Question

早上好

我想报告一年中排名前n位的客户，然后显示该年排名前n位的每个客户的表现。样本df：

import pandas as pd
dfTest = [
             ('Client', ['A','A','A','A',
                         'B','B','B','B',
                         'C','C','C','C',
                         'D','D','D','D']),
            ('Year_Month', ['2018-08', '2018-09', '2018-10','2018-11',
                             '2018-08', '2018-09', '2018-10','2018-11',
                             '2018-08', '2018-09', '2018-10', '2018-11',
                             '2018-08', '2018-09', '2018-10', '2018-11']),
            ('Volume', [100, 200, 300,400,
                        1, 2, 3,4,
                        10, 20, 30,40,
                        1000, 2000, 3000,4000]
            ),
            ('state', ['Done', 'Tied Done', 'Tied Done','Done',
                       'Passed', 'Done', 'Passed', 'Done',
                       'Rejected', 'Done', 'Passed', 'Done',
                       'Done', 'Done', 'Done', 'Done']
            )
          ]
df = pd.DataFrame.from_items(dfTest)
print(df)

   Client Year_Month  Volume      state
0       A    2018-08     100       Done
1       A    2018-09     200  Tied Done
2       A    2018-10     300  Tied Done
3       A    2018-11     400       Done
4       B    2018-08       1     Passed
5       B    2018-09       2       Done
6       B    2018-10       3     Passed
7       B    2018-11       4       Done
8       C    2018-08      10   Rejected
9       C    2018-09      20       Done
10      C    2018-10      30     Passed
11      C    2018-11      40       Done
12      D    2018-08    1000       Done
13      D    2018-09    2000       Done
14      D    2018-10    3000       Done
15      D    2018-11    4000       Done

现在确定顶部，例如两个（n）；有关完成交易的客户：

d = [
    ('Done_Volume', 'sum')
]
# first filter by substring and then aggregate of filtered df
mask = ((df['state'] == 'Done') | (df['state'] == 'Tied Done'))
df_Client_Done_Volume = df[mask].groupby(['Client'])['Volume'].agg(d)
print(df_Client_Done_Volume)

Client             
A              1000
B                 6
C                60
D             10000

print(df_Client_Done_Volume.nlargest(2, 'Done_Volume'))

        Done_Volume
Client             
D             10000
A              1000

因此，客户A和D是我表现最好的两个（n）演员。现在，我想将此列表或df反馈回原始数据，以检索Year_Month在顶部排名上升且客户端列为行的一年中的表现

Client  2018-08 2018-09 2018-10 2018-11
A       100     200     300     400
D       1000    2000    3000    4000

Answer 1

IIUC

s=df.loc[df.state.isin(['Done','Tied Done'])].drop('state',1)
s=s.pivot(*s.columns)

s.loc[s.sum(1).nlargest(2).index]
Year_Month  2018-08  2018-09  2018-10  2018-11
Client                                        
D            1000.0   2000.0   3000.0   4000.0
A             100.0    200.0    300.0    400.0

Answer 2

您需要pandas.pivot_table方法

这是我的建议：

def get_top_n_performer(df, n):
    df_done = df[df['state'].isin(['Done', 'Tied Done'])]
    aggs= {'Volume':['sum']}
    data = df_done.groupby('Client').agg(aggs)
    data = data.reset_index()
    data.columns = ['Client','Volume_sum']
    data = data.sort_values(by='Volume_sum', ascending=False) 
    return data.head(n)

ls= list(get_top_n_performer(df, 2).Client.values)

data = pd.pivot_table(df[df['Client'].isin(ls)], values='Volume', index=['Client'],
               columns=['Year_Month'])
data = data.reset_index()

print(data)

输出：

Year_Month Client  2018-08  2018-09  2018-10  2018-11
0               A      100      200      300      400
1               D     1000     2000     3000     4000

我希望这会有所帮助！

找到一年中排名前n位的客户，然后在一年中的每个月中收集这些客户的交易量

2 个答案: