编写一个有效的循环来比较 GPS 坐标

我想遍历 GPS 坐标的数据框并删除所有彼此接近的坐标。

pick first row
  clalulate the distance between selected and the next row
  if the distance is < mindist and current row is not the last row continue to next row
  else select the current row (leave it in dataframe) and if the selected row is not the last row
   repeat from the begining

结果应该是一个数据帧，其 GPS 点至少为mindist彼此远离

一种方法是：

 node_distances <- function(node_coords)
  {
  n <- nrow(node_coords)
  from <- 1:(n - 1)
  to <- 2:n
  return(c(0, geodist::geodist_vec(node_coords[from, ]$lon,node_coords[from, ]$lat, node_coords[to, ]$lon, node_coords[to, ]$lat, paired = TRUE, measure = "geodesic")))
}
distances %>% filter(dist < mindist)

但这种方法只测试 2 行，这意味着它会在文件中产生很大的间隙。

我开始编写嵌套循环，但他的决定是错误的，不起作用而且很慢：

node_distances_hack <- function(node_coords)
{
  n <- nrow(node_coords)
  for(i in 1:n) {
    print(node_coords[i,])
    a<-i
    distance_c<-0
    mindist<-50
    while(distance_c<mindist || a >= n){
      distance_c<-geodist::geodist_vec(node_coords[i,]$lat,node_coords[i,]$lon,node_coords[a,]$lat,node_coords[a,]$lon, paired = TRUE, measure = "cheap")
      a<-a+1
      }
  }
}

更好的方法是什么？ 先感谢您， BR

您可以使用以下命令来完成此操作，而无需循环geodist:::geodist_xy_vec获取每对点之间的距离，因为这会生成一个成对矩阵。考虑这个函数：

remove_close <- function(df, CLOSE = 10000)
{
  dist_mat <- geodist:::geodist_xy_vec(df$lon, df$lat, df$lon, df$lat, "cheap")
  diag(dist_mat) <- CLOSE + 1
  clashes <- which(dist_mat < CLOSE, arr.ind = TRUE)
  duplicates <- unique(t(apply(clashes, 1, sort)))[, 2]
  df[-duplicates, ]
}

library(ggplot2)

set.seed(69)

df <- data.frame(lat  = runif(1000, 51, 54),
                 lon = runif(1000, 8, 13))

ggplot(df, aes(lon, lat)) + geom_point()

ggplot(remove_close(df), aes(lon, lat)) + geom_point()

^{Created on 2020-07-22 by the reprex package https://reprex.tidyverse.org (v0.3.0)}