Question

所以我前段时间曾问过类似的问题（请参阅Creating a tenure column in Days in R），但我无法达到正确的结果，但是现在我已经尝试了另一种可能的方法来询问同一件事，这可能变得容易一些。

问题：我正在寻找一个专栏，告诉我客户任职的日期。这是一些模拟代码：

    Date<-c("01/01/2018", "12/02/2018", "10/03/2018", "22/03/2018", "29/03/2018", "01/04/2018", "02/04/2018","04/04/2018","07/04/2018","11/04/2018", "15/04/2018", "17/04/2018","19/04/2018","21/04/2018","22/04/2018", "29/04/2018", "01/05/2018","03/05/2018","08/05/2018", "10/05/2018", "12/05/2018")
    ClientID<-c("aaa","bbb","ccc","ddd", "eee", "fff", "ggg","aaa","bbb","ccc","ddd", "eee", "fff", "ggg","aaa","bbb","ccc","ddd", "eee", "fff", "ggg")
    df<-cbind(ClientID, Date)
    df<-as.data.frame(df)
    df$Date<-dmy(df$Date)
    df$yearDay<-df$Date
    df$yearDay<-yday(df$yearDay)

给你这样的东西：

    df

   ClientID       Date      yearDay
   aaa          2018-01-01       1
   bbb          2018-02-12      43
   ccc          2018-03-10      69
   ddd          2018-03-22      81
   eee          2018-03-29      88
   fff          2018-04-01      91
   ggg          2018-04-02      92
   aaa          2018-04-04      94
   bbb          2018-04-07      97
   ccc          2018-04-11     101
   ddd          2018-04-15     105
   eee          2018-04-17     107
   fff          2018-04-19     109
   ggg          2018-04-21     111
   aaa          2018-04-22     112
   bbb          2018-04-29     119
   ccc          2018-05-01     121
   ddd          2018-05-03     123
   eee          2018-05-08     128
   fff          2018-05-10     130
   ggg          2018-05-12     132

现在，我要执行的操作（但不确定如何执行）是在每个客户端ID的第二个实例中获取yearDay号，并在前一个实例中减去yearDay。然后在第三个实例中获取yearDay编号，并在前一个实例中减去yearDay。依此类推（我有超过四百万行数据）。答案应该让我留任期。看起来像这样：-

    ClientID       Date      yearDay     tenureDay
    aaa          2018-01-01       1          1
    bbb          2018-02-12      43          1
    ccc          2018-03-10      69          1
    ddd          2018-03-22      81          1
    eee          2018-03-29      88          1 
    fff          2018-04-01      91          1
    ggg          2018-04-02      92          1
    aaa          2018-04-04      94          93 
    bbb          2018-04-07      97          54
    ccc          2018-04-11     101          48
    ddd          2018-04-15     105          24
    eee          2018-04-17     107          19
    fff          2018-04-19     109          18
    ggg          2018-04-21     111          19

关于我将如何实现这一目标的任何想法？

预先感谢您！

Answer 1

您可以使用mutate()包中的arrange()，lag()，group_by()和dplyr的组合。

library(dplyr)

df %>%
  group_by(ClientID) %>%
  arrange(yearDay) %>%
  mutate(tenureDay = yearDay - lag(yearDay))

在R

1 个答案: