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如何在python中返回类中调用的列表

object list python

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Bonnie Zou · 技术社区 · 2 年前

与此抗争:

编写一个名为get\u pets\u string的函数。get\u pets\u字符串应该有一个参数,即所有者的实例。获取\u pets\u字符串应根据以下格式:
David Joyner's pets are: Boggle Joyner, Artemis Joyner

class Name:
    def __init__(self, first, last):
        self.first = first
        self.last = last

class Pet:
    def __init__(self, name, owner):
        self.name = name
        self.owner = owner
        
class Owner:
    def __init__(self, name):
        self.name = name
        self.pets = []

如果功能正常工作,则打印:

David Joyner's pets are: Boggle Joyner, Artemis Joyner
Audrey Hepburn's pets are: Pippin Hepburn

owner_1 = Owner(Name("David", "Joyner"))
owner_2 = Owner(Name("Audrey", "Hepburn"))

pet_1 = Pet(Name("Boggle", "Joyner"), owner_1)
pet_2 = Pet(Name("Artemis", "Joyner"), owner_1)
pet_3 = Pet(Name("Pippin", "Hepburn"), owner_2)

owner_1.pets.append(pet_1)
owner_1.pets.append(pet_2)
owner_2.pets.append(pet_3)

print(get_pets_string(owner_1))
print(get_pets_string(owner_2))

以下是我的代码:

def get_pets_string(owner):
    for obj in owner.pets:
       return owner.name.first + " " + owner.name.last + "'s pets are: " + obj.name.first + " " + obj.name.last

我的答案只能为每个主人打印一只宠物,如下所示:

David Joyner's pets are: Boggle Joyner
Audrey Hepburn's pets are: Pippin Hepburn

4 回复 | 直到 2 年前

Eric Jin sud 2 年前

您可以实现一个快速助手函数来更改字符串的显示方式,只需显示名称即可:

def repr_name(name):
    return name.first + " " + name.last

然后您可以同时显示所有宠物:

def get_pets_string(owner):
   return f"{repr_name(owner.name)}'s pets are: {', '.join([repr_name(pet.name) for pet in owner.pets])}"

这个 ', '.join([str(pet.name) for pet in owner.pets]) 将每个宠物的名字放入列表中,然后用逗号分隔每个宠物。

Chris 2 年前

您可以使用 str.join 在生成格式化字符串的生成器表达式上。

', '.join(f"{pet.name.first} {pet.name.first}" for pet in owner_1.pets)

这将产生:

'Boggle Boggle, Artemis Artemis'

因为这是家庭作业,我不想给你完整的解决方案,所以我将把这部分和其余部分放在一起。

Lucas Masiero 2 年前

问题是函数正在返回值。Return表示函数结束,因此它不会遍历其他pet。

函数应该如下所示。。。

def get_pets_string(owner):
    owner_string = f"{owner.name.first} {owner.name.last}'s pets are: "
    
    pets_string = ""
    for obj in owner.pets:
        pet_name = f"{obj.name.first} {obj.name.last}"
        pets_string += pet_name + " "
    
    return owner_string + pets_string

实际上,您可以通过在每个类上返回全名作为methon来重构代码

Copperfield 2 年前

问题是您在第一次迭代中会立即返回

def get_pets_string(owner):
    for obj in owner.pets:
       return owner.name.first + " " + owner.name.last + "'s pets are: " + obj.name.first + " " + obj.name.last
        #^ here is the problem, you need to iterate over your whole list but this stop after the first iteration

你需要累积构建你的结果,就像这样

def get_pets_string(owner):
    owner_name = owner.name.first + " " + owner.name.last + "'s pets are: "
    pets = []
    for obj in owner.pets:
       pets.append(obj.name.first + " " + obj.name.last) #gather all the pets names in a list
    return  owner_name + ", ".join(pets) #use join to put together all the pets name we build previously

Bonnie Zou 2 年前

非常感谢您对以上所有回复!!我非常感谢你的投入和知识。我确实对每个回复都发表了评论,但我不知道为什么我的评论消失了。。。所以我把它写在这里是为了感谢大家的帮助和建议!!