我有一个函数，可以返回价格等于或低于指定值的餐厅。这是我当前的代码：
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))
我得到的是正确餐厅的一长串清单，但重复出现了。
[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)])]
而对于正确的输出，它应该只打印两个餐厅一次。我该如何纠正，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)]
最佳答案
比赛结束后，停止在餐厅菜单上循环播放；为此使用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)))]
另一种方法是使用集合而不是列表。集只能容纳唯一的对象，因此多次添加餐厅不会产生任何效果：
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)
为此，您还需要使菜单使用元组，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)))
因此它们是完全不变的，这是使用集合的要求。
使用
set仅收集唯一的餐厅的另一个效果是，由于集合是无序的，因此返回的餐厅的顺序可能会更改。相反，它们以依赖于实现的顺序保存对象，从而有利于高效地测试已经存在的对象。演示对于只有两个廉价餐馆的简单示例，其顺序恰好与第一个版本返回的顺序相匹配：
>>> 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)))]
如果您不能将元组用于菜单序列，并且由于某些原因而不能使用
break技巧，则每次都必须使用（缓慢且昂贵的）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
这很慢且成本很高，因为Python会分别测试列表中的每个元素以查看
rest == element是否为真，而使用set的技巧称为哈希表，用于快速检查对象是否已经存在，通常只需要花费一些时间即可一次计算检查。关于python - 遍历namedtuple列表，选择餐厅，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/31634825/