使用geosphere计算GPS点之间的最小距离会给出不准确的值

Question

我正在尝试计算R中一系列GPS点的最近邻之间的平均距离。我找到了两个代码来获得这些值。但它似乎没有给出以米为单位的正确距离。当我与谷歌地图对比时，我发现它离我们太远了。

我找到了这个答案：R - Finding closest neighboring point and number of neighbors within a given radius, coordinates lat-long

library(geosphere)
sp.groups <- groups
coordinates(sp.groups) <- ~Long+Lat
class(sp.groups)
d<- distm(sp.groups)
min.d<- apply(d, 1, function(x) order(x, decreasing=F)[2])
min.d
mean(min.d)
groupdist<- cbind(groups, groups[min.d,], apply(d, 1, function(x) order(x, decreasing=F)[2]))
colnames(groupdist)<- c(colnames(groups), 'neighbor', 'n.lat', 'n.long','dist')

在这里使用了包rgeos，但它给出了相同的结果：Calculate the distance between two points of two datasets (nearest neighbor)

library(rgeos)
sp.groups <- groups
coordinates(sp.groups) <- ~Long+Lat
proj4string(sp.groups) <-CRS("+proj=utm +datum=WGS84")
class(sp.groups)
d<- gDistance(sp.groups, byid=TRUE)
min.d<- apply(d, 1, function(x) order(x, decreasing=F)[2])
min.d
mean(min.d)
groupdist<- cbind(groups, groups[min.d,], apply(d, 1, function(x) order(x, decreasing=F)[2]))
colnames(groupdist)<- c(colnames(groups), 'neighbor', 'n.lat', 'n.long','dist')

当我去谷歌地球上查看的时候，距离可能会很远。它甚至为最近的邻居提供了160-200m的不同值。此外，一些最近的邻居没有相同的距离值，请参见K11和K3，然后参见K3和K11。以下是我获得的结果，并添加了来自Google Maps的期望值：

Group Lat        Long      neighbor  n.lat      n.long    dist  GMaps
K1    -26.96538  21.80965  K34       -26.96503  21.80940  27    44
K10   -26.96575  21.81132  K1        -26.96538  21.80965  1     172
K11   -26.96249  21.81120  K3        -26.96387  21.81053  22    166
K24   -26.96033  21.81090  K11       -26.96249  21.81120  3     240
K3    -26.96387  21.81053  K11       -26.96249  21.81120  3     166
K34   -26.96503  21.80940  K1        -26.96538  21.80965  1     44

怎么啦？

我的数据

groups<-data.frame(Group = c('K1', 'K10', 'K11', 'K24', 'K3', 'K34'), 
Lat = c(-26.96538, -26.96575, -26.96249, -26.96033, -26.96387, -25.96503), 
Longitude = c(21.80965, 21.81132, 21.81120, 21.80190, 21.81053, 21.80940))

Answer 1

我不确定您在这里试图计算的是什么，但列distance只是指具有最小距离的点的位置/编号。我添加了一个具有实际最小距离的数字，看起来与您预期的结果相似。

library(geodist, include.only = NULL)
library(sp, include.only = NULL)

groups <- data.frame(Group = c('K1', 'K10', 'K11', 'K24', 'K3', 'K34'), 
                     Lat   = c(-26.96538, -26.96575, -26.96249, -26.96033, -26.96387, -25.96503), 
                     Long  = c(21.80965, 21.81132, 21.81120, 21.80190, 21.81053, 21.80940))

sp.groups <- groups
sp::coordinates(sp.groups) <- ~Long+Lat

# mindistance Matrix
d <- geodist::geodist(groups, measure = "cheap")


# position of minimum distance
diag(d) <- Inf
min.d <- max.col(-d)
min.d
#> [1] 2 1 5 5 3 4


groupdist <- cbind(groups, groups[min.d,], min.d)
colnames(groupdist) <- c(colnames(groups), 'neighbor', 'n.lat', 'n.long','closest_to')

# get minimum distance for each pair of coordinates
groupdist$distance <- d[cbind(seq_along(min.d), min.d)]
groupdist
#>     Group       Lat     Long neighbor     n.lat   n.long closest_to    distance
#> 2      K1 -26.96538 21.80965      K10 -26.96575 21.81132          2    171.1554
#> 1     K10 -26.96575 21.81132       K1 -26.96538 21.80965          1    171.1554
#> 5     K11 -26.96249 21.81120       K3 -26.96387 21.81053          5    167.2225
#> 5.1   K24 -26.96033 21.80190       K3 -26.96387 21.81053          5    944.4119
#> 3      K3 -26.96387 21.81053      K11 -26.96249 21.81120          3    167.2225
#> 4     K34 -25.96503 21.80940      K24 -26.96033 21.80190          4 110613.1362

由reprex包创建于2021-08-22 (v2.0.1)