问题描述
我有两个矩阵，一个矩阵长200K行，另一个矩阵20K.对于第一个矩阵中的每一行(这是一个点)，我试图找到第二个矩阵中的哪一行(也是一个点)最接近第一个矩阵中的点.这是我对样本数据集尝试的第一种方法:
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)
但是，使用distm函数时出现Error: cannot allocate vector of size 30.1 Gb错误.
However, I get a Error: cannot allocate vector of size 30.1 Gb error when I use the distm function.
关于该主题有几篇文章:
There have been several posts on this topic:
此人使用bigmemory来计算SAME数据帧中点之间的距离，但是我不确定如何适应它以计算两个不同矩阵中的点之间的距离... "> https://stevemosher.wordpress.com/2012/04/12/nick-stokes-distance-code-now-with-big-memory/
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/
该函数还可以用于计算SAME矩阵中各点之间的距离矩阵...
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
这与我要执行的操作几乎相同，但是他们实际上并没有提出一种适用于大数据的解决方案:我尝试了使用bigmemory的方法，但是却出现了Error in CreateFileBackedBigMatrix(as.character(backingfile), as.character(backingpath), : Problem creating filebacked matrix.错误，我认为是因为数据帧太大.
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.
有人解决这个问题吗?我愿意接受其他包装创意！
Has anyone come up with a good solution to this problem? I am open to other package ideas!
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)
})
推荐答案
您可以使用此R(cpp)函数:
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])
)
*/
N = 2e4花费0.5秒，N = 2e5花费4.2秒.我无法让您的代码可以进行比较.
It takes 0.5 sec for N = 2e4 and 4.2 sec for N = 2e5.I can't make your code work to compare.
这篇关于R:大数据区吗?计算两个矩阵之间的最小距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！