如何迭代具有两个不同标准的列表？

Question

大家好，我遇到了一个 for 循环问题，我无法在线找到解决方案。

假设我创建了两个类，一个用于人，一个用于动物。通过为动物类设置一个循环方法，它遍历一个人的年龄列表，我想找到每只具有相同年龄或至少最小年龄差距的宠物的主人（这个例子只是为了让代码看起来更简单）。我使用了min() 函数来找到年龄差距最小的人，这很好。

但是，如果我在选择过程中再添加一个标准呢？例如，我只想将动物分配给那些拥有少于 3 只宠物的人，这意味着即使一个人的年龄差距最小，如果该人已经拥有 3 只宠物，则无法将宠物分配给该人。在这种情况下，循环必须找到下一个拥有少于 3 只宠物的年龄差距最小的人。在我的情况下，必须将 A1 分配给 P1，因为它是拥有少于 3 只宠物的年龄差距最小的人。

到目前为止，这是我的代码：

class People:
    def __init__(self, name, age, pets_owned):
        self.name=name
        self.age=age
        self.pets_owned=pets_owned

P1=People("John",16, 1)
P2=People("Alex",10, 4)
P3=People("Anna", 20, 3)


People_List=[P1, P2, P3]
People_Age=[P1.age, P2.age, P3.age]

class Animal:
    def __init__(self, name, age, owner):
        self.name=name
        self.age=age
        self.owner=owner

    def find(self):
        closest_age = (min(People_Age, key=lambda x: abs(x - self.age)))
        for a in People_List:
            if a.age ==closest_age and a.pets_owned<3:
                self.owner=a.name
                a.pets_owned+=1
                break

            elif a.age==closest_age and a.pets_owned >=3:
                pass #this is where I`m stuck


        print(self.owner)


A1=Animal("Snoopy",7,"not_owned_yet")


A1.find()

Answer 1

如果您知道由于某个条件而不会包括某些人，我会预先过滤传入列表以排除这些人。基本上，根本不用 for 循环，只需过滤列表，然后找到最小值，然后添加宠物。

class People:
    def __init__(self, name, age, pets_owned):
        self.name=name
        self.age=age
        self.pets_owned=pets_owned

P1=People("John",16, 1)
P2=People("Alex",10, 4)
P3=People("Anna", 20, 3)


People_List=[P1, P2, P3]
People_Age=[P1.age, P2.age, P3.age]

class Animal:
    def __init__(self, name, age, owner):
        self.name=name
        self.age=age
        self.owner=owner

    def find(self):
        people_with_less_than_3 = filter(lambda x: x.pets_owned<3, People_List) # filter the list to only include people that have less than 3 pets
        try:
            person_with_closest_age = min(people_with_less_than_3, key=lambda x: abs(x.age - self.age)) # change this to return a person as well
        except:
            # do something if no person with < 3 pets
        self.owner = person_with_closest_age.name
        print(self.owner)


A1=Animal("Snoopy",7,"not_owned_yet")


A1.find()

Answer 2

您可以根据多个属性进行排序，方法是让您的排序键返回一个元组。在我的示例中，我们首先根据年龄差距（优先级较高）进行排序，然后根据拥有的宠物数量（优先级较低）进行排序。您不需要像我一样使用sorted，因为我只是用它来演示如何基于多个属性进行排序。您可能希望将min 与相同的密钥一起使用以获得最符合条件的人。您还需要修改 assign_new_owner 以实际分配新所有者而不是（使用 min）而不是打印人员：

class Person:

    def __init__(self, name, age, pets_owned):
        self.name = name
        self.age = age
        self.pets_owned = pets_owned

    def __str__(self):
        return f"{self.name}, age {self.age} owns {self.pets_owned} pet(s)."

class Animal:

    def __init__(self, name, age, owner=None):
        self.name = name
        self.age = age
        self.owner = owner

    def assign_new_owner(self, people):
        sorted_people = sorted(people, key=lambda p: (abs(p.age - self.age), p.pets_owned))
        for person in sorted_people:
            print(person)

def main():

    people = [
        Person("Alex", 16, 0),
        Person("Nigel", 15, 2),
        Person("Fred", 10, 3),
        Person("Tom", 10, 0),
        Person("Tyler", 15, 0),
        Person("Sam", 15, 1)
        ]

    animal = Animal("Snoopy", 10)
    animal.assign_new_owner(people)

    return 0

if __name__ == "__main__":
    import sys
    sys.exit(main())

输出：

 Tom, age 10 owns 0 pet(s).
 Fred, age 10 owns 3 pet(s).
 Tyler, age 15 owns 0 pet(s).
 Sam, age 15 owns 1 pet(s).
 Nigel, age 15 owns 2 pet(s).
 Alex, age 16 owns 0 pet(s).

编辑：使用 min 后，代码可能看起来更像这样：

class Person:

    def __init__(self, name, age, pets_owned):
        self.name = name
        self.age = age
        self.pets_owned = pets_owned

    def __str__(self):
        return f"{self.name}, age {self.age} owns {self.pets_owned} pet(s)."

class Animal:

    def __init__(self, name, age, owner=None):
        self.name = name
        self.age = age
        self.owner = owner

    def __str__(self):
        return f"{self.name}, age {self.age} is owned by {self.owner.name if self.owner else 'no one'}."

    def assign_new_owner(self, people):
        self.owner = min(people, key=lambda p: (abs(p.age - self.age), p.pets_owned))

def main():

    people = [
        Person("Alex", 16, 0),
        Person("Nigel", 15, 2),
        Person("Fred", 10, 3),
        Person("Tom", 10, 0),
        Person("Tyler", 15, 0),
        Person("Sam", 15, 1)
        ]

    animal = Animal("Snoopy", 10)
    print(animal)
    animal.assign_new_owner(people)
    print(animal)

    return 0

if __name__ == "__main__":
    import sys
    sys.exit(main())

输出：

Snoopy, age 10 is owned by no one.
Snoopy, age 10 is owned by Tom.