类变量在 for 循环中保持为空

Question

我想确保hotel_name 不存在于Hotel.hotels 列表中。 似乎当我开始喂食时，每次 for 循环都在一个空列表中查找。

请注意，如果我不使用 for 循环并且只使用

Hotel.hotels.append([number,hotel_name,city,total_number,empty_rooms])

它打印酒店列表

[[1, 'Crown Plaza', 'alex', 20, 2], [1, 'Crown Plaza', 'alex', 20, 2], [2, 'Radisson Blu', 'cairo', 24, 22], [3, 'Paradise Inn', 'dubai', 390, 200], [4, 'Four Seasons', 'alex', 1000, 400], [5, 'Address', 'dubai', 500, 200], [6, 'Fairmont', 'dubai', 1000, 100], [7, 'Rotana', 'dubai', 5000, 300]]
[Finished in 0.1s]

这是文件

class Hotel():
    """""""""
    this is hotel class file
    """
    hotels = []

    def __init__(self,number,hotel_name,city,total_number,empty_rooms):
        self.number = number
        self.hotel_name = hotel_name
        self.city = city
        self.total_number = total_number
        self.empty_rooms = empty_rooms
        for i in Hotel.hotels:
            if self.hotel_name not in i:
                Hotel.hotels.append([number,hotel_name,city,total_number,empty_rooms])
            # else:
            #     print "Hotel already exists!"

feed_dataBase_with_hotel = Hotel(1,"Crown Plaza","alex",20,2)
feed_dataBase_with_hotel = Hotel(1,"Crown Plaza","alex",20,2)# cull repeated
feed_dataBase_with_hotel = Hotel(2,"Radisson Blu","cairo",24,22)
feed_dataBase_with_hotel = Hotel(3,"Paradise Inn","dubai",390,200)
feed_dataBase_with_hotel = Hotel(4,"Four Seasons","alex",1000,400)
feed_dataBase_with_hotel = Hotel(5,"Address","dubai",500,200)
feed_dataBase_with_hotel = Hotel(6,"Fairmont","dubai",1000,100)
feed_dataBase_with_hotel = Hotel(7,"Rotana","dubai",5000,300)

print Hotel.hotels

非常感谢 Patrick 的回答 更新 如果我想从字典中构建一个列表怎么办，因为我想访问空房间并用另一个类更改它的值

class Hotel():
    """
    this is hotel class file
    """
    hotels = {}
    hotelList = []

    def __init__(self,number,hotel_name,city,total_number,empty_rooms):
        self.number = number
        self.hotel_name = hotel_name
        self.city = city
        self.total_number = total_number
        self.empty_rooms = empty_rooms

        # edited: no for needed
        if self.hotel_name in Hotel.hotels:
            print('Hotel {} Already exists!'.format(self.hotel_name))
            return # or raise ValueError & handle it

        Hotel.hotels[self.hotel_name] = self

        tempList = Hotel.hotels.items()
        for i in tempList:
            x = Hotel.hotels.items()[i][1]
            Hotel.hotelList.append(x)

更新

另一个预订类将使用我们在酒店类中使用的实例变量hotel_name

from hotel import Hotel
from customer import Customer
from notification import Notification

class Reservation():
    reservations =[]
    def reserve_room(self,hotel_name, customer_name):
        x = Hotel.hotels.values()
        for i in x:
            if Hotel.hotel_name in i:
                Reservation.reservations.append([hotel_name,customer_name])
                i[4] -=1

AttributeError: class Hotel 没有属性'hotel_name'

更新 来自Understanding getitem method 使用

def __getitem__(self, hotel_name):
          return self.hotel_name

问题已解决！

特别感谢Patrick

Answer 1

我建议改变一些东西来修复它：

• 您正在迭代一个可能的 1000 个 Hotels 的列表以查找是否有相同的名称，如果您使用 set 或 dict 这样做会更好，因为在这些查找中O(1)。 Lists 有O(n) 查找最坏情况时间，这意味着set/dict 是恒定时间，无论您有多少Hotels。列表越来越慢，您需要搜索的Hotels 越多。

• 你用新的酒店实例覆盖相同的变量，它们自己被创建和遗忘 - 你只将它们的值存储在你的 Hotel.hotels 列表中，而不是存储构造的 Hotel 本身。 Hotel 实例的整个构造毫无意义。

建议的更改：

• 将静态存储变成字典，让您获得快速查找的好处
• 存储您的 Hotel-instances 而不是您创建 Hotels 时使用的值
• 不要直接打印Hotel.hotelDict - 我介绍了一种方法，它只接受你的dicti的值并将它们排序打印 - 默认情况下我按Hotel.number排序

---

class Hotel():
    """""""""
    this is hotel class file
    """   
    hotelDict = {} # checking "in" for sets is much faster then list - a dict is similar
                   # fast in checking and can hold values - serves double duty here
    # sorted hotels by value, sorted by key

    @classmethod
    def getSortedHotels(cls, sortKey = lambda x:x.number):
        """Takes all values (Hotel's) from Hotel.hotelDict
        and prints them after sorting. Default sort key is Hotel.number"""
        return sorted(cls.hotelDict.values(), key=sortKey) 

    def __init__(self,number,hotel_name,city,total_number,empty_rooms):
        if hotel_name in Hotel.hotelDict:
            print("Hotel already exists: {}".format(hotel_name))
            return # or raise ValueError("Hotel already exists") and handle the error
# see https://stackoverflow.com/questions/3209233/how-to-replace-an-instance-in-init
# if you want to try to replace it using __new__() but not sure if its possible

        self.number = number
        self.hotel_name = hotel_name
        self.city = city
        self.total_number = total_number
        self.empty_rooms = empty_rooms
        Hotel.hotelDict[self.hotel_name] = self

    def __repr__(self):
        """Neater print output when printing them inside a list"""
        return "{:>3}) {} in {} has {} of wich {} are empty.".format(
        self.number,self.hotel_name,self.city,self.total_number,self.empty_rooms)
        # if you want a "simple" list-like output of your attributes:
        # comment the return above and uncomment:
        # return repr([self.number,self.hotel_name,self.city,
        #              self.total_number,self.empty_rooms])

    def __str__(self):
        """Neater print output when printing Hotel instances"""
        return self.__repr__()

测试：

feed_dataBase_with_hotel = Hotel(1,"Crown Plaza","alex",20,2)
feed_dataBase_with_hotel = Hotel(1,"Crown Plaza","alex",20,2) # cull repeated
feed_dataBase_with_hotel = Hotel(2,"Radisson Blu","cairo",24,22)
feed_dataBase_with_hotel = Hotel(3,"Paradise Inn","dubai",390,200)
feed_dataBase_with_hotel = Hotel(4,"Four Seasons","alex",1000,400)
feed_dataBase_with_hotel = Hotel(5,"Address","dubai",500,200)
feed_dataBase_with_hotel = Hotel(6,"Fairmont","dubai",1000,100)
feed_dataBase_with_hotel = Hotel(7,"Rotana","dubai",5000,300)

print Hotel.getSortedHotels() 

print Hotel(99,"NoWayInn","NoWhere",1,200)

输出：

Hotel already exists: Crown Plaza
[  1) Crown Plaza in alex has 20 of wich 2 are empty.,   
   2) Radisson Blu in cairo has 24 of wich 22 are empty.,   
   3) Paradise Inn in dubai has 390 of wich 200 are empty.,   
   4) Four Seasons in alex has 1000 of wich 400 are empty.,   
   5) Address in dubai has 500 of wich 200 are empty.,   
   6) Fairmont in dubai has 1000 of wich 100 are empty.,   
   7) Rotana in dubai has 5000 of wich 300 are empty.]

 99) NoWayInn in NoWhere has 1 of wich 200 are empty.

如果您希望您的酒店按名称排序，很简单：

print Hotel.getSortedHotels(sortKey=lambda x:x.hotel_name)

Answer 2

看看这一行for i in Hotel.hotels：这里

for i in Hotel.hotels:
    if self.hotel_name not in i:
          Hotel.hotels.append([number,hotel_name,city,total_number,empty_rooms])

并尝试获取在哪种情况下它将填充您的数组

Answer 3

问题是您在迭代一个空的“酒店”列表时 .append 。 首先在 for 循环中检查现有的 hotel_name，然后再追加。

for hotel in self.hotels:
    if self.hotel_name == hotel[1]:  # hotel_name is at index 1
        print('Already exists!')
        return  # return before appending
#now append
self.hotels.append([number,hotel_name,city,total_number,empty_rooms])

如果你不想退货，试试这个

for hotel in self.hotels:
    if self.hotel_name == hotel[1]:  # hotel_name is at index 1
        print('Already exists!')
        break
else:  # break did not happen
    self.hotels.append([number,hotel_name,city,total_number,empty_rooms])