问题
I have two matrices, one is 200K rows long, the other is 20K. For each row (which is a point) in the first matrix, I am trying to find which row (also a point) in the second matrix is closest to the point in the first matrix. This is the first method that I tried on a sample dataset:
#Test dataset
pixels.latlon=cbind(runif(200000,min=-180, max=-120), runif(200000, min=50, max=85))
grwl.latlon=cbind(runif(20000,min=-180, max=-120), runif(20000, min=50, max=85))
#calculate the distance matrix
library(geosphere)
dist.matrix=distm(pixels.latlon, grwl.latlon, fun=distHaversine)
#Pick out the indices of the minimum distance
rnum=apply(dist.matrix, 1, which.min)
However, I get a Error: cannot allocate vector of size 30.1 Gb error when I use the distm function.
There have been several posts on this topic:
This one uses bigmemory to calculate distances between points in the SAME dataframe, but I'm not sure how to adapt it to calculate distances between points in two different matrices...https://stevemosher.wordpress.com/2012/04/12/nick-stokes-distance-code-now-with-big-memory/
This one also works for calculating a distance matrix between points in the SAME matrix...Efficient (memory-wise) function for repeated distance matrix calculations AND chunking of extra large distance matrices
And this one is pretty much identical to what I want to do, but they didn't actually come up with a solution that worked for large data: R: distm with Big Memory I tried this method, which uses bigmemory, but get a Error in CreateFileBackedBigMatrix(as.character(backingfile), as.character(backingpath), :
Problem creating filebacked matrix. error, I think because the dataframe is too large.
Has anyone come up with a good solution to this problem? I am open to other package ideas!
Updated code which fixed the issue
pixels.latlon=cbind(runif(200000,min=-180, max=-120), runif(200000, min=50, max=85))
grwl.tibble = tibble(long=runif(20000,min=-180, max=-120), lat=runif(20000, min=50, max=85), id=runif(20000, min=0, max=20000))
rnum <- apply(pixels.latlon, 1, function(x) {
xlon=x[1]
xlat=x[2]
grwl.filt = grwl.tibble %>%
filter(long < (xlon+0.3) & long >(xlon-0.3) & lat < (xlat+0.3)&lat >(xlat-.3))
grwl.latlon.filt = cbind(grwl.filt$long, grwl.filt$lat)
dm <- distm(x, grwl.latlon.filt, fun=distHaversine)
rnum=apply(dm, 1, which.min)
id = grwl.filt$id[rnum]
return(id)
})
回答1:
You can use this R(cpp) function:
#include <Rcpp.h>
using namespace Rcpp;
double compute_a(double lat1, double long1, double lat2, double long2) {
double sin_dLat = ::sin((lat2 - lat1) / 2);
double sin_dLon = ::sin((long2 - long1) / 2);
return sin_dLat * sin_dLat + ::cos(lat1) * ::cos(lat2) * sin_dLon * sin_dLon;
}
int find_min(double lat1, double long1,
const NumericVector& lat2,
const NumericVector& long2,
int current0) {
int m = lat2.size();
double lat_k, lat_min, lat_max, a, a0;
int k, current = current0;
a0 = compute_a(lat1, long1, lat2[current], long2[current]);
// Search before current0
lat_min = lat1 - 2 * ::asin(::sqrt(a0));
for (k = current0 - 1; k >= 0; k--) {
lat_k = lat2[k];
if (lat_k > lat_min) {
a = compute_a(lat1, long1, lat_k, long2[k]);
if (a < a0) {
a0 = a;
current = k;
lat_min = lat1 - 2 * ::asin(::sqrt(a0));
}
} else {
// No need to search further
break;
}
}
// Search after current0
lat_max = lat1 + 2 * ::asin(::sqrt(a0));
for (k = current0 + 1; k < m; k++) {
lat_k = lat2[k];
if (lat_k < lat_max) {
a = compute_a(lat1, long1, lat_k, long2[k]);
if (a < a0) {
a0 = a;
current = k;
lat_max = lat1 + 2 * ::asin(::sqrt(a0));
}
} else {
// No need to search further
break;
}
}
return current;
}
// [[Rcpp::export]]
IntegerVector find_closest_point(const NumericVector& lat1,
const NumericVector& long1,
const NumericVector& lat2,
const NumericVector& long2) {
int n = lat1.size();
IntegerVector res(n);
int current = 0;
for (int i = 0; i < n; i++) {
res[i] = current = find_min(lat1[i], long1[i], lat2, long2, current);
}
return res; // need +1
}
/*** R
N <- 2000 # 2e6
M <- 500 # 2e4
pixels.latlon=cbind(runif(N,min=-180, max=-120), runif(N, min=50, max=85))
grwl.latlon=cbind(runif(M,min=-180, max=-120), runif(M, min=50, max=85))
# grwl.latlon <- grwl.latlon[order(grwl.latlon[, 2]), ]
library(geosphere)
system.time({
#calculate the distance matrix
dist.matrix = distm(pixels.latlon, grwl.latlon, fun=distHaversine)
#Pick out the indices of the minimum distance
rnum=apply(dist.matrix, 1, which.min)
})
find_closest <- function(lat1, long1, lat2, long2) {
toRad <- pi / 180
lat1 <- lat1 * toRad
long1 <- long1 * toRad
lat2 <- lat2 * toRad
long2 <- long2 * toRad
ord1 <- order(lat1)
rank1 <- match(seq_along(lat1), ord1)
ord2 <- order(lat2)
ind <- find_closest_point(lat1[ord1], long1[ord1], lat2[ord2], long2[ord2])
ord2[ind + 1][rank1]
}
system.time(
test <- find_closest(pixels.latlon[, 2], pixels.latlon[, 1],
grwl.latlon[, 2], grwl.latlon[, 1])
)
all.equal(test, rnum)
N <- 2e4
M <- 2e4
pixels.latlon=cbind(runif(N,min=-180, max=-120), runif(N, min=50, max=85))
grwl.latlon=cbind(long = runif(M,min=-180, max=-120), lat = runif(M, min=50, max=85))
system.time(
test <- find_closest(pixels.latlon[, 2], pixels.latlon[, 1],
grwl.latlon[, 2], grwl.latlon[, 1])
)
*/
It takes 0.5 sec for N = 2e4 and 4.2 sec for N = 2e5.
I can't make your code work to compare.
回答2:
This would use much less memory, as it does it one row at a time, rather than creating the full distance matrix (though it will be slower)
library(geosphere)
rnum <- apply(pixels.latlon, 1, function(x) {
dm <- distm(x, grwl.latlon, fun=distHaversine)
return(which.min(dm))
})
Much of the time is taken up with the complicated Haversine formula. As you are really only interested in finding the closest point, not in the exact distances, we could use a simpler distance measure. Here is an alternative using a formula based on this article http://jonisalonen.com/2014/computing-distance-between-coordinates-can-be-simple-and-fast/, and also using a quadratic approximation to the cosine (which is itself expensive to calculate)...
#quadratic cosine approximation using lm (run once)
qcos <- lm(y~x+I(x^2), data.frame(x=0:90, y=cos((0:90)*2*pi/360)))$coefficients
cosadj <- function(lat) qcos[1]+lat*(qcos[2]+qcos[3]*lat)
#define rough dist function
roughDist <- function(x,y){#x should be a single (lon,lat), y a (n*2) matrix of (lon,lat)
latDev <- x[2]-y[,2]
lonDev <- (x[1]-y[,1])*cosadj(abs(x[2]))
return(latDev*latDev+lonDev*lonDev) #don't need the usual square root or any scaling parameters
}
And then you can just replace Haversine with this new function...
rnum <- apply(pixels.latlon, 1, function(x) {
dm <- distm(x, grwl.latlon, fun=roughDist)
return(which.min(dm))
})
On my machine this runs about three times faster than the Haversine version.
来源：https://stackoverflow.com/questions/49863185/r-distm-for-big-data-calculating-minimum-distances-between-two-matrices