python - Django 和 Python 的基本匹配

Question

所以我有一个附加到UserProfile 的模型Education。该模型的工作原理如下:用户 曾就读于不同地区的几所学校。

我想做的是根据用户的去向进行“评分”。基本上，如果他们去了同一所学校，他们会得到10分，同一城市，5分，同一州，2分，依此类推。

我已经做了一些函数来尝试这个，但我失败了。有什么建议吗？

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

像这样怎么样？

从收集有关单个用户的信息的函数开始，并返回该用户的单个字典:

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

然后，有一个单独的函数，为两个用户中的每一个调用一次信息收集函数，并使用该信息计算点数。

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 函数的核心内容被卸载到它自己的辅助函数中。此函数查找两个属性列表的交集，并将共享属性的数量乘以某个点值。它通过将属性列表转换为集合然后使用 set intersection operator 来实现此目的。 .

因此，similarity_points(["MN","OR","PA", "NJ"],["AZ","NJ","PA"], 2) 会发现两场比赛(新泽西州和宾夕法尼亚州)，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，您可以像这样调用上面的代码:

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