Python：这是覆盖 __eq__ 和 __hash__ 的好方法吗？

Question

我是 Python 新手，我想确保我正确覆盖了 __eq__ 和 __hash__，以免以后造成痛苦的错误：

（我使用的是 Google App Engine。）

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)))

一切看起来都不错？

更新以反映 @Alex 的 cmets

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 的两个实体比较相等，但具有不同的哈希值

我将通过使相等性取决于课程和前向链接来解决第二个问题，但是对集合的更改（因此没有重复）和散列相同。即：

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一个对象的“真实价值”至关重要，否则它们应该从__eq__ 和__hash__ 中省略。

编辑：从__hash__ 对tuple 的调用中删除，这些调用充其量是多余的（并且可能具有破坏性，正如@Mark [[tx!!!]] 的评论所建议的那样）；正如@Phillips [[tx!!!]] 的评论所建议的，在散列中将set 更改为frozenset。