Django和Python的基础配对机制

Question

我有一个叫做 Education 的模型，它和 UserProfile 关联在一起。这个模型的工作原理很简单：User 可以在不同地方上过好几所学校。

我想做的是根据用户上过的学校来给他们打分。基本上，如果他们上的是同一所学校，就给10分；如果在同一个城市，就给5分；如果在同一个州，就给2分，依此类推。

我已经尝试写了一些函数来实现这个目标，但都没有成功。有没有什么建议？

info = {}
def edu_info(user1):
    user_1_cities = []
    user_1_schools = []
    user_1_state = []
    first_one = Education.objects.filter(owner=user1)
    for i in first_one:
        user_1_cities.append(str(i.city))
        user_1_schools.append(str(i.school))
        user_1_state.append(str(i.state))
        info[str(i.owner.username)] = {}
        info[str(i.owner.username)]['cities'] = user_1_cities
        info[str(i.owner.username)]['schools'] = user_1_schools
        info[str(i.owner.username)]['state'] = user_1_state
    return info

def check_match(user1, user2):
    match_score = {}    
    first_info = edu_info(user1)
    dict = edu_info(user2)
    for item in dict:
        cities = dict[item]['cities']
        #user2 = item
        #print cities
        for city in cities:
            if city in first_info['jmitchel3']['cities']:
                match_score['user'] = 'jmitchel3'
                match_score['user2'] = str(user2.user.username)
                match_score['city'] = city
                print "here! " + str(city)
            else:
                print "not here! " + str(city)

    return match_score




check_match(j,t)

Answer 1

这样做怎么样呢？

首先，写一个函数来收集单个用户的信息，并返回一个包含该用户信息的字典：

def edu_info(user1):
    user_1_cities = []
    user_1_schools = []
    user_1_state = []
    first_one = Education.objects.filter(owner=user1)
    for i in first_one:
        user_1_cities.append(str(i.city))
        user_1_schools.append(str(i.school))
        user_1_state.append(str(i.state))
    info = {}
    info['cities'] = user_1_cities
    info['schools'] = user_1_schools
    info['states'] = user_1_state
    return info

接着，再写一个单独的函数，它会调用你刚才的收集信息的函数，分别为两个用户收集信息，并计算他们的得分。

def check_match(user_1, user_2)
    info_1 = edu_info(user_1)
    info_2 = edu_info(user_2)
    school_points = similarity_points(info_1["schools"], info_2["schools"], 10)
    city_points = similarity_points(info_1["cities"], info_2["cities"], 5)
    state_points = similarity_points(info_1["states"], info_2["states"], 2)
    return school_points + city_points + state_points

check_match 函数的核心部分被放到了一个辅助函数里。这个辅助函数会找出两个属性列表的交集，并将共享的属性数量乘以某个分值。它通过将属性列表转换为集合，然后使用 集合交集运算符 来实现这个功能。

比如，similarity_points(["MN","OR","PA", "NJ"],["AZ","NJ","PA"], 2) 会找到两个匹配项（NJ 和 PA），因此返回 4。

def similarity_points(attr_1, attr_2, points)
    """Award a number of points for each shared attribute.

    attr_1 and attr_2 should be lists to compare. 
    """
    number_shared = len(set(attr_1) & set(attr_2))
    return number_shared * points

然后，你可以像这样调用上面的代码：

wilduck_jmitchel3_points = check_match("Wilduck", "jmitchel3")