如何将条件函数求和到R中的分组行答案

【问题标题】：how to sum conditional functions to grouped rows in R如何将条件函数求和到R中的分组行
【发布时间】：2021-05-18 04:49:12
【问题描述】：

我有以下数据框

customerid	payment_month	payment_date	bill_month	charges
1	January	22	January	30
1	February	15	February	21
1	March	2	March	33
1	May	4	April	43
1	May	4	May	23
1	June	13	June	32
2	January	12	January	45
2	February	15	February	56
2	March	2	March	67
2	April	4	April	65
2	May	4	May	54
2	June	13	June	68
3	January	25	January	45
3	February	26	February	56
3	March	30	March	67
3	April	1	April	65
3	June	1	May	54
3	June	1	June	68

（id数据要大得多）我想用下面的函数计算支付效率，

效率=（未逾期支付金额/总账单金额）*100

不迟于不迟于账单当月的 21 日付款。（在 1 月 22 日支付 1 月的账单被视为迟到）

我想用期望的输出来计算每个客户的效率

customerid	effectivity
1	59.90
2	100
3	37.46

我已经尝试使用以下代码来计算一个 id 并且它有效。但我想将其应用并分配给整个组 ID，并将其汇总为 1 列（有效性）和每个 ID 1 行。我曾尝试使用 group by、aggregation 和 ifelse 函数，但没有任何效果。我该怎么办？

df1 <- filter(df, (payment_month!=bill_month & id==1) | (payment_month==bill_month & payment_date > 21 & id==1) )
df2 <-filter(df, id==1001)
x <- sum(df1$charges)
x <- sum(df2$charges)
100-(x/y)*100

【问题讨论】：

标签： r

【解决方案1】：

使用dplyr的选项


library(dplyr)
df %>%
    group_by(customerid) %>%
    summarise(
        effectivity = sum(
            charges[payment_date <= 21 & payment_month == bill_month]) / sum(charges) * 100,
        .groups = "drop")
## A tibble: 3 x 2
#customerid effectivity
#       <int>       <dbl>
#1          1        59.9
#2          2       100  
#3          3        37.5

【讨论】：

谢谢！有效。我将条件更改为 payment_data >= 21 & payment_mont!=bill_month 并用 100 减去有效性以获得所需的数字并使其更精确。

【解决方案2】：


df %>% 
  group_by(customerid) %>% 
  mutate(totalperid = sum(charges)) %>% 
  mutate(pay_month_number = match(payment_month , month.name),
         bill_month_number = match(bill_month , month.name)) %>% 
  mutate(nolate = ifelse(pay_month_number > bill_month_number, TRUE, FALSE)) %>% 
  summarise(efficiency = case_when(nolate = TRUE ~ (charges/totalperid)*100))

【讨论】：