计算R中累积距离的矩阵

Question

I need an efficient way to calculate a matrix of distances between a series of points. The catch is that you can only get from point 'i' to point 'k' by passing through all points 'j' in between. As an example, imagine an island with 5 beaches and you want to calculate the distance between all beaches along the shoreline because you cannot cut across the island (including in both directions: clockwise or counter-clockwise).

Below are some example data. (Note: You will need to install the package 'geosphere' to use the 'distm' function, which calculates the distance between GPS coordinates along the surface of the Earth)

library("geosphere")

longitude = c(-119.003, -119.067, -119.121, -119.089, -119.003)
latitude = c(33.503, 33.539, 33.485, 33.413, 33.440)
long.lat.mat = as.matrix(cbind(longitude, latitude))

# Use "distm" to calculate Euclidean (straight-line) distances between sites (in km)
euclid.dist.mat = distm(long.lat.mat) / 1000

# Create an empty matrix of alongshore distances (from "rows" to "columns")
alongshore.dist.mat = matrix(ncol=dim(long.lat.mat)[1], nrow=dim(long.lat.mat)[1], data=NA)   

# Diagonal is zero. Adjacent sites are the same as Euclidean distance
diag(alongshore.dist.mat) = 0   
diag(alongshore.dist.mat[,-1]) = diag(euclid.dist.mat[,-1])
alongshore.dist.mat[1,dim(long.lat.mat)[1]] = euclid.dist.mat[1,dim(long.lat.mat)[1]]
alongshore.dist.mat[lower.tri(alongshore.dist.mat)] = t(alongshore.dist.mat)[lower.tri(t(alongshore.dist.mat))]

# > alongshore.dist.mat
#           [,1]      [,2]      [,3]      [,4]      [,5]
# [1,] 0.0000000 7.1650632        NA        NA 7.0131279
# [2,] 7.1650632 0.0000000 7.8265783        NA        NA
# [3,]        NA 7.8265783 0.0000000 8.5483605        NA
# [4,]        NA        NA 8.5483605 0.0000000 8.5365807
# [5,] 7.0131279        NA        NA 8.5365807 0.0000000

Now, how to fill in remaining cells? As an example:

alongshore.dist.mat[1,3] = 7.1650632 + 7.8265783 = 14.991642 

...representing site 1 -> site 2 -> site 3. By contrast:

alongshore.dist.mat[3,1] = 8.5483605 + 8.5365807 + 7.0131279 = 24.098069

...representing site 3 -> site 4 -> site 5 -> site 1.

I suspect that the "cumsum" function can be used efficiently, but not sure exactly how to set it up. I am hoping for a solution avoiding for-loops, as I in reality am working with data containing dozens of points.

Answer 1

You could first build a matrix of all edges between two locations:

dists <- expand.grid(x=1:5, y=1:5)
dists$weight <- alongshore.dist.mat[as.matrix(dists)]
dists <- subset(dists, x != y & !is.na(weight))
dists
#    x y   weight
# 2  2 1 7.165063
# 5  5 1 7.013128
# 6  1 2 7.165063
# 8  3 2 7.826578
# 12 2 3 7.826578
# 14 4 3 8.548360
# 18 3 4 8.548360
# 20 5 4 8.536581
# 21 1 5 7.013128
# 24 4 5 8.536581

Now you can build a graph and compute the all-pairs shortest paths:

library(igraph)
g <- graph.data.frame(dists, vertices=data.frame(x=1:5))
shortest.paths(g)
#           1         2         3         4         5
# 1  0.000000  7.165063 14.991642 15.549709  7.013128
# 2  7.165063  0.000000  7.826578 16.374939 14.178191
# 3 14.991642  7.826578  0.000000  8.548360 17.084941
# 4 15.549709 16.374939  8.548360  0.000000  8.536581
# 5  7.013128 14.178191 17.084941  8.536581  0.000000