如何使用data.table:=有效地计算坐标对之间的距离

Question

我想找到最有效(最快)的方法来计算lat长坐标对之间的距离.

已经提出了一个不太有效的解决方案(这里)使用sapply和spDistsN1{sp}.我相信如果一个人会在运营商spDistsN1{sp}内部data.table使用,:=但我无法做到这一点,这可以更快.有什么建议？

这是一个可重复的例子:

# load libraries
  library(data.table)
  library(dplyr)
  library(sp)
  library(rgeos)
  library(UScensus2000tract)

# load data and create an Origin-Destination matrix
  data("oregon.tract")

# get centroids as a data.frame
  centroids <- as.data.frame(gCentroid(oregon.tract,byid=TRUE))

# Convert row names into first column
  setDT(centroids, keep.rownames = TRUE)[]

# create Origin-destination matrix
  orig <- centroids[1:754, ]
  dest <- centroids[2:755, ]
  odmatrix <- bind_cols(orig,dest)
  colnames(odmatrix) <- c("origi_id", "long_orig", "lat_orig", "dest_id", "long_dest", "lat_dest")

我失败的尝试使用 data.table

odmatrix[ , dist_km := spDistsN1(as.matrix(long_orig, lat_orig), as.matrix(long_dest, lat_dest), longlat=T)]

这是一个有效的解决方案(但可能效率较低)

odmatrix$dist_km <- sapply(1:nrow(odmatrix),function(i)
  spDistsN1(as.matrix(odmatrix[i,2:3]),as.matrix(odmatrix[i,5:6]),longlat=T))

head(odmatrix)

>   origi_id long_orig lat_orig  dest_id long_dest lat_dest dist_km
>      (chr)     (dbl)    (dbl)    (chr)     (dbl)    (dbl)   (dbl)
> 1 oregon_0   -123.51   45.982 oregon_1   -123.67   46.113 19.0909
> 2 oregon_1   -123.67   46.113 oregon_2   -123.95   46.179 22.1689
> 3 oregon_2   -123.95   46.179 oregon_3   -123.79   46.187 11.9014
> 4 oregon_3   -123.79   46.187 oregon_4   -123.83   46.181  3.2123
> 5 oregon_4   -123.83   46.181 oregon_5   -123.85   46.182  1.4054
> 6 oregon_5   -123.85   46.182 oregon_6   -123.18   46.066 53.0709

Answer 1

我编写了自己的版本,geosphere::distHaversine以便更自然地适应data.table :=呼叫,它可能在这里有用

dt.haversine <- function(lat_from, lon_from, lat_to, lon_to, r = 6378137){
    radians <- pi/180
    lat_to <- lat_to * radians
    lat_from <- lat_from * radians
    lon_to <- lon_to * radians
    lon_from <- lon_from * radians
    dLat <- (lat_to - lat_from)
    dLon <- (lon_to - lon_from)
    a <- (sin(dLat/2)^2) + (cos(lat_from) * cos(lat_to)) * (sin(dLon/2)^2)
    return(2 * atan2(sqrt(a), sqrt(1 - a)) * r)
}

下面是关于如何执行对原有一些基准Rcpp::sourceCpp("distance_calcs.cpp"),并geosphere::distHaversine

#include <Rcpp.h>
using namespace Rcpp;

double inverseHaversine(double d){
  return 2 * atan2(sqrt(d), sqrt(1 - d)) * 6378137.0;
}

double distanceHaversine(double latf, double lonf, double latt, double lont,
                         double tolerance){
  double d;
  double dlat = latt - latf;
  double dlon =  lont - lonf;

  d = (sin(dlat * 0.5) * sin(dlat * 0.5)) + (cos(latf) * cos(latt)) * (sin(dlon * 0.5) * sin(dlon * 0.5));
  if(d > 1 && d <= tolerance){
    d = 1;
  }
  return inverseHaversine(d);
}

double toRadians(double deg){
  return deg * 0.01745329251;  // PI / 180;
}

// [[Rcpp::export]]
Rcpp::NumericVector rcpp_distance_haversine(Rcpp::NumericVector latFrom, Rcpp::NumericVector lonFrom, 
                        Rcpp::NumericVector latTo, Rcpp::NumericVector lonTo,
                        double tolerance) {

  int n = latFrom.size();
  NumericVector distance(n);

  double latf;
  double latt;
  double lonf;
  double lont;
  double dist = 0;

  for(int i = 0; i < n; i++){

    latf = toRadians(latFrom[i]);
    lonf = toRadians(lonFrom[i]);
    latt = toRadians(latTo[i]);
    lont = toRadians(lonTo[i]);
    dist = distanceHaversine(latf, lonf, latt, lont, tolerance);

    distance[i] = dist;
  }
  return distance;
}

当然,由于距离的计算方式采用两种不同的技术(geo和hasrsine),结果会略有不同.

Answer 2

感谢@ chinsoon12的评论,我发现一个很快速的解决方案相结合distGeo{geosphere}和data.table.在我的笔记本电脑中,快速解决方案的速度比替代方案快120倍.

让我们将数据集放大以比较速度性能.

# Multiplicate data observations by 1000 
  odmatrix <- odmatrix[rep(seq_len(nrow(odmatrix)), 1000), ]

缓慢解决

system.time(
           odmatrix$dist_km <- sapply(1:nrow(odmatrix),function(i)
             spDistsN1(as.matrix(odmatrix[i,2:3]),as.matrix(odmatrix[i,5:6]),longlat=T)) 
            )

 >   user  system elapsed 
 >   222.17    0.08  222.84 

快速解决方案

# load library
  library(geosphere)

# convert the data.frame to a data.table
  setDT(odmatrix)

system.time(
            odmatrix[ , dist_km2 := distGeo(matrix(c(long_orig, lat_orig), ncol = 2), 
                                            matrix(c(long_dest, lat_dest), ncol = 2))/1000]
           )

>   user  system elapsed 
>   1.76    0.03    1.79