How to get coordinates of a selected point in R plot? - r
I need to pass to R program coordinates of a point selected by the mouse pointer, to perform some calculations. I have problems getting it to work.
I know that this code should identify point on a plot:
plot(kk2$k2,kk2$k1)
identify(kk2$k2,kk2$k1)
But even that doesn't work. On a plot appears some meaningless number, while point has two coordinates. why?
How to fix at least that?
My goal is to have the point coordinates returned to R and perform some calculations on them. The dataset kk2 has only two columns - k1 and k2, nothing more.
The package "gatepoints" available on CRAN will allow you to draw a gate returning your points of interest.
If you are using RStudio it is better to plot in a separate x11 window by first opening a new x11 device:
X11()
Now plot your points, I've made up some simple data:
kk2 <- data.frame(k2=1:10, k1=1:10)
plot(kk2, col = "red", pch = 16)
Run the command below and then select your points by left clicking and right clicking to close the polygon:
selectedPoints <- fhs(kk2)
This will return:
selectedPoints
#> [1] "4" "5" "7"
#> attr(,"gate")
#> k2 k1
#> 1 6.099191 8.274120
#> 2 8.129107 7.048649
#> 3 8.526881 5.859404
#> 4 5.700760 6.716428
#> 5 5.605314 5.953430
#> 6 6.866882 3.764390
#> 7 3.313575 3.344069
#> 8 2.417270 5.217868
locator {graphics} R Documentation
Graphical Input
Description
Reads the position of the graphics cursor when the (first) mouse button is pressed.
![> pts <- locator(4)
> polygon(pts)
> png(); plot(1,1)
> pts <- locator(4)
> polygon(pts)
> dev.off()][1]
Try something like this, since identify returns the seq_along(x) for the point that you click near (what you refer to as 'some meaningless number'):
x <- rnorm(10)
y <- rnorm(10)
plot(x,y)
out <- sapply(list(x,y),"[",identify(x,y))
# do some clicking
out
# something like this is returned for the x/y points
# [,1] [,2]
#[1,] -0.62221766 -0.73838314
#[2,] -0.69896643 0.40186536
#[3,] 0.06077831 -1.63940474
#[4,] -0.09900270 0.00062011
The key is using the result as an index. This can then be used to identify the specific xy coordinates:
n <- 10
x <- runif(n)
y <- runif(n)
df <- data.frame(x=x, y=y)
plot(y ~ x, data=df)
df[identify(x, y, n=1),]
Related
Finding the peak of a mountain
so I've combined those 2 rasters and made them into one dem raster which contains elevation values:
dem1 = read_stars("srtm_43_06.tif")
dem2 = read_stars("srtm_44_06.tif")
pol = st_read("israel_borders.shp")
dem = st_mosaic(dem1, dem2)
dem = dem[, 5687:6287, 2348:2948]
names(dem) = "elevation"
dem = st_warp(src = dem, crs = 32636, method = "near", cellsize = 90)
Now I need to calculate a point geometry of the peak of the mountain by finding the centroid of the pixel that has the highest elevation in the image, does anyone know what functions I can use?
Building on Grzegorz Sapijaszko's example, here is an alternative path to the top of the mountain.
library(terra)
f <- system.file("ex/elev.tif", package="terra")
x <- rast(f)
If there is a single maximum, you can do
g <- global(x, which.max)
xyFromCell(x, g[,1])
# x y
#[1,] 6.020833 50.17917
Now, consider a situation with multiple maxima. I add three more cells with the maximum value.
x[c(1000, 2500, 5000)] <- 547
We can find the four highest peaks with:
g <- global(x, which.max)[[1]]
v <- x[g] |> unlist()
y <- ifel(x == v, v, NA)
p <- as.points(y)
crds(p)
#[1,] 6.020833 50.17917
#[2,] 6.154167 50.10417
#[3,] 5.987500 49.97083
#[4,] 6.237500 49.75417
You should not warp (project with terra) the raster data first because that changes the cell values and potentially the location of the highest peak. You should find the peaks with the original data, but then you can transform the results like this.
pp <- project(p, "EPSG:32636")
crds(pp)
# x y
#[1,] -1411008 5916157
#[2,] -1404896 5904422
#[3,] -1422145 5894509
#[4,] -1413735 5864236
With your files, you could start with something like
ff <- c("srtm_43_06.tif", "srtm_44_06.tif")
v <- vrt(ff)
g <- global(x, which.max)
And then continue as in the examples above.
Let's use terra, however similar approach can be applied by raster package as well. For testing purposes we will use raster supplied with terra package
library(terra)
#> terra 1.5.12
f <- system.file("ex/elev.tif", package="terra")
v <- rast(f)
plot(v)
You can check the details of your raster just typing the raster object name and pressing enter, you can check the min and max values with minmax() function form terra:
minmax(v)
#> elevation
#> [1,] 141
#> [2,] 547
Let's create another raster by copying original one, however checking if the value is the max value of elevation:
w <- v == minmax(v)[2]
plot(w)
Let's create a substitution matrix, and substitute all FALSE with NA and TRUE with 1:
mx <- matrix(c(FALSE, NA, TRUE, 1), ncol = 2, byrow = TRUE)
w <- classify(w, mx)
plot(v)
plot(as.polygons(w), add=TRUE)
Let's find centroids of those polygon(s):
pts <- centroids(as.polygons(w))
plot(pts, add=TRUE)
Let's see our coordinates:
as.data.frame(pts, geom = "WKT")
#> elevation geometry
#> 1 1 POINT (6.020833 50.179167)
Created on 2022-01-29 by the reprex package (v2.0.1)
3D with value interpolation in R (X, Y, Z, V)
Is there an R package that does X, Y, Z, V interpolation? I see that Akima does X, Y, V but I need one more dimension. Basically I have X,Y,Z coordinates plus the value (V) that I want to interpolate. This is all GIS data but my GIS does not do voxel interpolation So if I have a point cloud of XYZ coordinates with a value of V, how can I interpolate what V would be at XYZ coordinate (15,15,-12) ? Some test data would look like this: X <-rbind(10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50) Y <- rbind(10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50) Z <- rbind(-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29) V <- rbind(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,25,35,75,25,50,0,0,0,0,0,10,12,17,22,27,32,37,25,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,50,125,130,105,110,115,165,180,120,100,80,60,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
I had the same question and was hoping for an answer in R.
My question was: How do I perform 3D (trilinear) interpolation using regular gridded coordinate/value data (x,y,z,v)? For example, CT images, where each image has pixel centers (x, y) and greyscale value (v) and there are multiple image "slices" (z) along the thing being imaged (e.g., head, torso, leg, ...).
There is a slight problem with the given example data.
# original example data (reformatted)
X <- rep( rep( seq(10, 50, by=10), each=25), 3)
Y <- rep( rep( seq(10, 50, by=10), each=5), 15)
Z <- rep(c(-5, -17, -29), each=125)
V <- rbind(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,25,35,75,25,50,0,0,0,0,0,10,12,17,22,27,32,37,25,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,50,125,130,105,110,115,165,180,120,100,80,60,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
# the dimensions of the 3D grid described do not match the number of values
(length(unique(X))*length(unique(Y))*length(unique(Z))) == length(V)
## [1] FALSE
## which makes sense since 75 != 375
# visualize this:
library(rgl)
plot3d(x=X, y=Y, z=Z, col=terrain.colors(181)[V])
# examine the example data real quick...
df <- data.frame(x=X,y=Y,z=Z,v=V);
head(df);
table(df$x, df$y, df$z);
# there are 5 V values at each X,Y,Z coordinate... duplicates!
# redefine Z so there are 15 unique values
# making 375 unique coordinate points
# and matching the length of the given value vector, V
df$z <- seq(-5, -29, length.out=15)
head(df)
table(df$x, df$y, df$z);
# there is now 1 V value at each X,Y,Z coordinate
# that was for testing, now actually redefine the Z vector.
Z <- rep(seq(-5,-29, length.out = 15), 25)
# plot it.
library(rgl)
plot3d(x=X, y=Y, z=Z, col=terrain.colors(181)[V])
I couldn't find any 4D interpolation functions in the usual R packages, so I wrote a quick and dirty one. The following implements (without ANY error checking... caveat emptor!) the technique described at: https://en.wikipedia.org/wiki/Trilinear_interpolation
# convenience function #1:
# define a function that takes a vector of lookup values and a value to lookup
# and returns the two lookup values that the value falls between
between = function(vec, value) {
# extract list of unique lookup values
u = unique(vec)
# difference vector
dvec = u - value
vals = c(u[dvec==max(dvec[dvec<0])], u[dvec==min(dvec[dvec>0])])
return(vals)
}
# convenience function #2:
# return the value (v) from a grid data.frame for given point (x, y, z)
get_value = function(df, xi, yi, zi) {
# assumes df is data.frame with column names: x, y, z, v
subset(df, x==xi & y==yi & z==zi)$v
}
# inputs df (x,y,z,v), points to look up (x, y, z)
interp3 = function(dfin, xin, yin, zin) {
# TODO: check if all(xin, yin, zin) equals a grid point, if so just return the point value
# TODO: check if any(xin, yin, zin) equals a grid point, if so then do bilinear or linear interp
cube_x <- between(dfin$x, xin)
cube_y <- between(dfin$y, yin)
cube_z <- between(dfin$z, zin)
# find the two values in each dimension that the lookup value falls within
# and extract the cube of 8 points
tmp <- subset(dfin, x %in% cube_x &
y %in% cube_y &
z %in% cube_z)
stopifnot(nrow(tmp)==8)
# define points in a periodic and cubic lattice
x0 = min(cube_x); x1 = max(cube_x);
y0 = min(cube_y); y1 = max(cube_y);
z0 = min(cube_z); z1 = max(cube_z);
# define differences in each dimension
xd = (xin-x0)/(x1-x0); # 0.5
yd = (yin-y0)/(y1-y0); # 0.5
zd = (zin-z0)/(z1-z0); # 0.9166666
# interpolate along x:
v00 = get_value(tmp, x0, y0, z0)*(1-xd) + get_value(tmp,x1,y0,z0)*xd # 2.5
v01 = get_value(tmp, x0, y0, z1)*(1-xd) + get_value(tmp,x1,y0,z1)*xd # 0
v10 = get_value(tmp, x0, y1, z0)*(1-xd) + get_value(tmp,x1,y1,z0)*xd # 0
v11 = get_value(tmp, x0, y1, z1)*(1-xd) + get_value(tmp,x1,y1,z1)*xd # 65
# interpolate along y:
v0 = v00*(1-yd) + v10*yd # 1.25
v1 = v01*(1-yd) + v11*yd # 32.5
# interpolate along z:
return(v0*(1-zd) + v1*zd) # 29.89583 (~91.7% between v0 and v1)
}
> interp3(df, 15, 15, -12)
[1] 29.89583
Testing that same source's assertion that trilinear is simply linear(bilinear(), bilinear()), we can use the base R linear interpolation function, approx(), and the akima package's bilinear interpolation function, interp(), as follows:
library(akima)
approx(x=c(-11.857143,-13.571429),
y=c(interp(x=df[round(df$z,1)==-11.9,"x"], y=df[round(df$z,1)==-11.9,"y"], z=df[round(df$z,1)==-11.9,"v"], xo=15, yo=15)$z,
interp(x=df[round(df$z,1)==-13.6,"x"], y=df[round(df$z,1)==-13.6,"y"], z=df[round(df$z,1)==-13.6,"v"], xo=15, yo=15)$z),
xout=-12)$y
# [1] 0.2083331
Checked another package to triangulate:
library(oce)
Vmat <- array(data = V, dim = c(length(unique(X)), length(unique(Y)), length(unique(Z))))
approx3d(x=unique(X), y=unique(Y), z=unique(Z), f=Vmat, xout=15, yout=15, zout=-12)
[1] 1.666667
So 'oce', 'akima' and my function all give pretty different answers. This is either a mistake in my code somewhere, or due to differences in the underlying Fortran code in the akima interp(), and whatever is in the oce 'approx3d' function that we'll leave for another day.
Not sure what the correct answer is because the MWE is not exactly "minimum" or simple. But I tested the functions with some really simple grids and it seems to give 'correct' answers. Here's one simple 2x2x2 example:
# really, really simple example:
# answer is always the z-coordinate value
sdf <- expand.grid(x=seq(0,1),y=seq(0,1),z=seq(0,1))
sdf$v <- rep(seq(0,1), each=4)
> interp3(sdf,0.25,0.25,.99)
[1] 0.99
> interp3(sdf,0.25,0.25,.4)
[1] 0.4
Trying akima on the simple example, we get the same answer (phew!):
library(akima)
approx(x=unique(sdf$z),
y=c(interp(x=sdf[sdf$z==0,"x"], y=sdf[sdf$z==0,"y"], z=sdf[sdf$z==0,"v"], xo=.25, yo=.25)$z,
interp(x=sdf[sdf$z==1,"x"], y=sdf[sdf$z==1,"y"], z=sdf[sdf$z==1,"v"], xo=.25, yo=.25)$z),
xout=.4)$y
# [1] 0.4
The new example data in the OP's own, accepted answer was not possible to interpolate with my simple interp3() function above because:
(a) the grid coordinates are not regularly spaced, and
(b) the coordinates to lookup (x1, y1, z1) lie outside of the grid.
# for completeness, here's the attempt:
options(scipen = 999)
XCoor=c(78121.6235,78121.6235,78121.6235,78121.6235,78136.723,78136.723,78136.723,78136.8969,78136.8969,78136.8969,78137.4595,78137.4595,78137.4595,78125.061,78125.061,78125.061,78092.4696,78092.4696,78092.4696,78092.7683,78092.7683,78092.7683,78092.7683,78075.1171,78075.1171,78064.7462,78064.7462,78064.7462,78052.771,78052.771,78052.771,78032.1179,78032.1179,78032.1179)
YCoor=c(5213642.173,523642.173,523642.173,523642.173,523594.495,523594.495,523594.495,523547.475,523547.475,523547.475,523503.462,523503.462,523503.462,523426.33,523426.33,523426.33,523656.953,523656.953,523656.953,523607.157,523607.157,523607.157,523607.157,523514.671,523514.671,523656.81,523656.81,523656.81,523585.232,523585.232,523585.232,523657.091,523657.091,523657.091)
ZCoor=c(-3.0,-5.0,-10.0,-13.0,-3.5,-6.5,-10.5,-3.5,-6.5,-9.5,-3.5,-5.5,-10.5,-3.5,-5.5,-7.5,-3.5,-6.5,-11.5,-3.0,-5.0,-9.0,-12.0,-6.5,-10.5,-2.5,-3.5,-8.0,-3.5,-6.5,-9.5,-2.5,-6.5,-8.5)
V=c(2.4000,30.0,620.0,590.0,61.0,480.0,0.3700,0.0,0.3800,0.1600,0.1600,0.9000,0.4100,0.0,0.0,0.0061,6.0,52.0,0.3400,33.0,235.0,350.0,9300.0,31.0,2100.0,0.0,0.0,10.5000,3.8000,0.9000,310.0,0.2800,8.3000,18.0)
adf = data.frame(x=XCoor, y=YCoor, z=ZCoor, v=V)
# the first y value looks like a typo?
> head(adf)
x y z v
1 78121.62 5213642.2 -3.0 2.4
2 78121.62 523642.2 -5.0 30.0
3 78121.62 523642.2 -10.0 620.0
4 78121.62 523642.2 -13.0 590.0
5 78136.72 523594.5 -3.5 61.0
6 78136.72 523594.5 -6.5 480.0
x1=198130.000
y1=1913590.000
z1=-8
> interp3(adf, x1,y1,z1)
numeric(0)
Warning message:
In min(dvec[dvec > 0]) : no non-missing arguments to min; returning Inf
Whether the test data did or not make sense, I still needed an algorithm. Test data is just that, something to fiddle with and as a test data it was fine. I wound up programming it in python and the following code takes XYZ V and does a 3D Inverse Distance Weighted (IDW) interpolation where you can set the number of points used in the interpolation. This python recipe only interpolates to one point (x1, y1, z1) but it is easy enough to extend. import numpy as np import math #34 points XCoor=np.array([78121.6235,78121.6235,78121.6235,78121.6235,78136.723,78136.723,78136.723,78136.8969,78136.8969,78136.8969,78137.4595,78137.4595,78137.4595,78125.061,78125.061,78125.061,78092.4696,78092.4696,78092.4696,78092.7683,78092.7683,78092.7683,78092.7683,78075.1171,78075.1171,78064.7462,78064.7462,78064.7462,78052.771,78052.771,78052.771,78032.1179,78032.1179,78032.1179]) YCoor=np.array([5213642.173,523642.173,523642.173,523642.173,523594.495,523594.495,523594.495,523547.475,523547.475,523547.475,523503.462,523503.462,523503.462,523426.33,523426.33,523426.33,523656.953,523656.953,523656.953,523607.157,523607.157,523607.157,523607.157,523514.671,523514.671,523656.81,523656.81,523656.81,523585.232,523585.232,523585.232,523657.091,523657.091,523657.091]) ZCoor=np.array([-3.0,-5.0,-10.0,-13.0,-3.5,-6.5,-10.5,-3.5,-6.5,-9.5,-3.5,-5.5,-10.5,-3.5,-5.5,-7.5,-3.5,-6.5,-11.5,-3.0,-5.0,-9.0,-12.0,-6.5,-10.5,-2.5,-3.5,-8.0,-3.5,-6.5,-9.5,-2.5,-6.5,-8.5]) V=np.array([2.4000,30.0,620.0,590.0,61.0,480.0,0.3700,0.0,0.3800,0.1600,0.1600,0.9000,0.4100,0.0,0.0,0.0061,6.0,52.0,0.3400,33.0,235.0,350.0,9300.0,31.0,2100.0,0.0,0.0,10.5000,3.8000,0.9000,310.0,0.2800,8.3000,18.0]) def Distance(x1,y1,z1, Npoints): i=0 d=[] while i < 33: d.append(math.sqrt((x1-XCoor[i])*(x1-XCoor[i]) + (y1-YCoor[i])*(y1-YCoor[i]) + (z1-ZCoor[i])*(z1-ZCoor[i]) )) i = i + 1 distance=np.array(d) myIndex=distance.argsort()[:Npoints] weightedNum=0 weightedDen=0 for i in myIndex: weightedNum=weightedNum + (V[i]/(distance[i]*distance[i])) weightedDen=weightedDen + (1/(distance[i]*distance[i])) InterpValue=weightedNum/weightedDen return InterpValue x1=198130.000 y1=1913590.000 z1=-8 print(Distance(x1,y1,z1, 12))
How to create SpatialLine object
I am using sp package to create SpatialLines object and save it in the list of objects allLines. Later on I will need to compare SpatialLines to each other, but this goes beyond the current question. So far I only need to construct SpatialLines objects. This is the last code based on the answer of hrbrmstr: library(sp) allLines <- NULL x <- c(1,5,4,8) y <- c(1,3,4,7) xy <- cbind(x,y) xy.sp = sp::SpatialPoints(xy) spl <- SpatialLines(list(Lines(Line(xy.sp), ID="a"))) allLines <- rbind(allLines,spl) Error message: Error in (function (classes, fdef, mtable) : unable to find an inherited method for function ‘proj4string’ for signature ‘"NULL"’ How to solve this issue?
Is: library(sp) x <- c(1,5,4,8) y <- c(1,3,4,7) SpatialLines(list(Lines(Line(cbind(x,y)), ID="a"))) ## An object of class "SpatialLines" ## Slot "lines": ## [[1]] ## An object of class "Lines" ## Slot "Lines": ## [[1]] ## An object of class "Line" ## Slot "coords": ## x y ## [1,] 1 1 ## [2,] 5 3 ## [3,] 4 4 ## [4,] 8 7 ## ## ## ## Slot "ID": ## [1] "a" ## ## ## ## Slot "bbox": ## min max ## x 1 8 ## y 1 7 ## ## Slot "proj4string": ## CRS arguments: NA what you're looking for?
Getting back to your last question, try library(sp) as(xy.spdf, "SpatialLines") or, to create a Lines object (which may not be what you want), as(xy.spdf, "SpatialLines")#lines[[1]]
If you came to this question to find out how to make a group of lines (as implied by the name of the function, SpatialLines) you can find examples in the sp library, filed under "SpatialLines-class".
I found their example a little strange, so I edited it to make more sense for how I normally see the data.
## Make some line segments from points
## Note, l1a and l1b are a group of two lines
l1a <- rbind(c(1, 3), c(2,2) ,c(3,2))
l1b <- l1a + .05
l2 <- rbind(c(1,1), c(2,1.5), c(3,1))
## At this point it's just a matrix, and you can plot the points
plot(l1a, type="l", xlim=c(1,3.25), ylim=c(2,3.25), xlab="", ylab="")
lines(l1b)
## Make convert the matrix objects to line objects
Sl1a <- Line(l1a)
Sl1b <- Line(l1b)
Sl2 <- Line(l2)
## Group the individual lines into "lines"
S1 <- Lines(list(Sl1a, Sl1b), ID="a")
S2 <- Lines(list(Sl2), ID="b")
## Now combine the line groups into a "spatial line object"
Sl <- SpatialLines(list(S1,S2))
## Plot the group, then (for illustration) add each line
## separately with color to illustrate the groups
plot(Sl)
plot(SpatialLines(list(S1)), add=T, col="red")
plot(SpatialLines(list(S2)), add=T, col="blue")
## Examine the properties
summary(Sl)
plot(Sl, col = c("red", "blue"))
Both spatial line plots look like this:
Note the matrix object has named rows in the example. I don't see any benefit to doing this, and it's confusing because the names overlap but do not correspond with the IDs given.
Calculte the whole center of gravity/geometric center of a polygon list
I am looking for a method to calculate the center of gravity of each polygon in the list spatialpolygons:
I thought used a loop, but he gets me for the first polygon, I don't know the way, I am new to R, can someone please help me
Code:
for ( i in 1:length(polys1_T)) {
xx=mean(coordinates(polys1_T[[i]])[,1])
yy=mean(coordinates(polys1_T[[i]])[,2])
aa<-as.data.frame(cbind(xx,yy))
}
Edit:
Code:
inter1 <- read.table("c:/inter1.csv", header=TRUE)
# add a category (required for later rasterizing/polygonizing)
inter1 <- cbind(inter1,
cat
= rep(1L, nrow(inter1)), stringsAsFactors = FALSE)
# convert to spatial points
coordinates(inter1) <- ~long + lat
# gridify your set of points
gridded(inter1) <- TRUE
# convert to raster
r <- raster(inter1)
# convert raster to polygons
sp <- rasterToPolygons(r, dissolve = T)
plot(sp)
# addition transformation to distinguish well the set of polygons
polys <- slot(sp#polygons[[1]], "Polygons")
# plot
plot(sp, border = "gray", lwd = 2) # polygonize result
inter1.csv
result:
Polys is list of 9 polygons :is that it is possible to calculate the center of gravity for each polygon?
Give rgeos::gCentroid a look. You can apply it in many ways. If you have a SpatialPolygons object, say, from a call to readOGR, you can do:
map <- readOGR(dsn, layer)
centers <- data.frame(gCentroid(map, byid=TRUE))
to get all the centroids from it.
As an aside: while accurate—a more common term is "geometric center"/"centroid" vs "center of gravity"
EDIT
For plain, ol Polygons (the "hard" way, but slightly more accurate):
library(rgdal)
library(sp)
library(PBSmapping)
library(maptools)
do.call("rbind", lapply(polys, function(x) {
calcCentroid(SpatialPolygons2PolySet(SpatialPolygons(list(Polygons(list(x), ID=1)))))
}))[,3:4]
## X Y
## 1 5.8108434 20.16466
## 2 -3.2619048 29.38095
## 3 5.5600000 34.72000
## 4 3.8000000 32.57037
## 5 6.3608108 32.49189
## 6 -2.2500000 31.60000
## 7 -8.1733333 27.61333
## 8 0.3082011 27.44444
## 9 8.6685714 26.78286
and, to use your nearly-equivalent by-hand-method:
do.call("rbind", lapply(polys, function(x) {
data.frame(mean(coordinates(x)[,1]), mean(coordinates(x)[,2]))
}))
## mean.coordinates.x....1.. mean.coordinates.x....2..
## 1 5.819892 20.15484
## 2 -3.242593 29.37778
## 3 5.539474 34.71579
## 4 3.815517 32.56552
## 5 6.323034 32.47191
## 6 -2.230952 31.60000
## 7 -8.140476 27.61905
## 8 0.350000 27.40885
## 9 8.746825 26.92063
Each method gives you the centroid for each list element (and there are 9—not 5—in the example you provided).
If you ever have a huge list of these, consider using rbindlist from the data.table package (speedier + more memory efficient).
Create Spatial Data in R
I have a dataset of species and their rough locations in a 100 x 200 meter area. The location part of the data frame is not in a format that I find to be usable. In this 100 x 200 meter rectangle, there are two hundred 10 x 10 meter squares named A through CV. Within each 10 x 10 square there are four 5 x 5 meter squares named 1, 2, 3, and 4, respectively (1 is south of 2 and west of 3. 4 is east of 2 and north of 3). I want to let R know that A is the square with corners at (0 ,0), (10,0), (0,0), and (0,10), that B is just north of A and has corners (0,10), (0,20), (10,10), and (10,20), and K is just east of A and has corners at (10,0), (10,10), (20,0), and (20,10), and so on for all the 10 x 10 meter squares. Additionally, I want to let R know where each 5 x 5 meter square is in the 100 x 200 meter plot. So, my data frame looks something like this 10x10 5x5 Tree Diameter A 1 tree1 4 B 1 tree2 4 C 4 tree3 6 D 3 tree4 2 E 3 tree5 3 F 2 tree6 7 G 1 tree7 12 H 2 tree8 1 I 2 tree9 2 J 3 tree10 8 K 4 tree11 3 L 1 tree12 7 M 2 tree13 5 Eventually, I want to be able to plot the 100 x 200 meter area and have each 10 x 10 meter square show up with the number of trees, or number of species, or total biomass What is the best way to turn the data I have into spatial data that R can use for graphing and perhaps analysis?
Here's a start.
## set up a vector of all 10x10 position tags
tags10 <- c(LETTERS,
paste0("A",LETTERS),
paste0("B",LETTERS),
paste0("C",LETTERS[1:22]))
A function to convert (e.g.) {"J",3} to the center of the corresponding sub-square.
convpos <- function(pos10,pos5) {
## convert letters to major (x,y) positions
p1 <- as.numeric(factor(pos10,levels=tags10)) ## or use match()
p1.x <- ((p1-1) %% 10) *10+5 ## %% is modulo operator
p1.y <- ((p1-1) %/% 10)*10+5 ## %/% is integer division
## sort out sub-positions
p2.x <- ifelse(pos5 <=2,2.5,7.5) ## {1,2} vs {3,4} values
p2.y <- ifelse(pos5 %%2 ==1 ,2.5,7.5) ## odd {1,3} vs even {2,4} values
c(p1.x+p2.x,p1.y+p2.y)
}
usage:
convpos("J",2)
convpos(mydata$tenbytenpos,mydata$fivebyfivepos)
Important notes:
this is a proof of concept, I can pretty much guarantee I haven't got the correspondence of x and y coordinates quite right. But you should be able to trace through this line-by-line and see what it's doing ...
it should work correctly on vectors (see second usage example above): I switched from switch to ifelse for that reason
your column names (10x10) are likely to get mangled into something like X10.10 when reading data into R: see ?data.frame and ?check.names
Similar to what #Ben Bolker has done, here's a lookup function (though you may need to transpose something to make the labels match what you describe).
tenbyten <- c(LETTERS[1:26],
paste0("A",LETTERS[1:26]),
paste0("B",LETTERS[1:26]),
paste0("C",LETTERS[1:22]))
tenbyten <- matrix(rep(tenbyten, each = 2), ncol = 10)
tenbyten <- t(apply(tenbyten, 1, function(x){rep(x, each = 2)}))
# the 1234 squares
squares <- matrix(c(rep(c(1,2),10),rep(c(4,3),10)), nrow = 20, ncol = 20)
# stick together into a reference grid
my.grid <- matrix(paste(tenbyten, squares, sep = "-"), nrow = 20, ncol = 20)
# a lookup function for the site grid
coordLookup <- function(tbt, fbf, .my.grid = my.grid){
x <- col(.my.grid) * 5 - 2.5
y <- row(.my.grid) * 5 - 2.5
marker <- .my.grid == paste(tbt, fbf, sep = "-")
list(x = x[marker], y = y[marker])
}
coordLookup("BB",2)
$x
[1] 52.5
$y
[1] 37.5
If this isn't what you're looking for, then maybe you'd prefer a SpatialPolygonsDataFrame, which has proper polygon IDs, and you attach data to, etc. In that case just Google around for how to make one from scratch, and manipulate the row() and col() functions to get your polygon corners, similar to what's given in this lookup function, which only returns centroids.
Edit: getting SPDF started:
This is modified from the function example and can hopefully be a good start:
library(sp)
# really you have a 20x20 grid, counting the small ones.
# c(2.5,2.5) specifies the distance in any direction from the cell center
grd <- GridTopology(c(1,1), c(2.5,2.5), c(20,20)))
grd <- as.SpatialPolygons.GridTopology(grd)
# get centroids
coords <- coordinates(polys)
# make SPDF, with an extra column for your grid codes, taken from the above.
# you can add further columns to this data.frame(), using polys#data
polys <- SpatialPolygonsDataFrame(grd,
data=data.frame(x=coords[,1], y=coords[,2], my.ID = as.vector(my.grid),
row.names=getSpPPolygonsIDSlots(grd)))