R - Finding closest neighboring point and number of neighbors within a given radius, coordinates lat-long
Best option is to use libraries sp and rgeos, which enable you to construct spatial classes and perform geoprocessing.
library(sp)
library(rgeos)
Read the data and transform them to spatial objects:
mydata <- read.delim('d:/temp/testfile.txt', header=T)
sp.mydata <- mydata
coordinates(sp.mydata) <- ~long+lat
class(sp.mydata)
[1] "SpatialPointsDataFrame"
attr(,"package")
[1] "sp"
Now calculate pairwise distances between points
d <- gDistance(sp.mydata, byid=T)
Find second shortest distance (closest distance is of point to itself, therefore use second shortest)
min.d <- apply(d, 1, function(x) order(x, decreasing=F)[2])
Construct new data frame with desired variables
newdata <- cbind(mydata, mydata[min.d,], apply(d, 1, function(x) sort(x, decreasing=F)[2]))
colnames(newdata) <- c(colnames(mydata), 'neighbor', 'n.lat', 'n.long', 'n.area', 'n.canopy', 'n.avg.depth', 'distance')
newdata
pond lat long area canopy avg.depth neighbor n.lat n.long n.area n.canopy n.avg.depth
6 A10 41.95928 -72.14605 1500 66 60.61538 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222
3 AA006 41.96431 -72.12100 250 0 57.77778 Blacksmith 41.95508 -72.12380 361 77 71.31250
2 Blacksmith 41.95508 -72.12380 361 77 71.31250 AA006 41.96431 -72.12100 250 0 57.77778
5 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000
4 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444
5.1 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000
6.1 Boulder 41.91822 -72.14978 1392 98 43.53333 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222
distance
6 0.0085954872
3 0.0096462277
2 0.0096462277
5 0.0003059412
4 0.0003059412
5.1 0.0004548626
6.1 0.0374480316
Edit: if coordinates are in degrees and you would like to calculate distance in kilometers, use package geosphere
library(geosphere)
d <- distm(sp.mydata)
# rest is the same
This should provide better results, if the points are scattered across the globe and coordinates are in degrees
I add below an alternative solution using the newer sf package, for those interested and coming to this page now (as I did).
First, load the data and create the sf object.
# Using sf
mydata <- structure(
list(pond = c("A10", "AA006", "Blacksmith", "Borrow.Pit.1",
"Borrow.Pit.2", "Borrow.Pit.3", "Boulder"),
lat = c(41.95928, 41.96431, 41.95508, 41.95601, 41.95571, 41.95546,
41.918223),
long = c(-72.14605, -72.121, -72.123803, -72.15419, -72.15413,
-72.15375, -72.14978),
area = c(1500L, 250L, 361L, 0L, 0L, 0L, 1392L),
canopy = c(66L, 0L, 77L, 0L, 0L, 0L, 98L),
avg.depth = c(60.61538462, 57.77777778, 71.3125, 41.44444444,
37.7, 29.22222222, 43.53333333)),
class = "data.frame", row.names = c(NA, -7L))
library(sf)
data_sf <- st_as_sf(mydata, coords = c("long", "lat"),
# Change to your CRS
crs = "+proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs")
st_is_longlat(data_sf)
sf::st_distance calculates the distance matrix in meters using Great Circle distance when using lat/lon data.
dist.mat <- st_distance(data_sf) # Great Circle distance since in lat/lon
# Number within 1.5km: Subtract 1 to exclude the point itself
num.1500 <- apply(dist.mat, 1, function(x) {
sum(x < 1500) - 1
})
# Calculate nearest distance
nn.dist <- apply(dist.mat, 1, function(x) {
return(sort(x, partial = 2)[2])
})
# Get index for nearest distance
nn.index <- apply(dist.mat, 1, function(x) { order(x, decreasing=F)[2] })
n.data <- mydata
colnames(n.data)[1] <- "neighbor"
colnames(n.data)[2:ncol(n.data)] <-
paste0("n.", colnames(n.data)[2:ncol(n.data)])
mydata2 <- data.frame(mydata,
n.data[nn.index, ],
n.distance = nn.dist,
radius1500 = num.1500)
rownames(mydata2) <- seq(nrow(mydata2))
mydata2
pond lat long area canopy avg.depth neighbor n.lat n.long n.area n.canopy
1 A10 41.95928 -72.14605 1500 66 60.61538 Borrow.Pit.1 41.95601 -72.15419 0 0
2 AA006 41.96431 -72.12100 250 0 57.77778 Blacksmith 41.95508 -72.12380 361 77
3 Blacksmith 41.95508 -72.12380 361 77 71.31250 AA006 41.96431 -72.12100 250 0
4 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 Borrow.Pit.2 41.95571 -72.15413 0 0
5 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 Borrow.Pit.1 41.95601 -72.15419 0 0
6 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 Borrow.Pit.2 41.95571 -72.15413 0 0
7 Boulder 41.91822 -72.14978 1392 98 43.53333 Borrow.Pit.3 41.95546 -72.15375 0 0
n.avg.depth n.distance radius1500
1 41.44444 766.38426 3
2 71.31250 1051.20527 1
3 57.77778 1051.20527 1
4 37.70000 33.69099 3
5 41.44444 33.69099 3
6 37.70000 41.99576 3
7 29.22222 4149.07406 0
For obtaining the closest neighbor after calculating distance, you can use sort() with the partial = 2 argument. Depending on the amount of data, this could be much faster than using order as in the previous solution. The package Rfast is likely even faster but I avoid including additional packages here. See this related post for a discussion and benchmarking of various solutions: https://stackoverflow.com/a/53144760/12265198
In Rfast, there is a function called "dista" and computes the Euclidean or the Manhattan distances only (at the moment). It gives the option to compute the k-smallest distances. Alternatively it can return the indexes of the observations with the smallest distances. The cosinus distance is basically almost the same as the Eucledean distance (exluding a constant, 2 I think).