按值查找用户排名

Question

如何按值查找用户的排名。data = [        {            "playerId": 2,            "value": 196763.8        },        {           "playerId": 3,           "value": 196763.8        },        {           "playerId": 44,           "value": 196764        }]我试过：index = [ x['playerId'] for x in data ].index(3) #3 playerIdrank = index+1我得到的排名是：2对于玩家 Id 2 和 3，排名应为 1（因为值与 196763.8 相同），对于玩家 Id 44，排名应为 2

Answer 1

您可以使用 ，即：pandas

df = pd.DataFrame(data)

df['rank'] = df['value'].rank(ascending=True, method="min").astype(int)

df = df.sort_values(by=['rank']).reset_index(drop=True)

for i in range(1, len(df)):

    last, curr = df.loc[i-1, 'rank'], df.loc[i, 'rank']

    df.loc[i, 'rank'] = last + 1 if last < curr else last

print(df)

   playerId     value  rank

0         2  196763.8     1

1         3  196763.8     1

2        44  196764.0     2

Answer 2

您需要按以下条件对列表进行排序，并查找某些列表是否与其他列表具有相同的分数，并将它们合并为一个。valueplayer_id

这应该可以解决问题：

from collections import defaultdict

new_dict = defaultdict(list)

for x in data:

    new_dict[x['value']].append(x['playerId'])  

rank_list = [[k,v] for k,v in new_dict.items()]

# Set reverse=False if lower score is considered higher rank

rank_list.sort(key=lambda k: k[0], reverse=True) 

def get_rank(player_id):

    for i in range(len(rank_list)):

        if player_id in rank_list[i][1]:

            return i+1 # list is 0 indexed thus +1

print('Rank is %s for player id %s' % (get_rank(44), 44))

> Rank is 1 for player id 44

print('Rank is %s for player id %s' % (get_rank(2), 2))

> Rank is 2 for player id 2

print('Rank is %s for player id %s' % (get_rank(3), 3))

> Rank is 2 for player id 3