嵌套字典中元组值的排序

Question

我试着做一个函数，接受一个数据库和一年，它根据名字的数量计算名字的排名，然后更新数据库。你知道吗

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)}}

我必须把猫和狗分开排列，最流行的名字是排名1，所以按降序排列。你知道吗

我只允许使用基本表达式和语句，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和基本表达式和语句，我可以编写自己的函数并使用它们。你知道吗

我不能导入模块或使用lambda或任何东西，除了允许列表上的内容。这就是我被困住的原因。这是我的第一节编程课。我不知道如何对元组排序。我知道如何排序嵌套字典，有一个伟大的职位在这里，但我被困在这个元组。你能帮忙吗。你知道吗

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)