关于python：找到一年中排名前n位的客户，然后在一年中的每个月存储这些客户的数量

Find the top n clients for a year then bucket those client’s volume across each month the year

大家早安，

我想报告该年度的前 n 个客户，然后显示这些前 n 个客户中的每一个在一年中的表现。样本 df:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

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，比如说两个(n)；关于已完成交易的客户：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

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

所以客户 A 和 D 是我表现最好的两 (n) 个。
我现在想将此列表或 df 反馈到原始数据中，以检索它们在 Year_Month 上升到顶部且客户列为 rows

的一年中的表现

1
2
3

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

你需要 pandas.pivot_table 方法

这是我的建议：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

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：

1
2
3

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！

IIUC

1
2
3
4
5
6
7
8

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

原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/267936.html

关于python：找到一年中排名前n位的客户，然后在一年中的每个月存储这些客户的数量

Find the top n clients for a year then bucket those client’s volume across each month the year

相关推荐

发表回复