假设我有一些保险单的数据df。
library(tidyverse)
library(lubridate)
#Example data
d <- as.Date("2020-01-01", format = "%Y-%m-%d")
set.seed(50)
df <- data.frame(id = 1:10,
activation_dt = round(runif(10)*100,0) +d,
expiry_dt = d+round(runif(10)*100,0)+c(rep(180,5), rep(240,5)))
> df
id activation_dt expiry_dt
1 1 2020-03-12 2020-08-07
2 2 2020-02-14 2020-07-26
3 3 2020-01-21 2020-09-01
4 4 2020-03-18 2020-07-07
5 5 2020-02-21 2020-07-27
6 6 2020-01-05 2020-11-04
7 7 2020-03-11 2020-11-20
8 8 2020-03-06 2020-10-03
9 9 2020-01-05 2020-09-04
10 10 2020-01-12 2020-09-14
我想查看每个月有多少保单处于事件状态。我是通过以下方法完成的。
# Getting required result
df %>% arrange(activation_dt) %>%
pivot_longer(cols = c(activation_dt, expiry_dt),
names_to = "event",
values_to = "event_date") %>%
mutate(dummy = ifelse(event == "activation_dt", 1, -1)) %>%
mutate(dummy2 = floor_date(event_date, "month")) %>%
arrange(dummy2) %>% group_by(dummy2) %>%
summarise(dummy=sum(dummy)) %>%
mutate(dummy = cumsum(dummy)) %>%
select(dummy2, dummy)
# A tibble: 8 x 2
dummy2 dummy
<date> <dbl>
1 2020-01-01 4
2 2020-02-01 6
3 2020-03-01 10
4 2020-07-01 7
5 2020-08-01 6
6 2020-09-01 3
7 2020-10-01 2
8 2020-11-01 0
现在我遇到了如何处理缺失月份的问题，例如2020年4月至2020年6月等
最佳答案
data.table 解决方案:
- 生成月份序列
- 使用非等值连接查找每月有效的政策并对其进行计数
library(lubridate)
library(data.table)
setDT(df)
months <- seq(lubridate::floor_date(mindat,'month'),lubridate::floor_date(max(df$expiry_dt),'month'),by='month')
months <- data.table(months)
df[,c("activation_dt_month","expiry_dt_month"):=.(lubridate::floor_date(activation_dt,'month'),
lubridate::floor_date(expiry_dt,'month'))]
df[months, .(months),on = .(activation_dt_month<=months,expiry_dt_month>=months)][,.(nb=.N),by=months]
months nb
1: 2020-01-01 4
2: 2020-02-01 6
3: 2020-03-01 10
4: 2020-04-01 10
5: 2020-05-01 10
6: 2020-06-01 10
7: 2020-07-01 10
8: 2020-08-01 7
9: 2020-09-01 6
10: 2020-10-01 3
11: 2020-11-01 2
关于r - 应用滚动功能时如何显示缺失日期，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/64089397/