[英]Matrix of distances with Geosphere: avoid repeat calculus
I want to compute the distance among all points in a very large matrix using distm
from geosphere
. 我想使用来自地
geosphere
distm
来计算非常大的矩阵中所有点之间的距离。
See a minimal example: 看一个最小的例子:
library(geosphere)
library(data.table)
coords <- data.table(coordX=c(1,2,5,9), coordY=c(2,2,0,1))
distances <- distm(coords, coords, fun = distGeo)
The issue is that due to the nature of the distances I am computing, distm
gives me back a symmetric matrix, therefore, I could avoid to calculate more than half of the distances: 问题在于,由于我正在计算的距离的性质,
distm
给了我一个对称矩阵,因此,我可以避免计算超过一半的距离:
structure(c(0, 111252.129800202, 497091.059564718, 897081.91986428,
111252.129800202, 0, 400487.621661164, 786770.053508848, 497091.059564718,
400487.621661164, 0, 458780.072878927, 897081.91986428, 786770.053508848,
458780.072878927, 0), .Dim = c(4L, 4L))
May you help me to find a more efficient way to compute all those distances avoiding doing twice each one? 你可以帮我找一个更有效的方法来计算所有这些距离,避免每次做两次吗?
You can prepare a data frame of possible combinations without repetitions (with gtools
packages). 您可以准备可能组合的数据框而无需重复(使用
gtools
包)。 Then to compute distances for those pairs. 然后计算这些对的距离。 Here is the code:
这是代码:
library(gtools)
library(geosphere)
library(data.table)
coords <- data.table(coordX = c(1, 2, 5, 9), coordY = c(2, 2, 0, 1))
pairs <- combinations(n = nrow(coords), r = 2, repeats.allowed = F, v = c(1:nrow(coords)))
distances <- apply(pairs, 1, function(x) {
distm(coords[x[1], ], coords[x[2], ], fun = distGeo)
})
# Construct distances matrix
dist_mat <- matrix(NA, nrow = nrow(coords), ncol = nrow(coords))
dist_mat[upper.tri(dist_mat)] <- distances
dist_mat[lower.tri(dist_mat)] <- distances
dist_mat[is.na(dist_mat)] <- 0
print(dist_mat)
The results: 结果:
[,1] [,2] [,3] [,4]
[1,] 0.0 111252.1 497091.1 400487.6
[2,] 111252.1 0.0 897081.9 786770.1
[3,] 497091.1 400487.6 0.0 458780.1
[4,] 897081.9 786770.1 458780.1 0.0
If you want to compute all pairwise distances for points x
, it is better to use distm(x)
rather than distm(x,x)
. 如果要计算点
x
所有成对距离,最好使用distm(x)
而不是distm(x,x)
。 The distm
function returns the same symmetric matrix in both cases but when you pass it a single argument it knows that the matrix is symmetric, so it won't do unnecessary computations. distm
函数在两种情况下都返回相同的对称矩阵,但是当您传递一个参数时,它知道矩阵是对称的,因此它不会进行不必要的计算。
You can time it. 你可以计时。
library("geosphere")
n <- 500
xy <- matrix(runif(n*2, -90, 90), n, 2)
system.time( replicate(100, distm(xy, xy) ) )
# user system elapsed
# 61.44 0.23 62.79
system.time( replicate(100, distm(xy) ) )
# user system elapsed
# 36.27 0.39 38.05
You can also look at the R code for geosphere::distm
to check that it treats the two cases differently. 您还可以查看
geosphere::distm
的R代码,以检查它geosphere::distm
以不同方式处理这两种情况。
Aside: Quick google search finds parallelDist
: Parallel Distance Matrix Computation on CRAN. 除此之外:快速谷歌搜索找到
parallelDist
:CRAN上的并行距离矩阵计算。 The geodesic distance is an option. 测地距离是一种选择。
Using combn()
from base R might be slightly simpler and probably faster than loading additional packages. 使用基础R中的
combn()
可能稍微简单一些,并且可能比加载其他包更快。 Then, distm()
uses distGeo()
as a source, so using the latter should be even faster. 然后,
distm()
使用distGeo()
作为源,因此使用后者应该更快。
coords <- as.data.frame(coords) # this won't work with data.tables though
cbind(t(combn(1:4, 2)), unique(geosphere::distGeo(coords[combn(1:4, 2), ])))
# [,1] [,2] [,3]
# [1,] 1 2 111252.1
# [2,] 1 3 497091.1
# [3,] 1 4 897081.9
# [4,] 2 3 786770.1
# [5,] 2 4 400487.6
# [6,] 3 4 458780.1
We could check it out with a benchmark. 我们可以用基准测试一下。
Unit: microseconds
expr min lq mean median uq max neval cld
distm 555.690 575.846 597.7672 582.352 596.1295 904.718 100 b
distGeo 426.335 434.372 450.0196 441.516 451.8490 609.524 100 a
Looks good. 看起来不错。
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