Generate regularly spaced points in polygon - r

Question

Is there a way to generate regularly spaced (e.g., 500 meters apart) points within a polygon using R? I have been trying to use the sp package but can't seem to define a set of points that are spaced a certain distance apart from one another. My aim is to generate the points, then extract their lat/long coordinates into a new dataframe. Any help would be much appreciated! Thanks

Answer 1

Quite straight forward and almost out-of-the-box.
As OP did not share data, buckle up, put your seats in a vertical position and let us fly to Paris. There, we will adapt a geosphere function, and with its help we will divide up Paris' shape into lon / lat coordinates that are 500 meters apart each (vertically and horizontally).
# Load necessary libraries.
library(raster)
library(geosphere)
library(tidyverse)
library(sp)
# This is an adapted version of geosphere's destPoint() function that works with
# changing d (distance).
destPoint_v <- function (x, y, b, d, a = 6378137, f = 1/298.257223563, ...)
{
r <- list(...)$r
if (!is.null(r)) {
return(.old_destPoint(x, y, b, d, r = r))
}
b <- as.vector(b)
d <- as.vector(d)
x <- as.vector(x)
y <- as.vector(y)
p <- cbind(x, y, b, d)
r <- .Call("_geodesic", as.double(p[, 1]), as.double(p[, 2]),
as.double(p[, 3]), as.double(p[, 4]),
as.double(a), as.double(f),
PACKAGE = "geosphere")
r <- matrix(r, ncol = 3, byrow = TRUE)
colnames(r) <- c("lon", "lat", "finalbearing")
return(r[, 1:2, drop = FALSE])
}
# Data can be downloaded from
# http://osm13.openstreetmap.fr/~cquest/openfla/export/communes-20190101-shp.zip
# or
# https://www.data.gouv.fr/en/datasets/decoupage-administratif-communal-francais-issu-d-openstreetmap/
# ("Export simple de janvier 2019 (225Mo)")
# Load shapefile.
# shp <- raster::shapefile("Dropbox/work/crema/communes-20190101-shp/communes-20190101.shp")
# Extract Paris.
paris <- shp[shp$nom == "Paris", ]
# Set distance of points in meters.
dist <- 500
# Extract bounding box from Paris' SpatialPolygonDataFrame.
bbox <- raster::extent(paris)
# Calculate number of points on the vertical axis.
ny <- ceiling(geosphere::distGeo(p1 = c(bbox#xmin, bbox#ymin),
p2 = c(bbox#xmin, bbox#ymax)) / dist)
# Calculate maximum number of points on the horizontal axis.
# This needs to be calculated for the lowermost and uppermost horizontal lines
# as the distance between latitudinal lines varies when the longitude changes.
nx <- ceiling(max(geosphere::distGeo(p1 = c(bbox#xmin, bbox#ymin),
p2 = c(bbox#xmax, bbox#ymin)) / dist,
geosphere::distGeo(p1 = c(bbox#xmin, bbox#ymax),
p2 = c(bbox#xmax, bbox#ymax)) / dist))
# Create result data frame with number of points on vertical axis.
df <- data.frame(ny = 1:ny)
# Calculate coordinates along the vertical axis.
pts <- geosphere::destPoint(p = c(bbox#xmin, bbox#ymin),
b = 0, d = dist * (1:ny - 1))
df$x <- pts[, 1]
df$y <- pts[, 2]
# Add points on horizontal axis.
df <- tidyr::crossing(nx = 1:nx, df)
# Calculate coordinates.
pts <- destPoint_v(df$x, df$y, b = 90, 500 * (df$nx - 1))
# Turn coordinates into SpatialPoints.
pts <- SpatialPoints(cbind(pts[, 1], pts[, 2]), proj4string = CRS(proj4string(paris)))
# Cut to boundaries of Paris.
result <- raster::intersect(pts, paris)
# Plot result.
plot(result)
title("Paris in Points")
Kind of looks like a fish, doesn't it?

Answer 2

Here is a way to do assuming you have a lonlat polygon by first transforming it to a planar crs (not as nifty as Roman's solution with destPoint).
Packages and example data
library(raster)
library(rgdal)
p <- shapefile(system.file("external/lux.shp", package="raster"))[1,]
Transform to planar crs (pick one that matches your data!)
putm <- spTransform(p, "+proj=utm +zone=32 +datum=WGS84")
Create a raster with 500 m resolution, rasterize the polygon and transform to points
r <- raster(putm, res=500)
r <- rasterize(putm, r)
pts <- rasterToPoints(r, spatial=TRUE)
Transform the points to lon/lat and plot the results
pts_lonlat <- spTransform(pts, "+proj=longlat +datum=WGS84")
result <- coordinates(pts_lonlat)
plot(p)
points(result, pch="+", cex=.5)
(looks like an elephant)