I am trying to find the orthogonal distance between a set of location coordinates and a set of lines (roads or rivers). The set of points are in the form of latitude/longitude pairs, and the lines are in a shapefile (.shp). Plotting them on a map is not a problem, using either maptools or PBSmapping. But my basic problem is to find the minimum distance one has to travel from a location to reach a road or a river. Is there any way to do this in R?
If I understand correctly, you can do this simply enough with gDistance in the rgeos package.
Read in the lines as SpatialLines/DataFrame and points as SpatialPoints/DataFrame and then loop over each point calculating the distance each time:
require(rgeos)
## untested code
shortest.dists <- numeric(nrow(sp.pts))
for (i in seq_len(nrow(sp.pts)) {
shortest.dists[i] <- gDistance(sp.pts[i,], sp.lns)
}
Here sp.pts is the Spatial points object, and sp.lns is the Spatial lines object.
You must loop so that you only compare a single coordinate in sp.pts with the entirety of all lines geometries in sp.lns, otherwise you get the distance from an aggregate value across all points.
Since your data are in latitude/longitude you should transform both the lines and points to a suitable projection since the gDistance function assumes Cartesian distance.
MORE DISCUSSION AND EXAMPLE (edit)
It would be neat to get the nearest point on the line/s rather than just the distance, but this opens another option which is whether you need the nearest coordinate along a line, or an actual intersection with a line segment that is closer than any existing vertex. If your vertices are dense enough that the difference doesn't matter, then use spDistsN1 in the sp package. You'd have to extract all the coordinates from every line in the set (not hard, but a bit ugly) and then loop over each point of interest calculating the distance to the line vertices - then you can find which is the shortest and select that coordinate from the set of vertices, so you can have the distance and the coordinate easily. There's no need to project either since the function can use ellipsoidal distances with longlat = TRUE argument.
library(maptools)
## simple global data set, which we coerce to Lines
data(wrld_simpl)
wrld_lines <- as(wrld_simpl, "SpatialLinesDataFrame")
## get every coordinate as a simple matrix (scary but quick)
wrld_coords <- do.call("rbind", lapply(wrld_lines#lines, function(x1) do.call("rbind", lapply(x1#Lines, function(x2) x2#coords[-nrow(x2#coords), ]))))
Check it out interactively, you'll have to modify this to save the coords or minimum distances. This will plot up the lines and wait for you to click anywhere in the plot, then it will draw a line from your click to the nearest vertex on a line.
## no out of bounds clicking . . .
par(mar = c(0, 0, 0, 0), xaxs = "i", yaxs = "i")
plot(wrld_lines, asp = "")
n <- 5
for (i in seq_len(n)) {
xy <- matrix(unlist(locator(1)), ncol = 2)
all.dists <- spDistsN1(wrld_coords, xy, longlat = TRUE)
min.index <- which.min(all.dists)
points(xy, pch = "X")
lines(rbind(xy, wrld_coords[min.index, , drop = FALSE]), col = "green", lwd = 2)
}
The geosphere package has the dist2line function that does this for lon/lat data. It can use Spatial* objects or matrices.
line <- rbind(c(-180,-20), c(-150,-10), c(-140,55), c(10, 0), c(-140,-60))
pnts <- rbind(c(-170,0), c(-75,0), c(-70,-10), c(-80,20), c(-100,-50),
c(-100,-60), c(-100,-40), c(-100,-20), c(-100,-10), c(-100,0))
d <- dist2Line(pnts, line)
d
Illustration of the results
plot( makeLine(line), type='l')
points(line)
points(pnts, col='blue', pch=20)
points(d[,2], d[,3], col='red', pch='x')
for (i in 1:nrow(d)) lines(gcIntermediate(pnts[i,], d[i,2:3], 10), lwd=2)
Looks like this can be done in the sf package using the st_distance function.
You pass your two sf objects to the function. Same issue as with the other solutions in that you need to iterate over your points so that the function calculates the distance between every point to every point on the roadways. Then take the minimum of the resulting vector for the shortest distance.
# Solution for one point
min(st_distance(roads_sf, points_sf[1, ]))
# Iterate over all points using sapply
sapply(1:nrow(points_sf), function(x) min(st_distance(roads_sf, points_sf[x, ])))
Related
I have a series of points in an area whose 'footprint' shape is highly irregular:
I'd like to determine all of the coordinates within the footprint's vertices. The end goal is to determine which data points lay outside this footprint.
Does anyone have an efficient way to go about doing this??
My best idea to approaching this is to draw a polygon based on the green area's vertices and then use said polygon's coordinates to determine 'outlier' points' (though, I'm not sure how to do that yet -- one step at a time!).
However, when I try creating a convex hull, it obviously creates problems because of the irregular shape of my green space. [Anyone know of a way to create CONCAVE hulls?]
Alternatively, is there a way to draw polygons manually using a 'click the graph' type method?
...Again, if you have a better solution to my problem than using polygons, please by all means suggest that solution!
Alternatively, is there a way to draw polygons manually using a 'click
the graph' type method?
Here's one idea. First, some random points:
library(manipulate)
library(sp)
set.seed(1)
par(pch = 19, cex=.5)
x <- runif(1000)
y <- runif(1000)
Now, draw and capture the polygon:
coords <- data.frame()
manipulate({
plot(y~x)
res <- manipulatorMouseClick()
coords <<- rbind(coords, data.frame(x=res$userX, y=res$userY))
if (length(coords)) lines(coords)
})
And determine which points are inside/outside of it (see ?point.in.polygon):
res <- point.in.polygon(x, y, coords$x, coords$y)!=0
plot(y~x, col = res + 1L)
lines(coords)
I use extract() to get bilinear interpolated points out of my raster-object. Is there a way to show intermediate steps of the interpolation? I would be interested in the coordinates and values of the nearest four grid-points and the distances to my interpolation point.
With r = raster(...) and spdf = SpatialPointsDataFrame() my function call is the following:
out <- extract(r, spdf, method="bilinear")
You can run the code in raster:::.bilinearValue line by line, with raster = r; xyCoords = coordinates(spdf); layer=1, n=1
When extracting values of a raster along a SpatialLine in R, how to relate these values to the actual distance along this line?
Suppose I want to extract the value of the R logo along the following line:
library(raster)
r <- raster(system.file("external/rlogo.grd", package="raster"))
x=c(5, 95)
y=c(20, 50)
line = SpatialLines(list(Lines(Line(cbind(x,y)), ID="a")))
plot(r)
plot(line, add=TRUE)
I can extract the values and plot them - but how to replace the x values (1:length(vals) below) by the actual distance (starting e.g. at 0 from the left side of the line)?
vals <- extract(r, line)[[1]]
plot(1:length(vals), vals, type='o')
I could combine the extraction of the cells with xyFromCell to get the coordinates of the extracted cells as suggested here, but it is not clear to me how to go further.
I'm not sure what you're exactly asking, but if you looking for distances between the leftmost coordinate of the line segment and the centres of the cells which the line passes through, then you can find the distances like this:
x <- extract(r, l, cellnumbers=TRUE)[[1]]
xy <- xyFromCell(r, x[,1]) # get cell coordinates where the line passes
start <- xy[which.min(xy[,1]),] # leftmost coordinate of the line
d <- apply(xy, 1, function(x, start) sqrt(sum((x-start)^2)), start=start) # find distances between the line segment start and the cells
plot(1:length(d), d, type='o')
Here is a solution (partly on the basis of #jvj's input) through an attempt to compute the orthogonal projections of the cell centres provided by raster::extract on the line and then compute the distances along the line.
(This is an R-beginners script, likely easily improvable, but seems to work (and is of course only for rasters with projection respecting distances))
vals <- extract(r, line, cellnumbers=TRUE)[[1]]
cellsxy <- xyFromCell(r, vals[,1]) # coordinates of intersected cells (likely not ON the line)
linexy = spsample(line, 1000, "regular") # get the line as points
linexy <- matrix(cbind(linexy$x, linexy$y), ncol=2) # easier than Spatial object for later
orthoproj <- c() # to store the orthogonal projections of cells centres on the line
for (i in 1:nrow(cellsxy)) {
xypt = cellsxy[i,]
min.index <- which.min(spDistsN1(linexy, xypt))
orthopt <- linexy[min.index, ] # orthogonal projections = smaller distance
orthoproj <- c(orthoproj, c(orthopt[1], orthopt[2]))
}
orthoproj <- matrix(orthoproj, ncol=2, byrow=T)
orthoproj <- data.frame(x=orthoproj[,1], y=orthoproj[,2])
orthoproj <- orthoproj[order(orthoproj[,1]),] # reorder with increasing distance
orthoproj <- data.frame(x=orthoproj$x, y=orthoproj$y)
start <- linexy[which.min(linexy[,1]),] # leftmost coordinate of the line
dists <- apply(orthoproj, 1,
function(xy, start) sqrt(sum((xy-start)^2)),
start=start) # distances between 'start' and the orthogonal projections
plot(dists, rev(vals[,2]), type='o') # !! beware: order of 'vals' and 'dists'
# depending on the order in which cellnumbers are returned
# in raster::extract and the shape of your line !!
I'have a SpatialPointsDataFrame load with
pst<-readOGR("/data_spatial/coast/","points_coast")
And I would like to get a SpatialLines in output, I have find somthing
coord<-as.data.frame(coordinates(pst))
Slo1<-Line(coord)
Sli1<-Lines(list(Slo1),ID="coastLine")
coastline <- SpatialLines(list(Sli1))
class(coastline)
it seems to work but when I try plot(coastline) , I have a line that should not be there ...
Some one can help me ? The shapefile is here !
I have looked at the shapefile. There is an id column, but if you plot the data, it seems that the id is not ordered north-south or something. The extra lines are created because the point order is not perfect, connecting points that are next to each other in the table, but far from each other in terms of space. You could try to figure out the correct ordering of the data by calculating distances between points and then ordering on distance.
A workaround is to remove those lines that are longer than a certain distance, e.g. 500 m.. First, find out where distance between consecutive coordinates is larger than this distance: the breaks. Then take a subset of coordinates between two breaks and lastly create Lines for that subset. You end up with a coastline consisting of several (breaks-1) segments and without the erroneous ones.
# read data
library(rgdal)
pst<-readOGR("/data_spatial/coast/","points_coast")
coord<-as.data.frame(coordinates(pst))
colnames(coord) <- c('X','Y')
# determine distance between consective coordinates
linelength = LineLength(as.matrix(coord),sum=F)
# 'id' of long lines, plus first and last item of dataset
breaks = c(1,which(linelength>500),nrow(coord))
# check position of breaks
breaks = c(1,which(linelength>500),nrow(coord))
# plot extent of coords and check breaks
plot(coord,type='n')
points(coord[breaks,], pch=16,cex=1)
# create vector to be filled with lines of each subset
ll <- vector("list", length(breaks)-1)
for (i in 1: (length(breaks)-1)){
subcoord = coord[(breaks[i]+1):(breaks[i+1]),]
# check if subset contains more than 2 coordinates
if (nrow(subcoord) >= 2){
Slo1<-Line(subcoord)
Sli1<-Lines(list(Slo1),ID=paste0('section',i))
ll[[i]] = Sli1
}
}
# remove any invalid lines
nulls = which(unlist(lapply(ll,is.null)))
ll = ll[-nulls]
lin = SpatialLines(ll)
# add result to plot
lines(lin,col=2)
# write shapefile
df = data.frame(row.names=names(lin),id=1:length(names(lin)))
lin2 = SpatialLinesDataFrame(sl=lin, data=df)
proj4string(lin2) <- proj4string(pst)
writeOGR(obj=lin2, layer='coastline', dsn='/data_spatial/coast', driver='ESRI Shapefile')
I would like to identify linear features, such as roads and rivers, on raster maps and convert them to a linear spatial object (SpatialLines class) using R.
The raster and sp packages can be used to convert features from rasters to polygon vector objects (SpatialPolygons class). rasterToPolygons() will extract cells of a certain value from a raster and return a polygon object. The product can be simplified using the dissolve=TRUE option, which calls routines in the rgeos package to do this.
This all works just fine, but I would prefer it to be a SpatialLines object. How can I do this?
Consider this example:
## Produce a sinuous linear feature on a raster as an example
library(raster)
r <- raster(nrow=400, ncol=400, xmn=0, ymn=0, xmx=400, ymx=400)
r[] <- NA
x <-seq(1, 100, by=0.01)
r[cellFromRowCol(r, round((sin(0.2*x) + cos(0.06*x)+2)*100), round(x*4))] <- 1
## Quick trick to make it three cells wide
r[edge(r, type="outer")] <- 1
## Plot
plot(r, legend=FALSE, axes=FALSE)
## Convert linear feature to a SpatialPolygons object
library(rgeos)
rPoly <- rasterToPolygons(r, fun=function(x) x==1, dissolve=TRUE)
plot(rPoly)
Would the best approach be to find a centre line through the polygon?
Or is there existing code available to do this?
EDIT: Thanks to #mdsumner for pointing out that this is called skeletonization.
Here's my effort. The plan is:
densify the lines
compute a delaunay triangulation
take the midpoints, and take those points that are in the polygon
build a distance-weighted minimum spanning tree
find its graph diameter path
The densifying code for starters:
densify <- function(xy,n=5){
## densify a 2-col matrix
cbind(dens(xy[,1],n=n),dens(xy[,2],n=n))
}
dens <- function(x,n=5){
## densify a vector
out = rep(NA,1+(length(x)-1)*(n+1))
ss = seq(1,length(out),by=(n+1))
out[ss]=x
for(s in 1:(length(x)-1)){
out[(1+ss[s]):(ss[s+1]-1)]=seq(x[s],x[s+1],len=(n+2))[-c(1,n+2)]
}
out
}
And now the main course:
simplecentre <- function(xyP,dense){
require(deldir)
require(splancs)
require(igraph)
require(rgeos)
### optionally add extra points
if(!missing(dense)){
xy = densify(xyP,dense)
} else {
xy = xyP
}
### compute triangulation
d=deldir(xy[,1],xy[,2])
### find midpoints of triangle sides
mids=cbind((d$delsgs[,'x1']+d$delsgs[,'x2'])/2,
(d$delsgs[,'y1']+d$delsgs[,'y2'])/2)
### get points that are inside the polygon
sr = SpatialPolygons(list(Polygons(list(Polygon(xyP)),ID=1)))
ins = over(SpatialPoints(mids),sr)
### select the points
pts = mids[!is.na(ins),]
dPoly = gDistance(as(sr,"SpatialLines"),SpatialPoints(pts),byid=TRUE)
pts = pts[dPoly > max(dPoly/1.5),]
### now build a minimum spanning tree weighted on the distance
G = graph.adjacency(as.matrix(dist(pts)),weighted=TRUE,mode="upper")
T = minimum.spanning.tree(G,weighted=TRUE)
### get a diameter
path = get.diameter(T)
if(length(path)!=vcount(T)){
stop("Path not linear - try increasing dens parameter")
}
### path should be the sequence of points in order
list(pts=pts[path+1,],tree=T)
}
Instead of the buffering of the earlier version I compute the distance from each midpoint to the line of the polygon, and only take points that are a) inside, and b) further from the edge than 1.5 of the distance of the inside point that is furthest from the edge.
Problems can arise if the polygon kinks back on itself, with long segments, and no densification. In this case the graph is a tree and the code reports it.
As a test, I digitized a line (s, SpatialLines object), buffered it (p), then computed the centreline and superimposed them:
s = capture()
p = gBuffer(s,width=0.2)
plot(p,col="#cdeaff")
plot(s,add=TRUE,lwd=3,col="red")
scp = simplecentre(onering(p))
lines(scp$pts,col="white")
The 'onering' function just gets the coordinates of one ring from a SpatialPolygons thing that should only be one ring:
onering=function(p){p#polygons[[1]]#Polygons[[1]]#coords}
Capture spatial lines features with the 'capture' function:
capture = function(){p=locator(type="l")
SpatialLines(list(Lines(list(Line(cbind(p$x,p$y))),ID=1)))}
Thanks to #klewis at gis.stackexchange.com for linking to this elegant algorithm for finding the centre line (in response to a related question I asked there).
The process requires finding the coordinates on the edge of a polygon describing the linear feature and performing a Voronoi tessellation of those points. The coordinates of the Voronoi tiles that fall within the polygon of the linear feature fall on the centre line. Turn these points into a line.
Voronoi tessellation is done really efficiently in R using the deldir package, and intersections of polygons and points with the rgeos package.
## Find points on boundary of rPoly (see question)
rPolyPts <- coordinates(as(as(rPoly, "SpatialLinesDataFrame"),
"SpatialPointsDataFrame"))
## Perform Voronoi tessellation of those points and extract coordinates of tiles
library(deldir)
rVoronoi <- tile.list(deldir(rPolyPts[, 1], rPolyPts[,2]))
rVoronoiPts <- SpatialPoints(do.call(rbind,
lapply(rVoronoi, function(x) cbind(x$x, x$y))))
## Find the points on the Voronoi tiles that fall inside
## the linear feature polygon
## N.B. That the width parameter may need to be adjusted if coordinate
## system is fractional (i.e. if longlat), but must be negative, and less
## than the dimension of a cell on the original raster.
library(rgeos)
rLinePts <- gIntersection(gBuffer(rPoly, width=-1), rVoronoiPts)
## Create SpatialLines object
rLine <- SpatialLines(list(Lines(Line(rLinePts), ID="1")))
The resulting SpatialLines object:
You can get the boundary of that polygon as SpatialLines by direct coercion:
rLines <- as(rPoly, "SpatialLinesDataFrame")
Summarizing the coordinates down to a single "centre line" would be possible, but nothing immediate that I know of. I think that process is generally called "skeletonization":
http://en.wikipedia.org/wiki/Topological_skeleton
I think ideal solution would be to build such negative buffer which dynamically reach the minimum width and doesn't break when value is too large; keeps continued object and eventually, draws a line if the value is reached. But unfortunately, this may be very compute demanding because this would be done probably in steps and checks if the value for particular point is enough to have a point (of our middle line). Possible it's ne need to have infinitive number of steps, or at least, some parametrized value.
I don't know how to implement this for now.