Question

我想计算几个GPS点之间的距离。我试过了

distm(c(lon1,lat1), c(lon2,lat2), fun = distHaversine)

适用于一点，但不适用于数据框中的列。

所以我按照这里的推荐尝试了：

Calculate distance between 2 lat longs

但是我对这两个计算得到了不同的结果：

df <- read.table(sep=",", col.names=c("lat1", "lon1", "lat2", "lon2"),text="
7.348687,53.36575,7.348940,53.36507 
7.348940, 53.36507,7.350939,53.36484")


# as recommended in the link above
distHaversine(df[,2:1], df[,4:3])

[1]  80.18433 223.97181

# with distm
distm(c(7.348687,53.36575), c(7.348940,53.36507), fun = distHaversine)

         [,1]
[1,] 77.54033

distm(c(7.348940, 53.36507), c(7.350939,53.36484), fun = distHaversine)

         [,1]
 [1,] 135.2317

那么如何计算数据框列中两个GPS点之间的正确距离（dist（c（lon1，lat1），c（lon2，lat2），fun = distHaversine））？我仔细检查了距离，以至于我知道这种距离是正确的。

提前致谢。

Answer 1

鉴于您要在新列中存储的输出是：

77.54033 135.23165

试试这个

df$distance<-distHaversine(df[,1:2], df[,3:4])

哪个应该返回

> df
      lat1     lon1     lat2     lon2  distance
1 7.348687 53.36575 7.348940 53.36507  77.54033
2 7.348940 53.36507 7.350939 53.36484 135.23165

Answer 2

究竟是什么问题？您是否已经拥有distHaversine()所需的所有距离？

是否要将距离添加为数据框中的列？你走了：

f$dist <- distm(x = df[, c('lon1', 'lat1')], 
                y = df[, c('lon2', 'lat2')],
                fun = distHaversine
                )

Answer 3

我找到了问题的另一种解决方案

for (i in 1:2) {

  a<-df$lon1[i]
  b<-df$lat1[i]
  c<-df$lon2[i]
  d<-df$lat2[i]

  df$distance[i]<-distm(c(a,b),c(c,d), fun = distHaversine)
  }

write.xlsx(df, file = "C:/Users/distances.xlsx")

计算数据框

3 个答案: