使用类访问对象列表的属性

Question

我想建立一个基本上是对象数组的类。 我希望该类向我返回其成员的属性列表。 例如，从类中：

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

dog = Animal(1,2)
cat = Animal(3,4)

我想创建一个类Group来做：

my_group = Group([dog,cat])
print(my_group.height) #return [1,3] or (1,3) or array([1,3])

我已经想到要这样做：

import inspect

def get_attribute(instance):
    """Return all attributes of an instance (except all internal built-in)"""

    attributes = inspect.getmembers(instance, lambda a:not(inspect.isroutine(a)))
    return [a for a in attributes if not(a[0].startswith('__') and a[0].endswith('__'))]

class Group():
    def __init__(self, members):
        self.members = members

        attributes = [a[0] for a in get_attribute(list(members)[0])] #assuming members are all of the same class

        for a in attributes:
            setattr(self, a, [getattr(m, a) for m in self.members])

然后，我可以将Group用于其他类，例如：

class Human():
    def __init__(self, name, weight):
        self.name = name
        self.weight = weight

class Family(Group):
    pass 

alan = Human("alan", 1)
bob =  Human("bob", 2)
my_family = Family([alan, bob])
print(my_family.weight) #returns [1, 2]

这可行，但是如果成员数量很高，则效率很低，因为它遍历每个成员。 基本上，我的代码可以工作，但是我想使用诸如map之类的功能使其速度更快。

Answer 1

使用class属性：

class Animal():
    all_data = {"height":[],"age":[]}
    def __init__(self, height, age):
        self.height = height
        self.age = age
        Animal.all_data["age"].append(age)
        Animal.all_data["height"].append(height)

dog = Animal(1,2)
cat = Animal(3,4)


print(Animal.all_data["height"])
print(Animal.all_data["age"])

Answer 2

也许您可以定义一个方法使其保持动态：

class Animal():

    def __init__(self, height, age):
        self.height = height
        self.age = age

dog = Animal(1,2)
cat = Animal(3,4)

class Group():
    def __init__(self, members):
        self.members = members

    def get_all_attr(self, a):
        return [getattr(m,a) for m in self.members]

my_group = Group([dog,cat])
print(my_group.get_all_attr("height")) #returns [1,3] 

我认为除非每次使用懒惰初始化，否则您都无法避免不遍历成员。