问题
I have two data sets, A and B, which give locations of different points in the UK as such:
A = data.frame(reference = c(C, D, E), latitude = c(55.32043, 55.59062, 55.60859), longitude = c(-2.3954998, -2.0650243, -2.0650542))
B = data.frame(reference = c(C, D, E), latitude = c(55.15858, 55.60859, 55.59062), longitude = c(-2.4252843, -2.0650542, -2.0650243))
A has 400 rows and B has 1800 rows. For all the rows in A, I would like to find the shortest distance in kilometers between a point in A and each of the three closest points in B, as well as the reference and coordinates in lat and long of these points in B.
I tried using this post
R - Finding closest neighboring point and number of neighbors within a given radius, coordinates lat-long
However, even when I follow all the instructions, mainly using the command distm from the package geosphere, the distance comes up in a unit that can't possibly be kilometers. I don't see what to change in the code, especially since I am not familiar at all with the geo packages.
回答1:
Here is solution using a single loop and vectorizing the distance calculation (converted to km).
The code is using base R's rank function to order/sort the list of calculated distances.
The indexes and the calculated distances of the 3 shortest values are store back in data frame A.
library(geosphere)
A = data.frame(longitude = c(-2.3954998, -2.0650243, -2.0650542), latitude = c(55.32043, 55.59062, 55.60859))
B = data.frame(longitude = c(-2.4252843, -2.0650542, -2.0650243), latitude = c(55.15858, 55.60859, 55.59062))
for(i in 1:nrow(A)){
#calucate distance against all of B
distances<-geosphere::distGeo(A[i,], B)/1000
#rank the calculated distances
ranking<-rank(distances, ties.method = "first")
#find the 3 shortest and store the indexes of B back in A
A$shortest[i]<-which(ranking ==1) #Same as which.min()
A$shorter[i]<-which(ranking==2)
A$short[i]<-which(ranking ==3)
#store the distances back in A
A$shortestD[i]<-distances[A$shortest[i]] #Same as min()
A$shorterD[i]<-distances[A$shorter[i]]
A$shortD[i]<-distances[A$short[i]]
}
A
longitude latitude shortest shorter short shortestD shorterD shortD
1 -2.395500 55.32043 1 3 2 18.11777 36.633310 38.28952
2 -2.065024 55.59062 3 2 1 0.00000 2.000682 53.24607
3 -2.065054 55.60859 2 3 1 0.00000 2.000682 55.05710
As M Viking pointed out, for the geosphere package the data must be arranged Lon then Lat.
回答2:
geosphere library has several functions to help you. distGeo returns meters.
Note the data must be arranged Lon then Lat.
library(geosphere)
A = data.frame(longitude = c(-2.3954998, -2.0650243, -2.0650542), latitude = c(55.32043, 55.59062, 55.60859))
B = data.frame(longitude = c(-2.4252843, -2.0650542, -2.0650243), latitude = c(55.15858, 55.60859, 55.59062))
geosphere::distGeo(A, B)
# > geosphere::distGeo(A, B)
# [1] 18117.765 2000.682 2000.682
Vector of distances in meters
回答3:
I add below a solution using the spatialrisk package. The key functions in this package are written in C++ (Rcpp), and are therefore very fast.
The function spatialrisk::points_in_circle() calculates the observations within radius from a center point. Note that distances are calculated using the Haversine formula. Since each element of the output is a data frame, purrr::map_dfr is used to row-bind them together:
purrr::map2_dfr(A$latitude, A$longitude,
~spatialrisk::points_in_circle(B, .y, .x,
lon = longitude,
lat = latitude,
radius = 1e6)[1:3,],
.id = "id_A")
id_A reference latitude longitude distance_m
1 1 C 55.15858 -2.425284 18115.958
2 1 E 55.59062 -2.065024 36603.447
3 1 D 55.60859 -2.065054 38260.562
4 2 E 55.59062 -2.065024 0.000
5 2 D 55.60859 -2.065054 2000.412
6 2 C 55.15858 -2.425284 53219.597
7 3 D 55.60859 -2.065054 0.000
8 3 E 55.59062 -2.065024 2000.412
9 3 C 55.15858 -2.425284 55031.092
回答4:
I know this is a long way but, in this question, there exists a formula for calculation the distance on your own. So if we convert those codes into the R we can do the same by just using base R.
Function :
rad = function(x) {
return(x * pi / 180)
}
getDistance = function(p1, p2) {
R = 6378137 # Earth’s mean radius in meter
dLat = rad(p2[1] - p1[1])
dLong = rad(p2[2] - p1[2])
a = ( sin(dLat / 2) * sin(dLat / 2) +
cos(rad(p1[1])) * cos(rad(p2[1])) *
sin(dLong / 2) * sin(dLong / 2) )
c = 2 * atan2(sqrt(a),sqrt(1 - a))
d = R * c
return(d) # returns the distance in meter
}
Example :
p1 <- c(55.32043 , -2.395500)
p3 <- c(55.15858 , -2.425284)
getDistance(p1,p3)
18115.96
Thus, once we can call those two functions, we can calculate any distance between two locations. So,
output <-lapply( 1:nrow(A), function(i)
lapply(1:nrow(B), function(j)
cbind(A[i,],B[j,],Distance=getDistance(as.numeric(A[i,-1]),as.numeric(B[j,-1])))
))
do.call(rbind,lapply(1:3,function(i) do.call(rbind,output[[i]])))
gives,
reference latitude longitude reference latitude longitude Distance
1 C 55.32043 -2.395500 C 55.15858 -2.425284 18115.958
2 C 55.32043 -2.395500 D 55.60859 -2.065054 38260.562
3 C 55.32043 -2.395500 E 55.59062 -2.065024 36603.447
23 D 55.59062 -2.065024 C 55.15858 -2.425284 53219.597
21 D 55.59062 -2.065024 D 55.60859 -2.065054 2000.412
22 D 55.59062 -2.065024 E 55.59062 -2.065024 0.000
33 E 55.60859 -2.065054 C 55.15858 -2.425284 55031.092
31 E 55.60859 -2.065054 D 55.60859 -2.065054 0.000
32 E 55.60859 -2.065054 E 55.59062 -2.065024 2000.412
来源：https://stackoverflow.com/questions/57525670/find-closest-points-from-data-set-b-to-point-in-data-set-a-using-lat-long-in-r