Python 找到一年内排名前n的客户,然后将这些客户';一年中每个月的销售量
早上好 我想报告今年前n名客户的情况,然后展示这些前n名客户全年的表现。样本df:Python 找到一年内排名前n的客户,然后将这些客户';一年中每个月的销售量,python,pandas,dataframe,pandas-groupby,Python,Pandas,Dataframe,Pandas Groupby,早上好 我想报告今年前n名客户的情况,然后展示这些前n名客户全年的表现。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']),
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2(n);已完成交易的客户:
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是我的前两位(n)执行者。
现在,我想将这个列表或df反馈到原始数据中,以检索它们在一年中的性能,其中年/月在顶部上升,客户机列为行
Client 2018-08 2018-09 2018-10 2018-11
A 100 200 300 400
D 1000 2000 3000 4000
IIUC
IIUC
你需要方法
以下是我的建议:
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
我希望这有帮助 你需要一种方法
以下是我的建议:
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
我希望这有帮助 非常感谢@Wen Ben。“s.sum(1).nlagest(2).指数在一整年中都是总和?”文本:你能帮我回答以下问题吗?当然,午饭后让我试试。@panda让我看看that@Wen-本谢谢你非常感谢@Wen Ben。“s.sum(1).nlagest(2).指数在一整年中都是总和?”文本:你能帮我回答以下问题吗?当然,午饭后让我试试。@panda让我看看that@Wen-Ben Thank youThanks@CHAMI Soufiane,这在我的大数据集上返回了正确的结果。我很高兴这有帮助!谢谢@CHAMI Soufiane,这在我的大数据集上返回了正确的结果。我很高兴这有帮助!