Question

我有几个客户，每个月的25号都向客户收费。我想找出他们的合同终止前的最后一个结算日期。以下是数据框中的示例：

> data = [['Arthur','2019-03-01'],['Bart','2019-02-26'],['Cindy','2019-02-18'],['Douglas','2019-03-31']]

> df = pd.DataFrame(data, columns = ['Name','Termination Date'])  
> df

此外，以下是预期的输出：

> df['Last Billing Date'] =['2019-02-25','2019-02-25','2019-01-25','2019-03-25'] 
> df

Answer 1

这是一种方法

s=df['Termination Date'].apply(lambda x : x.replace(day=25))
df['New']=np.where(df['Termination Date']>=s,s,s-pd.DateOffset(months=1))

df
      Name Termination Date        New
0   Arthur       2019-03-01 2019-02-25
1     Bart       2019-02-26 2019-02-25
2    Cindy       2019-02-18 2019-01-25
3  Douglas       2019-03-31 2019-03-25

Answer 2

一个简单的解决方案是，如果某天在25之前减去一个月：

import datetime

def last_billing(termination_dt):
    if isinstance(termination_dt, str): # check if not in datetime format
        termination_dt = datetime.datetime.strptime(termination_dt, '%Y-%m-%d')
    if termination_dt.day < 25:
        return termination_dt.replace(day=25, month=termination_dt.month-1)
    return termination_dt.replace(day=25)

df['Last Billing Date'] = df['Termination Date'].apply(last_billing)

    Name    Termination Date    Last Billing Date
0   Arthur  2019-03-01          2019-02-25
1   Bart    2019-02-26          2019-02-25
2   Cindy   2019-02-18          2019-01-25
3   Douglas 2019-03-31          2019-03-25

如果性能存在问题，请vectorize使用该功能

import numpy as np

@np.vectorize
def last_billing(termination_dt):
    if isinstance(termination_dt, str):
        termination_dt = datetime.datetime.strptime(termination_dt, '%Y-%m-%d')
    if termination_dt.day < 25:
        return termination_dt.replace(day=25, month=termination_dt.month-1)
    return termination_dt.replace(day=25)

df['Last Billing Date'] = last_billing(df['Termination Date'])

时间比较：

%timeit df['Last Billing Date'] = df['Termination Date'].apply(last_billing)
## 113 ms ± 365 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit df['Last Billing Date'] = last_billing(df['Termination Date'])
## 108 ms ± 397 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 3

如果要以矢量化方式执行此操作：

 df['Termination Date'] = pd.to_datetime(df['Termination Date']) 

 before_25 = df['Termination Date'].dt.day < 25

 df.loc[before_25, 'Termination Date'] = df.loc[before_25, 'Termination Date'] + pd.DateOffset(months=-1)

 df['Termination Date'].apply(lambda dt: dt.replace(day=25)).values

如何在使用python终止合同之前找出客户的帐单日期？

3 个答案: