我如何改变_ult _uuu_uu_uu）神奇的方法，使这个算法能够工作？python

Question

Python排序问题

我有一个排序算法问题，它包括对子类Priority Person的对象进行排序（该对象将被放置在一个列表中，目标是更改子类Priority Person的lt方法以对列表上的.sort（）方法产生影响），如果优先权人的对象存在由参数True描述的缺陷，则应将其视为小于获得参数False的优先权人对象（这意味着此人没有任何缺陷），如果两者都有缺陷，或者两者都没有缺陷，则应考虑所涉及的优先权人的名字的词典顺序。优先权人子类别的主要类别是我上面所示的类别人：

class Person(object):

    def __init__(self, name):
        """Create a person"""
        self.name = name
        try:
            lastBlank = name.rindex(' ')
            self.lastName = name[lastBlank + 1:]
        except:
            self.lastName = name
        self.birthday = None

    def __lt__(self, other):
        """Returns True if self's name is lexicographically
           less than other's name, and False otherwise"""
        return self.name < other.name

    def __str__(self):
        """Returns self's name"""
        return self.name

class PriorityPerson(Person): # the __lt__ part of this class does not work and I don't know why?

   def __init__(self,name,deficiencia):
        super().__init__(name)
        self.deficiente = deficiencia
    
    
    def __lt__(self, other):
        if self.deficiente and other.deficiente and self.name< other.name:
            return self.name < other.name
        elif self.deficiente and other.deficiente and other.name < self.name:
            return other.name< self.name
        elif self.deficiente and not other.deficiente and self.name < other.name:
            return self.name < other.name
        elif self.deficiente and not other.deficiente and other.name < self.name:
            return self.name< other.name
        elif not self.deficiente and other.deficiente and self.name < other.name:
            return other.name < self.name
        elif not self.deficiente and other.deficiente and other.name < self.name:
            return other.name < self.name
        elif not self.deficiente and not other.deficiente and self.name < other.name:
            return  self.name < other.name
        elif not self.deficiente and not other.deficiente and other.name < self.name:
            return other.name < self.name

实施示例：

p1 = PriorityPerson("John Smith", False)
p2 = PriorityPerson("Sam Mendes", False)
p3 = PriorityPerson("Grandmother Anne", True)
p4 = PriorityPerson("Stephen Hawking", True)
p5 = PriorityPerson("Betty Large", True)

listaPessoas = [p1,p2,p3,p4, p5]
listaPessoas.sort()
for p in listaPessoas:
   print(p)

正确输出：

Betty Large
Grandmother Anne
Stephen Hawking
John Smith
Sam Mendes

我的输出错误：

Betty Large
Stephen Hawking
Grandmother Anne
Sam Mendes
John Smith

任何帮助都将不胜感激。 多谢各位

Answer 1

我希望这能奏效

class Person(object):
    def __init__(self, name):
        """Create a person"""
        self.name = name
        try:
            lastBlank = name.rindex(' ')
            self.lastName = name[lastBlank + 1:]
        except:
            self.lastName = name
        self.birthday = None

    def __lt__(self, other):
        """Returns True if self's name is lexicographically
        less than other's name, and False otherwise"""
        return self.name < other.name

    def __str__(self):
        """Returns self's name"""
        return self.name


class PriorityPerson(Person):
    def __init__(self, name, deficiencia):
        super().__init__(name)
        self.deficiente = deficiencia

    def __lt__(self, other):
        if self.deficiente and not other.deficiente:
            # If self is VIP and other is not, just return True
            return True
        elif not self.deficiente and other.deficiente:
            # If other is VIP and self is not, just return False
            return False

        # On equal priority, normal compare
        return super().__lt__(other)

p1 = PriorityPerson("John Smith", False)
p2 = PriorityPerson("Sam Mendes", False)
p3 = PriorityPerson("Grandmother Anne", True)
p4 = PriorityPerson("Stephen Hawking", True)
p5 = PriorityPerson("Betty Large", True)

listaPessoas = [p1,p2,p3,p4, p5]
listaPessoas.sort()
for p in listaPessoas:
   print(p)

输出：

Betty Large
Grandmother Anne
Stephen Hawking
John Smith
Sam Mendes

Answer 2

您基本上希望按照优先级和名称进行排序。因此，您必须首先比较不同优先级，并且只有相同优先级的名称：

def __lt__(self, other):
    if self.deficiente != other.deficiente: # yes, you can compare bools!
        return self.deficiente
    else:
        return self.name < other.name