Python：這是覆蓋__eq__和__hash__的好方法嗎？

Question

我是新來的Python，我想確保我推翻__eq__和__hash__正確，以免以後造成痛苦的錯誤：（我使用谷歌應用程序引擎）

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()) 

稍微複雜一點的類型：

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))) 

一切好看嗎？

更新，以反映@亞歷克斯的意見

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

第一個是罰款。第二個是有問題的，原因有二：具有相同.courses但不同.forwardLinks在.courses

兩個實體

1. 有可能是重複的會比較平等的，但有不同的散列

我會解決的第二一個通過平等取決於課程和前向鏈接，但是這兩個集合的變化（因此不重複），並且對於哈希也是如此。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))) 

這當然是假設前向鏈路是爲對象的「實際價值」是至關重要的，否則他們應該從兩個__eq__和__hash__被省略。

編輯：從__hash__調用tuple這充其量冗餘（並可能損壞，通過@馬克[[TX !!!]]評論所建議的）中除去;正如@Phillips [[tx !!!]]的評論所建議的那樣，在散列中將set更改爲frozenset。