根据R中的条件计算日期之间的平均差

Posted 2021-03-28

我有这个数据集：

Date New_Renew
 2019-01-10 22:11:16  Renewing
 2019-02-23 00:21:48  Renewing
 2019-03-05 05:26:17  Renewing
 2019-04-18 15:05:10       NEW
 2019-04-18 15:07:52       NEW
 2019-04-26 11:32:25  Renewing
 2019-05-03 14:15:25  Renewing
 2019-05-08 21:10:08       NEW
 2019-05-16 13:35:57  Renewing
 2019-05-24 13:18:23  Renewing
 2019-06-01 12:42:21  Renewing
 2019-06-17 18:08:09  Renewing
 2019-06-26 13:40:29  Renewing
 2019-12-13 17:57:43  Renewing
 2020-01-03 11:49:14  Renewing
 2020-01-11 11:46:51  Renewing
 2020-01-14 21:08:08       NEW
 2020-01-18 21:14:30       NEW
 2020-01-21 16:08:37       NEW
 2020-01-28 11:41:44  Renewing
 2020-01-30 13:34:21  Renewing
 2020-02-03 13:29:37  Renewing
 2020-02-18 17:15:52  Renewing
 2020-02-20 13:37:52  Renewing
 2020-02-24 12:55:25  Renewing
 2020-02-26 21:13:38       NEW
 2020-03-04 13:23:41  Renewing
 2020-03-09 16:48:36  Renewing

我想要的是，当New_Renew变量等于NEW时，计算与NEW相关联的日期之间的差的平均值。简而言之，用户多久执行一次新交易。

答案

library(data.table)
library(xts)
library(lubridate)
library(tbl2xts)


DT <- read.table(text = 'Date, New_Renew
 2019-01-10 22:11:16,Renewing
 2019-02-23 00:21:48,Renewing
 2019-03-05 05:26:17,Renewing
 2019-04-18 15:05:10,NEW
 2019-04-18 15:07:52,NEW
 2019-04-26 11:32:25,Renewing
 2019-05-03 14:15:25,Renewing
 2019-05-08 21:10:08,NEW
 2019-05-16 13:35:57,Renewing
 2019-05-24 13:18:23,Renewing
 2019-06-01 12:42:21,Renewing
 2019-06-17 18:08:09,Renewing
 2019-06-26 13:40:29,Renewing
 2019-12-13 17:57:43,Renewing
 2020-01-03 11:49:14,Renewing
 2020-01-11 11:46:51,Renewing
 2020-01-14 21:08:08,NEW
 2020-01-18 21:14:30,NEW
 2020-01-21 16:08:37,NEW
 2020-01-28 11:41:44,Renewing
 2020-01-30 13:34:21,Renewing
 2020-02-03 13:29:37,Renewing
 2020-02-18 17:15:52,Renewing
 2020-02-20 13:37:52,Renewing
 2020-02-24 12:55:25,Renewing
 2020-02-26 21:13:38,NEW
 2020-03-04 13:23:41,Renewing
 2020-03-09 16:48:36,Renewing', 
                 sep = ',', 
                 header = T)

df <- xts(DT, order.by = ymd_hms(DT$Date))

new_items <- which(DT$New_Renew=="NEW")

dif <- DT

dif$difference <- NA

renewal <- 0

for (i in 1:nrow(df)){

  if (df[i,2]=='Renewing' & renewal == 0){
    renewal <- i
  } else if (df[i,2]=='Renewing' & renewal != 0){
    next
  } else if (df[i, 2]=='NEW' & renewal != 0) {
    dif[i, 'difference'] <- index(df[i, 2]) - index(df[renewal, 2])
  } else {
    dif[i, 'difference'] <- index(df[i, 2]) - index(df[renewal, 2])
    renewal <- 0
  }

}

mean_diff <- mean(dif$difference, na.rm = T)

另一答案

使用aggregate和diff。 60*24将产生秒数转换为天数。

aggregate(Date ~ New_Renew, dat, function(x) mean(diff(x))/(60*24))
#  New_Renew         Date
# 1       NEW 52.38292438 
# 2  Renewing  0.01471444