Python：这是覆盖eq和hash的好方法吗？

Question

我是Python新手，我想确保我正确地重写了__eq__和__hash__，以免以后引起痛苦的错误：

（我正在使用谷歌应用引擎。）

class Course(db.Model):
    dept_code = db.StringProperty()
    number = db.IntegerProperty()
    title = db.StringProperty()
    raw_pre_reqs = db.StringProperty(multiline=True)
    original_description = db.StringProperty()

    def getPreReqs(self):
        return pickle.loads(str(self.raw_pre_reqs))

    def __repr__(self):
        title_msg = self.title if self.title else "Untitled"
        return "%s %s: %s" % (self.dept_code, self.number, title_msg)

    def __attrs(self):
        return (self.dept_code, self.number, self.title, self.raw_pre_reqs, self.original_description)

    def __eq__(self, other):
        return isinstance(other, Course) and self.__attrs() == other.__attrs()

    def __hash__(self):
        return hash(self.__attrs())

稍微复杂一点的类型：

class DependencyArcTail(db.Model):
    ''' A list of courses that is a pre-req for something else '''
    courses = db.ListProperty(db.Key)

    ''' a list of heads that reference this one '''
    forwardLinks = db.ListProperty(db.Key)

    def __repr__(self):
        return "DepArcTail %d: courses='%s' forwardLinks='%s'" % (id(self), getReprOfKeys(self.courses), getIdOfKeys(self.forwardLinks))

    def __eq__(self, other):
        if not isinstance(other, DependencyArcTail):
            return False

        for this_course in self.courses:
            if not (this_course in other.courses):
                return False

        for other_course in other.courses:
            if not (other_course in self.courses):
                return False

        return True

    def __hash__(self):
        return hash((tuple(self.courses), tuple(self.forwardLinks)))

一切看起来都很好？

更新以反映@Alex的评论

class DependencyArcTail(db.Model):
    ''' A list of courses that is a pre-req for something else '''
    courses = db.ListProperty(db.Key)

    ''' a list of heads that reference this one '''
    forwardLinks = db.ListProperty(db.Key)

    def __repr__(self):
        return "DepArcTail %d: courses='%s' forwardLinks='%s'" % (id(self), getReprOfKeys(self.courses), getIdOfKeys(self.forwardLinks))

    def __eq__(self, other):
        return isinstance(other, DependencyArcTail) and set(self.courses) == set(other.courses) and set(self.forwardLinks) == set(other.forwardLinks)

    def __hash__(self):
        return hash((tuple(self.courses), tuple(self.forwardLinks)))

Answer 1

第一个很好。第二个问题有两个原因：

1. .courses中可能有重复项
2. 具有相同.courses但不同.forwardLinks的两个实体将比较相等但具有不同散列

我将通过使相等依赖于进程和前向链接来修复第二个链接，但这两个链接都对集合进行了更改（因此没有重复），对于散列也是如此。一、 e.：

def __eq__(self, other):
    if not isinstance(other, DependencyArcTail):
        return False

    return (set(self.courses) == set(other.courses) and
            set(self.forwardLinks) == set(other.forwardLinks))

def __hash__(self):
    return hash((frozenset(self.courses), frozenset(self.forwardLinks)))

这当然是假设前向链接对对象的“真实值”至关重要，否则应该从__eq__和__hash__中省略它们。

编辑：从__hash__调用tuple中删除，这些调用充其量是多余的（而且可能是有害的，正如@Mark[[tx]的注释所建议的那样！！！]])；根据@Phillips[[tx]的注释建议，将哈希中的set更改为frozenset！！！]].