我遇到的问题是我必须计算形状之间的欧几里德距离矩阵，其范围从20,000到60,000点，产生10-20GB的数据量。我必须运行这些计算中的每一个数千次，所以20GB x 7,000（每个计算是一个不同的点云）。形状可以是2D或3D。
已编辑（更新的问题）
是否有更有效的方法来计算前后距离而不使用两个独立的嵌套循环？
我知道我可以保存数据矩阵并计算最小值 每个方向的距离，但是存在巨大的内存问题 大点云。
有没有办法加快计算速度和/或清理代码以缩短时间？
数据示例，其中每列表示形状的尺寸，每行是形状中的一个点：
first_configuration <- matrix(1:6,2,3)
second_configuration <- matrix(6:11,2,3)
colnames(first_configuration) <- c("x","y","z")
colnames(second_configuration) <- c("x","y","z")
此代码计算坐标之间的欧几里德距离：
m <- nrow(first_configuration)
n <- nrow(second_configuration)
D <- sqrt(pmax(matrix(rep(apply(first_configuration * first_configuration, 1, sum), n), m, n, byrow = F) + matrix(rep(apply(second_configuration * second_configuration, 1, sum), m), m, n, byrow = T) - 2 * first_configuration %*% t(second_configuration), 0))
D
输出：
[,1] [,2]
[1,] 8.660254 10.392305
[2,] 6.928203 8.660254
编辑：包括hausdorff平均代码
d1 <- mean(apply(D, 1, min))
d2 <- mean(apply(D, 2, min))
average_hausdorff <- mean(d1, d2)
EDIT（Rcpp解决方案）： 这是我尝试在Rcpp中实现它，因此矩阵永远不会保存到内存中。现在工作但很慢。
sourceCpp(code=
#include <Rcpp.h>
#include <limits>
using namespace Rcpp;
// [[Rcpp::export]]
double edist_rcpp(NumericVector x, NumericVector y){
double d = sqrt( sum( pow(x - y, 2) ) );
return d;
}
// [[Rcpp::export]]
double avg_hausdorff_rcpp(NumericMatrix x, NumericMatrix y){
int nrowx = x.nrow();
int nrowy = y.nrow();
double new_low_x = std::numeric_limits<int>::max();
double new_low_y = std::numeric_limits<int>::max();
double mean_forward = 0;
double mean_backward = 0;
double mean_hd;
double td;
//forward
for(int i = 0; i < nrowx; i++) {
for(int j = 0; j < nrowy; j++) {
NumericVector v1 = x.row(i);
NumericVector v2 = y.row(j);
td = edist_rcpp(v1, v2);
if(td < new_low_x) {
new_low_x = td;
}
}
mean_forward = mean_forward + new_low_x;
new_low_x = std::numeric_limits<int>::max();
}
//backward
for(int i = 0; i < nrowy; i++) {
for(int j = 0; j < nrowx; j++) {
NumericVector v1 = y.row(i);
NumericVector v2 = x.row(j);
td = edist_rcpp(v1, v2);
if(td < new_low_y) {
new_low_y = td;
}
}
mean_backward = mean_backward + new_low_y;
new_low_y = std::numeric_limits<int>::max();
}
//hausdorff mean
mean_hd = (mean_forward / nrowx + mean_backward / nrowy) / 2;
return mean_hd;
}
)
EDIT（RcppParallel解决方案）： 绝对比串行Rcpp解决方案更快，而且肯定是R解决方案。如果有人有关于如何改进我的RcppParallel代码以减少一些额外时间的提示，那将非常感激！
sourceCpp(code=
#include <Rcpp.h>
#include <RcppParallel.h>
#include <limits>
// [[Rcpp::depends(RcppParallel)]]
struct minimum_euclidean_distances : public RcppParallel::Worker {
//Input
const RcppParallel::RMatrix<double> a;
const RcppParallel::RMatrix<double> b;
//Output
RcppParallel::RVector<double> medm;
minimum_euclidean_distances(const Rcpp::NumericMatrix a, const Rcpp::NumericMatrix b, Rcpp::NumericVector medm) : a(a), b(b), medm(medm) {}
void operator() (std::size_t begin, std::size_t end) {
for(std::size_t i = begin; i < end; i++) {
double new_low = std::numeric_limits<double>::max();
for(std::size_t j = 0; j < b.nrow(); j++) {
double dsum = 0;
for(std::size_t z = 0; z < b.ncol(); z++) {
dsum = dsum + pow(a(i,z) - b(j,z), 2);
}
dsum = pow(dsum, 0.5);
if(dsum < new_low) {
new_low = dsum;
}
}
medm[i] = new_low;
}
}
};
// [[Rcpp::export]]
double mean_directional_hausdorff_rcpp(Rcpp::NumericMatrix a, Rcpp::NumericMatrix b){
Rcpp::NumericVector medm(a.nrow());
minimum_euclidean_distances minimum_euclidean_distances(a, b, medm);
RcppParallel::parallelFor(0, a.nrow(), minimum_euclidean_distances);
double results = Rcpp::sum(medm);
results = results / a.nrow();
return results;
}
// [[Rcpp::export]]
double max_directional_hausdorff_rcpp(Rcpp::NumericMatrix a, Rcpp::NumericMatrix b){
Rcpp::NumericVector medm(a.nrow());
minimum_euclidean_distances minimum_euclidean_distances(a, b, medm);
RcppParallel::parallelFor(0, a.nrow(), minimum_euclidean_distances);
double results = Rcpp::max(medm);
return results;
}
)
使用大小为37,775和36,659的大点云的基准：
//Rcpp serial solution
system.time(avg_hausdorff_rcpp(ll,rr))
user system elapsed
409.143 0.000 409.105
//RcppParallel solution
system.time(mean(mean_directional_hausdorff_rcpp(ll,rr), mean_directional_hausdorff_rcpp(rr,ll)))
user system elapsed
260.712 0.000 33.265
答案 0 :(得分：2)
我尝试使用JuliaCall来计算Hausdorff的平均距离。
JuliaCall将Julia嵌入R.
我只在JuliaCall中尝试使用串行解决方案。它似乎比问题中的RcppParallel和Rcpp序列解决方案更快，但我没有基准数据。因为并行计算的能力是在Julia中构建的。 Julia中的并行计算版本应该没有太大困难。在发现之后我会更新我的答案。
以下是我写的julia文件：
# Calculate the min distance from the k-th point in as to the points in bs
function min_dist(k, as, bs)
n = size(bs, 1)
p = size(bs, 2)
dist = Inf
for i in 1:n
r = 0.0
for j in 1:p
r += (as[k, j] - bs[i, j]) ^ 2
## if r is already greater than the upper bound,
## then there is no need to continue doing the calculation
if r > dist
continue
end
end
if r < dist
dist = r
end
end
sqrt(dist)
end
function avg_min_dist_from(as, bs)
distsum = 0.0
n1 = size(as, 1)
for k in 1:n1
distsum += min_dist_from(k, as, bs)
end
distsum / n1
end
function hausdorff_avg_dist(as, bs)
(avg_min_dist_from(as, bs) + avg_min_dist_from(bs, as)) / 2
end
这是使用julia函数的R代码：
first_configuration <- matrix(1:6,2,3)
second_configuration <- matrix(6:11,2,3)
colnames(first_configuration) <- c("x","y","z")
colnames(second_configuration) <- c("x","y","z")
m <- nrow(first_configuration)
n <- nrow(second_configuration)
D <- sqrt(matrix(rep(apply(first_configuration * first_configuration, 1, sum), n), m, n, byrow = F) + matrix(rep(apply(second_configuration * second_configuration, 1, sum), m), m, n, byrow = T) - 2 * first_configuration %*% t(second_configuration))
D
d1 <- mean(apply(D, 1, min))
d2 <- mean(apply(D, 2, min))
average_hausdorff <- mean(d1, d2)
library(JuliaCall)
## the first time of julia_setup could be quite time consuming
julia_setup()
## source the julia file which has our hausdorff_avg_dist function
julia_source("hausdorff.jl")
## check if the julia function is correct with the example
average_hausdorff_julia <- julia_call("hausdauff_avg_dist",
first_configuration,
second_configuration)
## generate some large random point clouds
n1 <- 37775
n2 <- 36659
as <- matrix(rnorm(n1 * 3), n1, 3)
bs <- matrix(rnorm(n2 * 3), n2, 3)
system.time(julia_call("hausdauff_avg_dist", as, bs))
我的笔记本电脑上的时间不到20秒，请注意这是JuliaCall的串行版本的性能！我使用相同的数据来测试问题中的RCpp串行解决方案，运行时间超过10分钟。我的笔记本电脑上没有RCpp并行，所以我无法尝试。正如我所说，Julia具有内置的并行计算能力。