=250), Dish(name='Poached salmon', price=18.5, calories=550), Dish(name='Rack of lamb', price=24.0, calories=850), Dish(name='Marjolaine cake', price=8.5, calories=950)))]

如果您不能使用元组作为菜单序列，并且由于某种原因无法使用break技巧，则每次都必须使用（慢速且昂贵的）列表成员资格测试：

def Collection_is_cheap(C, price):
    result = []
    for rest in C:
        for dish in rest.menu:
            if dish.price <= price and rest not in result:
                result.append(rest)
    return result

这是缓慢且昂贵的，因为Python将分别测试列表中的每个元素以查看rest == element是否为真，而使用集合时，使用称为哈希表的技巧来快速检查如果一个对象已经存在，通常只需要进行一次计算检查。

迭代命名元组列表，选择餐馆

1 个答案: