R - 如何计算两组坐标点之间的距离？

Question

我有一个与此 SO 帖子非常相似的问题：

Geographic / geospatial distance between 2 lists of lat/lon points (coordinates)

这是一个经过编辑的坐标示例集，用于说明我的情况：

require(tidyverse)

list1 <- data.frame(longitude = c(72, 74, 76, 78, 79, 82), 
                    latitude = c(20.5, 19, 19.5, 20, 22, 21),
                    area = "A")
list2 <- data.frame(longitude = c(71, 73, 75, 77, 79, 78.5, 72), 
                    latitude = c(21.5, 22, 20.5, 23, 23.5, 24, 24), 
                    area = "B")

df <- bind_rows(list1, list2)

ggplot(data = df) +
    geom_point(aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list1, aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list1[c(2,6),], aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list1[c(1,4),], aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list2[c(1,7),], aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list2[c(7,6),], aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list2[c(6,5),], aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list2[c(5,3),], aes(x = longitude, y = latitude, color = area)) +
    geom_line(data = list2[c(3,1),], aes(x = longitude, y = latitude, color = area))

所以我需要计算两个坐标点列表之间的最小距离。我已经能够完成这项工作，但我注意到我需要更高效的东西 - 数据太大了。

我想到的一种可能性是形成这些区域的非重叠多边形并计算一组点到相邻多边形的距离。有没有办法形成这些多边形？凸面外壳不是一种选择，因为这些区域非常粗糙。

另一种选择是在区域之间形成一条线。

编辑：我在图中添加了一些线条来说明多边形。

Answer 1

也许这就是你要找的？

#load libraries
library(dplyr)
library(sf)

#create row_id's, and make it a simple (spatial) feature
list1.sf <- list1 %>% 
  mutate( id = row_number() ) %>% 
  st_as_sf( coords = c("longitude", "latitude"), crs = 4326 )
list2.sf <- list2 %>% 
  mutate( id = row_number() ) %>% 
  st_as_sf( coords = c("longitude", "latitude"), crs = 4326 )

#find nearest points in list2 for each id in list1, and as a bonus, calculate the distance to this point
list1.sf %>% 
  dplyr::group_by( id ) %>%
  dplyr::mutate( np = sf::st_nearest_feature( geometry, list2.sf ),
                 dist_np = as.numeric( sf::st_distance( geometry, list2.sf[np,] ) ) )


# Simple feature collection with 6 features and 4 fields
# geometry type:  POINT
# dimension:      XY
# bbox:           xmin: 72 ymin: 19 xmax: 82 ymax: 22
# epsg (SRID):    4326
# proj4string:    +proj=longlat +datum=WGS84 +no_defs
# # A tibble: 6 x 5
# # Groups:   id [6]
# area     id    geometry    np dist_np
# * <fct> <int> <POINT [°]> <int>   <dbl>
# 1 A         1   (72 20.5)     1 151880.
# 2 A         2     (74 19)     3 196361.
# 3 A         3   (76 19.5)     3 152335.
# 4 A         4     (78 20)     3 318287.
# 5 A         5     (79 22)     5 166111.
# 6 A         6     (82 21)     5 415019.

Answer 2

你可以计算欧几里得距离 我改变了一点数据集。我放弃最后一列。

x1 <- data.frame(longitude = c(72, 74, 76, 78, 79, 82), 
                    latitude = c(20.5, 19, 19.5, 20, 22, 21))
x2 <- data.frame(longitude = c(71, 73, 75, 77, 79, 78.5, 72), 
                    latitude = c(21.5, 22, 20.5, 23, 23.5, 24, 24))

euc.dist <- function(x1, x2) sqrt(sum((x1 - x2) ^ 2))


dist <- NULL
for(i in 1:nrow(x1)) dist[i] <- euc.dist(x1[i,],x2[i,])
dist