Django和Python的基本匹配

Question

所以我有一个模型Education，它附加到UserProfile。这个模型就像一个假设一样工作：User在不同地区的几所学校就读过。在

我想做的是根据用户去了哪里做个“评分”。基本上，如果他们上的是同一所学校，他们得10分，同样的城市，5分，同样的州，2分，以此类推。在

我做了一些函数来尝试这个，但是我失败了。有什么建议吗？在

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

这样的怎么样？在

从一个函数开始，该函数收集单个用户的信息，并为该用户返回一个字典：

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，该函数为两个用户中的每个用户调用一次信息收集函数，并使用该信息计算点数。在

^{pr2}$

check_match函数的核心部分被卸载到它自己的helper函数中。此函数用于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。它将属性列表转换为集合，然后使用set intersection operator。在

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

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，您可以调用上述代码，如下所示：

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