如何使用data.table有效地计算坐标对之间的距离：= [英] How to efficiently calculate distance between pair of coordinates using data.table :=

问题描述

我想找到最有效（最快）的方法来计算两对lat长坐标之间的距离。

I want to find the most efficient (fastest) method to calculate the distances between pairs of lat long coordinates.

一个不那么高效的解决方案已经提出（此处）使用 sapply 和 spDistsN1 {sp} 。我相信这可以做得更快，如果 spDistsN1 {sp} 里面 data.table c $ c>：= 运算符，但是我还是不能这样做。有任何建议吗？

A not so efficient solution has been presented (here) using sapply and spDistsN1{sp}. I believe this could be made much faster if one would use spDistsN1{sp} inside data.table with the := operator but I haven't been able to do that. Any suggestions?

以下是可重复范例：

# 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")

使用 data.table

My failed attempt using data.table

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

这是一个可行的解决方案

Here is a solution that works (but probably less efficiently)

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

推荐答案

I写了我自己的版本 geosphere :: distHaversine ，以便它更自然地适合 data.table ：= 调用，它可能在这里使用

I wrote my own version of geosphere::distHaversine so that it would more naturally fit into a data.table := call, and it might be of use here

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)
}

如何对原始 geosphere :: distHaversine 和 geosphere :: distGeo

dt1 <- copy(odmatrix); dt2 <- copy(odmatrix); dt3 <- copy(odmatrix)

library(microbenchmark)

microbenchmark(

    dtHaversine = {
        dt1[, dist := dt.haversine(lat_orig, long_orig, lat_dest, long_dest)]
    }   ,

    haversine = {
        dt2[ , dist := distHaversine(matrix(c(long_orig, lat_orig), ncol = 2), 
                                     matrix(c(long_dest, lat_dest), ncol = 2))]
    },

    geo = {
        dt3[ , dist := distGeo(matrix(c(long_orig, lat_orig), ncol = 2), 
                               matrix(c(long_dest, lat_dest), ncol = 2))]
    }
)

# Unit: microseconds
#         expr     min       lq     mean   median       uq      max neval
# dtHaversine 370.300 396.6210 434.5841 411.4305 463.9965  906.797   100
#   haversine 651.974 681.1745 776.6127 706.2760 731.3480 1505.765   100
#         geo 647.699 679.8285 743.4914 706.0465 742.1605 1272.310   100

当然，在两种不同的技术（geo& haversine），结果将略有不同。

Naturally, due to the way the distances are calculated in the two different techniques (geo & haversine), the results will differ slightly.

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