重叠(相交)时间间隔和xts

时间:2018-08-16 15:44:11

标签: r xts overlap intersect

有两个时间数据集:来自雨水收集器的数据-时间间隔tistartend和雨水p(每周期的雨水总量,单位为mm)< / p>

ti <- data.frame(
             start = c("2017-06-05 19:30:00", "2017-06-06 12:00:00"),
               end = c("2017-06-05 23:30:00", "2017-06-06 14:00:00"),
                 p = c(16.4, 4.4)
      )

ti[,1] <- as.POSIXct(ti[, 1])
ti[,2] <- as.POSIXct(ti[, 2])

测量站的时间序列tstime和参数q,即排水量(立方米/秒)

ts <- data.frame(stringsAsFactors=FALSE,
              time = c("2017-06-05 16:00:00", "2017-06-05 19:00:00",
                       "2017-06-05 21:00:00", "2017-06-05 23:00:00",
                       "2017-06-06 9:00:00", "2017-06-06 11:00:00", "2017-06-06 13:00:00",
                       "2017-06-06 16:00:00", "2017-06-06 17:00:00"),
                 q = c(0.78, 0.84, 0.9, 0.78, 0.78, 0.78, 0.78, 1.22, 1.25)
      )
ts[,1] <- as.POSIXct(ts[,1])

我需要将时间序列与时间间隔相交,并在ts中在TRUE/FALSE中创建一个新列,如果该行在下雨间隔(TRUE)中而不是({{1 }})就像这样:

FALSE

您对如何应用这种简单操作有任何想法吗?

1 个答案:

答案 0 :(得分:2)

使用sqldf

library(sqldf)
sqldf('select ts.*, case when ti.p is not null then 1 else 0 end as rain 
      from ts
      left join ti
      on start <= time and
         time <= end')

结果:

                 time    q rain
1 2017-06-05 16:00:00 0.78    0
2 2017-06-05 19:00:00 0.84    0
3 2017-06-05 21:00:00 0.90    1
4 2017-06-05 23:00:00 0.78    1
5  2017-06-06 9:00:00 0.78    0
6 2017-06-06 11:00:00 0.78    0
7 2017-06-06 13:00:00 0.78    1
8 2017-06-06 16:00:00 1.22    0
9 2017-06-06 17:00:00 1.25    0