求每月不相等值的平均值，并根据某些条件进行分配

Question

我目前正在努力将我的数据转换成有用的数据集。我需要平均分配从第一个月到最后一个月的付款。问题在于支付不一致且不平等。此外，还有一些已全额支付的款项，应从第一笔付款加上根据协议数据框架适用的期限开始分配

我的表格如下：

表1：付款

^{tb1}$

以下是付款数据框的代码：

payments = pd.DataFrame.from_dict({'cust_id': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 2, 9: 3, 10: 3}, 
                               'agreement_id': {0: 'A', 1: 'A', 2: 'A', 3: 'A', 4: 'B', 5: 'B', 6: 'B', 7: 'B',
                                                8: 'C', 9: 'D', 10: 'D'},
                               'date': {0: '12/1/20', 1: '2/2/21', 2: '2/3/21', 3: '5/1/21', 4: '1/2/21', 5: '1/9/21',
                                        6: '3/1/21', 7: '4/23/21', 8: '1/21/21', 9: '3/4/21', 10: '5/3/21'},
                               'payment': {0: 200, 1: 200, 2: 100, 3: 200, 4: 50, 5: 20, 6: 80, 7: 90, 8: 600, 9: 150, 10: 150}})

表2：协议

^{tb2}$

以下是协议数据框的代码：

agreement = pd.DataFrame.from_dict({'agreement_id': {0: 'A', 1: 'B', 2: 'C', 3: 'D'}, 'activation': {0: '12/1/20', 1: '1/2/21', 2: '1/21/21', 3: '3/4/21'}, 'term_months': {0: 24, 1: 6, 2: 6, 3: 6}, 'total_fee': {0: 4800, 1: 300, 2: 600, 3: 300}})

我想要的结果如下：

^{tb3}$

或者，以代码形式：

    cust_id agreement_id     date  payment
0         1            A  12/1/20   116.67
1         1            A   1/1/21   116.67
2         1            A   2/1/21   116.67
3         1            A   3/1/21   116.67
4         1            A   4/1/21   116.67
5         1            A   5/1/21   116.67
6         1            B   1/1/21    60.00
7         1            B   2/1/21    60.00
8         1            B   3/1/21    60.00
9         1            B   4/1/21    60.00
10        2            C   1/1/21   100.00
11        2            C   2/1/21   100.00
12        2            C   3/1/21   100.00
13        2            C   4/1/21   100.00
14        2            C   5/1/21   100.00
15        2            C   6/1/21   100.00
16        3            D   3/1/21    50.00
17        3            D   4/1/21    50.00
18        3            D   5/1/21    50.00
19        3            D   6/1/21    50.00
20        3            D   7/1/21    50.00
21        3            D   8/1/21    50.00

激活日期与首次付款日期相同

我尝试使用以下代码（由AlexK建议）创建另一列，但它仅适用于总付款少于总费用的情况。但是，当总付款等于总费用时，我需要相应地分配付款，从付款开始到月底（开始加上月期限）

payments['date'] = pd.to_datetime(payments['date'])
resampled_payments = (payments
   .set_index('date')
   .groupby(['cust_id', 'agreement_id'])
   .resample('MS')
   .agg({'payment': sum})
   .reset_index()
)

resampled_payments['avg_monthly_payment'] = (resampled_payments
   .groupby(['cust_id', 'agreement_id'])['payment']
   .transform('mean')
)

Answer 1

这里是R解决方案（因为您也用R标记了它）

#load libraries
library(tidyverse)
library(lubridate)

pymts <- read.table(text = "cust_id agreement_id    date    payment
1   A   12/1/20 200
1   A   2/2/21  200
1   A   2/3/21  100
1   A   5/1/21  200
1   B   1/2/21  50
1   B   1/9/21  20
1   B   3/1/21  80
1   B   4/23/21 90
2   C   1/21/21 600
3   D   3/4/21  150
3   D   5/3/21  150", header = T)

agmt <- read.table(text = "agreement_id activation  term_months total_fee
A   12/1/20 24  4800
B   1/21/21 6   600
C   1/21/21 6   600
D   3/4/21  6   300", header = T)

#final code

final<- pymts %>% mutate(date = as.Date(date, "%m/%d/%y")) %>%
  left_join(agmt %>% mutate(activation = as.Date(activation, "%m/%d/%y")), by = "agreement_id") %>%
  group_by(cust_id, agreement_id) %>%
  mutate(d = n(),
         date = floor_date(date, "month")) %>%
  complete(date = seq.Date(from = min(date), by = "month", length.out = ifelse(sum(payment) == first(total_fee),
                                                                                                    first(term_months), 
                                                                                                   (year(max(date)) -
                                                                                                      year(min(date)))*12 +
                                                                                                      month(max(date)) - 
                                                                                                      month(min(date)) +1))) %>%
  mutate(payment = sum(payment, na.rm = T)) %>%
  filter(!duplicated(date)) %>%
  mutate(payment = payment/n()) %>%
  select(1:4) %>% ungroup()


final
# A tibble: 22 x 4
   cust_id agreement_id date       payment
     <int> <chr>        <date>       <dbl>
 1       1 A            2020-12-01    117.
 2       1 A            2021-01-01    117.
 3       1 A            2021-02-01    117.
 4       1 A            2021-03-01    117.
 5       1 A            2021-04-01    117.
 6       1 A            2021-05-01    117.
 7       1 B            2021-01-01     60 
 8       1 B            2021-02-01     60 
 9       1 B            2021-03-01     60 
10       1 B            2021-04-01     60 
# ... with 12 more rows

Answer 2

考虑到您的数据帧，这应该是可行的

from dateutil.relativedelta import relativedelta

# Transofrm column to date
payments['date']= pd.to_datetime(payments['date'])
agreement['activation']= pd.to_datetime(agreement['activation'])

final =pd.merge(payments,agreement,on='agreement_id',how='left')

# set date to beginning of month
final['date'] = pd.to_datetime(final.date).dt.to_period('M').dt.to_timestamp()


def set_date_range(df):
    if df['payment'].sum() == df['total_fee'].iloc[0]:
        return pd.date_range(min(g['date']), periods=df['term_months'].iloc[0], freq='M')
    else:
        return pd.date_range(min(g['date']),
                             max(g['date'])+relativedelta(months=+1), freq='M' )

# Create dataframe with dates
seq_df = pd.DataFrame()
for i,g in final.groupby(['cust_id', 'agreement_id']):
    seq_df = pd.concat([seq_df,
                        pd.DataFrame({'cust_id': i[0], 'agreement_id': i[1],  'date': set_date_range(g)})])

# Set date to beginnig of month
seq_df['date'] = pd.to_datetime(seq_df.date).dt.to_period('M').dt.to_timestamp()

final = (pd.concat([final, seq_df], sort=True)
              .sort_values(['cust_id', 'agreement_id', 'date'])
              .reset_index(drop=True)
              .reindex(columns=final.columns))

final['payment'] = final.groupby(by=['cust_id', 'agreement_id'])["payment"].transform("sum")

final = final.drop_duplicates(['cust_id', 'agreement_id', 'date'])

final['n'] = final.groupby(by=['cust_id', 'agreement_id'])["cust_id"].transform("count")
final['payment_due'] = final['payment']/final['n']
final[['cust_id','agreement_id','date', 'payment_due']]

我无法准确地复制管道表单tidyverse，但输出应该匹配。最困难的部分是seq_df的创建，但应该很好（针对更通用的用例对其进行双重测试）