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In R, I have a collection of points (~15,000) within a polygon.

For every point, I want to calculate the location on the polygon boundary nearest to the point's location. I'm not bothered about the point's distance from the boundary (i.e. gDistance etc.).

I have a method that works (below). I used nearestPointOnLine from maptools but this only works for 1 point. I wrapped it up in apply but with my spdf being very large, this is taking a while;

require(maptools)
require(sp)
require(rgeos)

# create a dummy polygon: coordinates, promote to Polygons and SpatialPolygons
coordsPol = cbind(c(1,2,3,4,4),c(3,2,2,1,3))
verts <- rbind(coordsPol, coordsPol[1, ])
pol <- SpatialPolygons(list(Polygons(list(Polygon(verts)), ID="a")))

# create spatialpointsdataframe (they all fall inside the polygon)
p = data.frame(x=c(1.5,2.5,3,3.5,3.8),y=c(2.75,2.25,2.55,2.9,1.6),ID=c("a","b","c","d","e"))
coordinates(p) <- ~x+y

# Extract the point on the polygon boundary that is nearest to each spdf point.
# The polygon needs to be a matrix (ggplot::fortify) and maptools::nearestPointOnLine
# only works 1 point at a time so requires apply

v2.x <- apply(p@coords,1,function(x) nearestPointOnLine(as.matrix(fortify(pol)[,c(1,2)]),x))[1,]
v2.y <- apply(p@coords,1,function(x) nearestPointOnLine(as.matrix(fortify(pol)[,c(1,2)]),x))[2,]

# The output from nearestpointOnLine can be converted to a SpatialPoints object
v2m <- matrix(cbind(v2.x,v2.y),ncol=2)
v2p <- SpatialPoints(v2m)
  • Are there any functions (i thought rgdal would have one) that take a polygon and many points and give resulting nearest location points on a boundary (instead of apply and nearestPointOnLine)?
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  • There's a lot of questions here, and it might be better if you can post it as separate questions, each with sample data. You see, I might know how to do one of these things but I won't answer because I can't provide a complete solution.
    – Spacedman
    Commented Feb 12, 2018 at 15:48
  • The third one is quite easy - if you know the total perimeter of the loop. If the clockwise distance is greater than half the perimeter, then the anticlockwise distance is shorter and is "perimeter minus clockwise distance"
    – Spacedman
    Commented Feb 12, 2018 at 15:54
  • thank you, of course it is. i shall separate the two other main issues out into 2 qu's
    – Sam
    Commented Feb 12, 2018 at 15:55

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