Please look at the following small working example:
#### Pseudo data
nobs1 <- 4000
nobs2 <- 5000
mylon1 <- runif(nobs1, min=0, max=1)-76
mylat1 <
Here's an option that decreases the runtime to ~2 seconds on my machine because part of it is vectorized.
A direct comparison with the original solution follows.
Test data:
nobs1 <- 4000
nobs2 <- 5000
mylon1 <- runif(nobs1, min=0, max=1)-76
mylat1 <- runif(nobs1, min=0, max=1)+37
mylon2 <- runif(nobs2, min=0, max=1)-76
mylat2 <- runif(nobs2, min=0, max=1)+37
Original solution:
#### define a distance function
thedistance <- function(lon1, lat1, lon2, lat2) {
R <- 6371 # Earth mean radius [km]
delta.lon <- (lon2 - lon1)
delta.lat <- (lat2 - lat1)
a <- sin(delta.lat/2)^2 + cos(lat1) * cos(lat2) * sin(delta.lon/2)^2
c <- 2 * asin(min(1,sqrt(a)))
d = R * c
return(d)
}
ptm <- proc.time()
#### Calculate distances between locations
# Initiate the resulting distance vector
ndistance <- nobs1*nobs2 # The number of distances
mydistance <- vector(mode = "numeric", length = ndistance)
k=1
for (i in 1:nobs1) {
for (j in 1:nobs2) {
mydistance[k] = thedistance(mylon1[i],mylat1[i],mylon2[j],mylat2[j])
k=k+1
}
}
proc.time() - ptm
User System elapsed
148.243 0.681 148.901
My approach:
# modified (vectorized) distance function:
thedistance2 <- function(lon1, lat1, lon2, lat2) {
R <- 6371 # Earth mean radius [km]
delta.lon <- (lon2 - lon1)
delta.lat <- (lat2 - lat1)
a <- sin(delta.lat/2)^2 + cos(lat1) * cos(lat2) * sin(delta.lon/2)^2
c <- 2 * asin(pmin(1,sqrt(a))) # pmin instead of min
d = R * c
return(d)
}
ptm2 <- proc.time()
lst <- vector("list", length = nobs1)
for (i in seq_len(nobs1)) {
lst[[i]] = thedistance2(mylon1[i],mylat1[i],mylon2,mylat2)
}
res <- unlist(lst)
proc.time() - ptm2
User System elapsed
1.988 0.331 2.319
Are the results all equal?
all.equal(mydistance, res)
#[1] TRUE