Scalar field visualisation in R - r

I have a table with 3 columns
x y f
-101.0 -101.0 0.0172654144157
...
x and y are coordinates. f is value.
I want to make a 2d picture, where x and y are coordinates and f is a colour. But I need this picture to be not a number of coloured points, but a continuous schedule.
Help me someone please


There are a couple of simple ways to do this if you have a regular grid with your data. Try:
require(ggplot2)
require(lattice)
# make some data
s = 100
i = 0.5
x0 <- 27
y0 <- 34
df <- expand.grid(x=seq(0,s,i), y=seq(0,s,i))
df <- transform(df, f = cos( 10*pi * sqrt((x - x0)^2 + (y-y0)^2)))
# try as points
ggplot(df,aes(x,y,color=f)) + geom_point()
# or as tile
ggplot(df,aes(x,y,fill=f)) + geom_tile()
# or even easier with lattice
levelplot(f ~ x * y, df)
Output examples:

Related

R - How to select or separate the bottom half of cyclic, irregular data?

Below are data (50 points) obtained with cyclic voltammetry. I need to analyze just a portion of it: the portion below between the leftmost to the rightmost point (for example, the data from the green point to the blue point in the plot below).
x <- c(-0.4982, -0.3770, -0.2545, -0.1323, -0.0096, 0.1127, 0.2353, 0.3577, 0.4802,
0.6024, 0.7251, 0.8470, 0.9696, 0.9109, 0.7887, 0.6662, 0.5441, 0.4213, 0.2990,
0.1763, 0.0541, -0.0685, -0.1906, -0.3133, -0.4356, -0.4395, -0.3172, -0.1946,
-0.0723, 0.0501, 0.1724, 0.2950, 0.4175, 0.5400, 0.6623, 0.7848, 0.9070, 0.9735,
0.8510, 0.7290, 0.6063, 0.4840, 0.3614, 0.2391, 0.1165, -0.0090, -0.1316, -0.2539,
-0.3765)
y <- c(-3.24903226, -1.26193548, -0.51612903, -0.09741935, 0.21161290, 0.45870968,
0.69096774, 1.26387097, 4.03225806, 4.77806452, 4.55677419, 3.88129032,
3.36645161, 2.23677419, 1.37741935, 0.74516129, 0.22645161, -0.23161290,
-0.74129032, -1.66387097, -3.84709677, -6.11225806, -7.21741935, -6.37548387,
-4.11225806, -1.88516129, -0.78967742, -0.24709677, 0.08774194, 0.35096774,
0.57612903, 0.90193548, 2.16322581, 4.85935484, 5.02387097, 4.33870968,
3.66258065, 2.88645161, 1.77870968, 1.06000000, 0.48451613, 0.00193548,
-0.46193548, -1.10967742, -2.53161290, -5.13741935, -6.94903226, -7.14387097,
-5.36580645)
The image shows the complete dataset (black line). The points are the x and y sample data. Of course, I can't simple slice the y below the leftmost x, that is y[y < y[which.min(x)]] because I'll lost almost half of the data I want.
Any idea? Thanks

Would it work to take the points where x is decreasing?
x2 = x[x < lag(x) | x > lead(x)]
y2 = y[x < lag(x) | x > lead(x)]
plot(x2, y2)
df <- data.frame(x, y)
df2 <- data.frame(x = x[x < lag(x) | x > lead(x)],
y = y[x < lag(x) | x > lead(x)]) # edit added lead to get right edge
library(ggplot2)
ggplot(df, aes(x, y)) +
geom_point() +
geom_path() +
geom_point(data = df2, color = "red", size = 3)

#NOT DUPLICATED!!# How can I get kernel density value from geom_density output in ggplot in R? [duplicate]

I would like to know what is geom_density() exactly doing, so I justify the graph and if there is any way of extracting the function or points that generates for each of the curves being plotted.
Thanks

Typing get("compute_group", ggplot2::StatDensity) (or, formerly, get("calculate", ggplot2:::StatDensity)) will get you the algorithm used to calculate the density. (At root, it's a call to density() with kernel="gaussian" the default.)
The points used in the plot are invisibly returned by print.ggplot(), so you can access them like this:
library(ggplot2)
m <- ggplot(movies, aes(x = rating))
m <- m + geom_density()
p <- print(m)
head(p$data[[1]], 3)
# y x density scaled count PANEL group ymin ymax
# 1 0.0073761 1.0000 0.0073761 0.025917 433.63 1 1 0 0.0073761
# 2 0.0076527 1.0176 0.0076527 0.026888 449.88 1 1 0 0.0076527
# 3 0.0078726 1.0352 0.0078726 0.027661 462.81 1 1 0 0.0078726
## Just to show that those are the points you are after,
## extract and use them to create a lattice xyplot
library(gridExtra)
library(lattice)
mm <- xyplot(y ~x, data=p$data[[1]], type="l")

As suggested in other answers, you can access the ggplot points using print.ggplot(). However, print()-ing code also prints the ggplot object, which may not be desired.
You can get extract the ggplot object data, without printing the plot, using ggplot_build():
library(ggplot2)
library(ggplot2movies)
m <- ggplot(movies, aes(x = rating))
m <- m + geom_density()
p <- ggplot_build(m) # <---- INSTEAD OF `p <- print(m)`
head(p$data[[1]], 3)
# y x density scaled count n PANEL group ymin
# 1 0.007376115 1.000000 0.007376115 0.02591684 433.6271 58788 1 -1 0
# 2 0.007652653 1.017613 0.007652653 0.02688849 449.8842 58788 1 -1 0
# 3 0.007872571 1.035225 0.007872571 0.02766120 462.8127 58788 1 -1 0
# Just to show that those are the points you are after, extract and use them
# to create a lattice xyplot
library(lattice)
m2 <- xyplot(y ~x, data=p$data[[1]], type="l")
library(gridExtra)
grid.arrange(m, m2, nrow=1)

3D surface plot in R, given x,y,z coordinates

I have the following data-set, and need to plot a surface based on this set of data (of 60 3D points). Here, X, Y is the horizontal plane coordinates, and Z is the vertical / height coordinate.
p = read.csv("points.csv")
PTS X Y Z
1 101 481897.9 5456408 94.18695
2 102 481888.8 5456417 94.30702
3 103 481877.0 5456410 94.29034
4 104 481879.9 5456425 94.25546
5 105 481872.7 5456424 94.09370
After looking through several posts and trying to use functions in several libraries, I still cannot figure out a way to properly plot the surface. I've tried the following:
library(plotly)
plot_ly( y= Y, x = X, z = Z, data=p, type = "surface") #returns empty graphic frame
PX = data.matrix(p$X)
PY = data.matrix(p$Y)
PZ = data.matrix(p$Z)
library(plot3D)
surf3D(PX, PY, PZ)
#returns: Error in if (is.na(var)) ispresent <- FALSE else if (length(var) == 1) if (is.logical(var)) if (!var) ispresent <- FALSE :
argument is of length zero
library(lattice)
wireframe(p$Z ~ p$X*p$Y, data = p) #returns just a cube
library(rgl)
surface3d(p$X,p$Y,p$Z)
#returns: Error in rgl.surface(x = c(481897.916, 481888.8482, 481876.9524, 481879.9393, : y' length != 'x' rows * 'z' cols;
#although there are 60 data points in the form (X,Y,Z) in the data set, with no points missing any coordinate
I must have been doing something horribly wrong here. Would anyone mind to point out what the mistake is?

You cannot make a 3D surface plot with this data because to do it you have to have Z value for each (X,Y) couple, like this :
X1 X2 X3 ... Xn
Y1 Z11 Z12 Z13 ... Z1n
Y2 Z21 Z22 Z23 ... Z2n
Y3 Z31 Z32 Z33 ... Z3n
. .
. .
. .
Ym Zm1 Zm2 Zm3 ... Zmn
For example you don't have Z value for (481897.9,5456417) couple.
So, all you can do is a scatter3d plot :
plot_ly(data = p,x = X,y = Y, z = Z,type = "scatter3d",showlegend = FALSE)

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))

R - What algorithm does geom_density() use and how to extract points/equation of curves?

I would like to know what is geom_density() exactly doing, so I justify the graph and if there is any way of extracting the function or points that generates for each of the curves being plotted.
Thanks

Typing get("compute_group", ggplot2::StatDensity) (or, formerly, get("calculate", ggplot2:::StatDensity)) will get you the algorithm used to calculate the density. (At root, it's a call to density() with kernel="gaussian" the default.)
The points used in the plot are invisibly returned by print.ggplot(), so you can access them like this:
library(ggplot2)
m <- ggplot(movies, aes(x = rating))
m <- m + geom_density()
p <- print(m)
head(p$data[[1]], 3)
# y x density scaled count PANEL group ymin ymax
# 1 0.0073761 1.0000 0.0073761 0.025917 433.63 1 1 0 0.0073761
# 2 0.0076527 1.0176 0.0076527 0.026888 449.88 1 1 0 0.0076527
# 3 0.0078726 1.0352 0.0078726 0.027661 462.81 1 1 0 0.0078726
## Just to show that those are the points you are after,
## extract and use them to create a lattice xyplot
library(gridExtra)
library(lattice)
mm <- xyplot(y ~x, data=p$data[[1]], type="l")

As suggested in other answers, you can access the ggplot points using print.ggplot(). However, print()-ing code also prints the ggplot object, which may not be desired.
You can get extract the ggplot object data, without printing the plot, using ggplot_build():
library(ggplot2)
library(ggplot2movies)
m <- ggplot(movies, aes(x = rating))
m <- m + geom_density()
p <- ggplot_build(m) # <---- INSTEAD OF `p <- print(m)`
head(p$data[[1]], 3)
# y x density scaled count n PANEL group ymin
# 1 0.007376115 1.000000 0.007376115 0.02591684 433.6271 58788 1 -1 0
# 2 0.007652653 1.017613 0.007652653 0.02688849 449.8842 58788 1 -1 0
# 3 0.007872571 1.035225 0.007872571 0.02766120 462.8127 58788 1 -1 0
# Just to show that those are the points you are after, extract and use them
# to create a lattice xyplot
library(lattice)
m2 <- xyplot(y ~x, data=p$data[[1]], type="l")
library(gridExtra)
grid.arrange(m, m2, nrow=1)

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