如何在嵌套字典中对元组值进行排名

Question

我正在尝试创建一个接受数据库和年份的函数，它根据名称计算名称的排名，然后更新数据库。

database = {('Spot','DOG'): {2013: (612, 1), 2014: (598, 3)},
            ('Princess','CAT'): {2013: (713, 2)},
            ('Smokey', 'CAT'): {2013: (523, 1), 2014: (514, 1)},
            ('Fido', 'DOG'): {2013: (558, 2), 2014: (655, 1)},
            ('Sparky','DOG'): {2104: (572, 2)}}

我必须将猫和狗分别排名，最流行的名字是排名 1，所以降序排列。

我只能使用基本的表达式和语句，len, range, enumerate, int, get, items, keys, values, pop, copy, popitem, update, append, insert, extend,min, max, index, split, join, replace, sorted, sort, reversed, reverse, sets 和基本的表达式和语句，我可以编写自己的函数并使用它们。

我不能导入模块或使用 lambda 或除允许列表中的内容之外的任何东西。这就是为什么我被卡住了。这是我的第一堂编程课。我不知道如何对元组进行排序。我知道如何对嵌套字典进行排序，这里有一篇很棒的帖子，但我被困在这个元组上。能否请你帮忙。

def rank_names_for_one_year(db, year):
    list = []
    list_2 = []
    for k in db:
        if 'MALE' in k:
            for key in db[k]:
                if year == key:
                    for n in db[k][year]:
                        list.append(n)
                        break
        if 'FEMALE' in k:
            for key in db[k]:
                if year == key:
                    for n in db[k][year]:
                        list_2.append(n)
                        break
    list.sort()
    new_list = [()]
    list_2.sort()
    new_list_2 = [()]

    for k in db:
        if 'MALE' in k:
            for i in range(len(list)):
                new_list.append((list[i],i+1))
            for key in db[k]:
                if year == key:
                    for n in db[k][year]:
                        for i in range(len(new_list)):
                            for j in range(len(new_list[i])):
                                if new_list[i][j] == n:
                                    db[k][year] = new_list[i]
        if 'FEMALE' in k:
            for i in range(len(list_2)):
                new_list_2.append((list_2[i],i+1))
            for key in db[k]:
                if year == key:
                    for n in db[k][year]:
                        for i in range(len(new_list_2)):
                            for j in range(len(new_list_2[i])):
                                if new_list_2[i][j] == n:
                                    db[k][year] = new_list_2[i]

Answer 1

database = {('TEST','DOG'):{2013:(612,1), 2014:(598,3)}, ('Spot','DOG'):{2013:(612,1), 2014:(598,3)},('Princess','CAT'):{2013:(713,2)},
            ('Princess1','CAT'):{2013:(713,2)},('Smokey', 'CAT'):{2013:(523,1), 2014:(514,1)},('Smokey1', 'CAT'):{2013:(523,1), 2014:(514,1)},('Fido', 'DOG'):{2013:(558, 2), 2014:(655, 1)},('Sparky','DOG'):{2104:(572,2)}}

def rank_names_for_one_year(db, year):
    temp = {'DOG' : [], 'CAT' : []}
    for k, v in db.items():
        if year not in v:
            continue
        temp[k[1]].append((v[year][0], k[0]))

    for animal_type, v in temp.items():
        rank = 0
        countPrev = -1
        for i, (count, name) in enumerate(reversed(sorted(v))):
            if countPrev != count:
                rank = i + 1
            countPrev = count
            db[(name, animal_type)][year] = (count, rank)


rank_names_for_one_year(database, 2013)