我确实有带有 gps 数据的数据框,我需要计算两行之间的距离(当前和上一个)
id time lat long heartrate altitude
1 20130424.tcx 2013-04-24T04:53:22Z 50.024818 14.522724 151 <NA>
2 20130424.tcx 2013-04-24T04:53:26Z 50.024818 14.522724 96 <NA>
3 20130424.tcx 2013-04-24T04:53:30Z 50.024818 14.522724 104 <NA>
4 20130424.tcx 2013-04-24T04:53:34Z 50.024818 14.522724 107 <NA>
5 20130424.tcx 2013-04-24T04:53:38Z 50.024818 14.522724 108 <NA>
6 20130424.tcx 2013-04-24T04:53:42Z 50.024818 14.522724 112 372.0
7 20130424.tcx 2013-04-24T04:53:46Z 50.024818 14.522724 151 372.0
8 20130424.tcx 2013-04-24T04:53:47Z 50.024677 14.522874 151 356.0
9 20130424.tcx 2013-04-24T04:53:50Z 50.024677 14.522874 118 356.0
10 20130424.tcx 2013-04-24T04:53:54Z 50.024677 14.522874 118 356.0
11 20130424.tcx 2013-04-24T04:53:58Z 50.024464 14.522917 147 358.0
12 20130424.tcx 2013-04-24T04:54:02Z 50.024464 14.522917 144 358.0
13 20130424.tcx 2013-04-24T04:54:06Z 50.024269 14.522853 150 367.0
14 20130424.tcx 2013-04-24T04:54:10Z 50.024269 14.522853 152 367.0
15 20130424.tcx 2013-04-24T04:54:13Z 50.024002 14.522874 152 380.0
我能够将数据连接到自身并获取每一行的前一行(可能有更简单的解决方案):
library(sqldf)
mydft = mydf[-nrow(mydf),]
mydft$id = seq_along(mydft$id) +1
mydf$id = seq_along(mydf$id)
mydft2 <- sqldf("select a.*, b.lat as lat2, b.long as long2 from mydf a left join mydft b using (id)")
我现在如何计算与列的距离lat, long, lat2, long2
?我尝试过这里描述的方法:
R <- 6371 # Earth mean radius [km]
mydft2$delta.long <- (mydft2$long2 - mydft2$long)
mydft2$delta.lat <- (mydft2$lat2 - mydft2$lat)
mydft2$a <- sin(mydft2$delta.lat/2)^2 + cos(mydft2$lat) * cos(mydft2$lat2) * sin(mydft2$delta.long/2)^2
mydft2$c <- 2 * asin(min(1,sqrt(mydft2$a)))
mydft2$d = R * c
但这仅返回列表 NA。