How to create a Matlab pumpkin in R? - r

I am trying to replicate the following visual with the following Matlab code:
% Pumpkin
[X,Y,Z]=sphere(200);
R=1-(1-mod(0:.1:20,2)).^2/12;
x=R.*X; y=R.*Y; z=Z.*R;
c=hypot(hypot(x,y),z)+randn(201)*.03;
surf(x,y,(.8+(0-(1:-.01:-1)'.^4)*.3).*z,c, 'FaceColor', 'interp', 'EdgeColor', 'none')
% Stem
s = [ 1.5 1 repelem(.7, 6) ] .* [ repmat([.1 .06],1,10) .1 ]';
[t, p] = meshgrid(0:pi/15:pi/2,0:pi/20:pi);
Xs = -(.4-cos(p).*s).*cos(t)+.4;
Zs = (.5-cos(p).*s).*sin(t) + .55;
Ys = -sin(p).*s;
surface(Xs,Ys,Zs,[],'FaceColor', '#008000','EdgeColor','none');
% Style
colormap([1 .4 .1; 1 1 .7])
axis equal
box on
material([.6 1 .3])
lighting g
camlight
I am working on the bottom but have not gotten very far (see here for reference). The code that I have is:
library(pracma)
library(rgl)
sphere <- function(n) {
dd <- expand.grid(theta = seq(0, 2*pi, length.out = n+1),
phi = seq(-pi, pi, length.out = n+1))
with(dd,
list(x = matrix(cos(phi) * cos(theta), n+1),
y = matrix(cos(phi) * sin(theta), n+1),
z = matrix(sin(phi), n+1))
)
}
# Pumpkin
sph<-sphere(200)
X<-sph[[1]]
Y<-sph[[2]]
Z<-sph[[3]]
R<- 1-(1-seq(from=0, to=20,by=0.1))^2/12
x<-R * X
y<-R * Y
z<-Z * R
c<-hypot(hypot(x,y),z)+rnorm(201)*0.3
persp3d(x,y,(0.8+(0-seq(from=1, to=-1, by=-0.01)^4)*0.3)*z,col=c)
and it gives me the following.
What is it that's going wrong in my present code? What would be a suggested fix?

As #billBokeey mentioned, there's a missing mod modulo operator function for periodic scaling factors.
In addition, the scaling on the z-axis 0.8 + (0-seq(from=1, to=-1, by=-0.01)^4) * 0.3 doesn't go well with the output from your sphere function. We maybe use Z[1,] to replace seq(from=1, to=-1, by=-0.01). phi = seq(-pi, pi, length.out = n+1)) shoud be seq(-pi/2, pi/2, length.out = n+1)) instead.
Finally, the color c needs to be convert to RGB code for persp3d.
Here's the result look like from the code below.
library(rgl)
sphere <- function(n) {
dd <- expand.grid(theta = seq(0, 2*pi, length.out = n+1),
phi = seq(-pi/2, pi/2, length.out = n+1))
with(dd,
list(x = matrix(cos(phi) * cos(theta), n+1),
y = matrix(cos(phi) * sin(theta), n+1),
z = matrix(sin(phi), n+1))
)
}
# Unit ball
sph <- sphere(200)
X <- sph[[1]]
Y <- sph[[2]]
Z <- sph[[3]]
# scaling
R <- 1 - (1 - seq(from=0, to=20, by=0.1) %% 2) ^ 2 / 12 # Modulo Operator %%
R2 <- 0.8 + (0 - seq(from=1, to=-1, by=-0.01)^4)*0.2 # didn't match with the order of z from sphere function
#R2 <- 0.8 - Z[1,]^4 * 0.2
x <- R * X # scale rows for wavy side
y <- R * Y # scale rows for wavy side
z <- t(R2 * t(Z)) # scale columns by transpose for flat oval shape
# color according to distance to [0,0,0]
hypot_3d <- function(x, y, z) {
return(sqrt(x^2 + y^2 + z^2))
}
c_ <- hypot_3d(x,y,z) + rnorm(201) * 0.03
color_palette <- terrain.colors(20) # color look-up table
col <- color_palette[ as.numeric(cut(c_, breaks = 20)) ] # assign color to 20 levels of c_
persp3d(x, y, z, color = col, aspect=FALSE)

Related

Understanding "levels" in r contour function of bivariate distribution

I have trouble understanding how to set the levels in the plot of a bivariate distribution in r. The documentation states that I can choose the levels by setting a
numeric vector of levels at which to draw contour lines
Now I would like the contour to show the limit containing 95% of the density or mass. But if, in the example below (adapted from here) I set the vector as a <- c(.95,.90) the code runs without error but the plot is not displayed. If instead, I set the vector as a <- c(.01,.05) the plot is displayed. But I am not sure I understand what the labels "0.01" and "0.05" mean with respect to the density.
library(mnormt)
x <- seq(-5, 5, 0.25)
y <- seq(-5, 5, 0.25)
mu1 <- c(0, 0)
sigma1 <- matrix(c(2, -1, -1, 2), nrow = 2)
f <- function(x, y) dmnorm(cbind(x, y), mu1, sigma1)
z <- outer(x, y, f)
a <- c(.01,.05)
contour(x, y, z, levels = a)
But I am not sure I understand what the labels "0.01" and "0.05" mean with respect to the density.
It means the points where the density is equal 0.01 and 0.05. From help("contour"):
numeric vector of levels at which to draw contour lines.
So it is the function values at which to draw the lines (contours) where the function is equal to those levels (in this case the density). Take a simple example which may help is x + y:
y <- x <- seq(0, 1, length.out = 50)
z <- outer(x, y, `+`)
par(mar = c(5, 5, 1, 1))
contour(x, y, z, levels = c(0.5, 1, 1.5))
Now I would like the contour to show the limit containing 95% of the density or mass.
In your example, you can follow my answer here and draw the exact points:
# input
mu1 <- c(0, 0)
sigma1 <- matrix(c(2, -1, -1, 2), nrow = 2)
# we start from points on the unit circle
n_points <- 100
xy <- cbind(sin(seq(0, 2 * pi, length.out = n_points)),
cos(seq(0, 2 * pi, length.out = n_points)))
# then we scale the dimensions
ev <- eigen(sigma1)
xy[, 1] <- xy[, 1] * 1
xy[, 2] <- xy[, 2] * sqrt(min(ev$values) / max(ev$values))
# then rotate
phi <- atan(ev$vectors[2, 1] / ev$vectors[1, 1])
R <- matrix(c(cos(phi), sin(phi), -sin(phi), cos(phi)), 2)
xy <- tcrossprod(R, xy)
# find the right length. You can change .95 to which ever
# quantile you want
chi_vals <- qchisq(.95, df = 2) * max(ev$values)
s <- sqrt(chi_vals)
par(mar = c(5, 5, 1, 1))
plot(s * xy[1, ] + mu1[1], s * xy[2, ] + mu1[2], lty = 1,
type = "l", xlab = "x", ylab = "y")
The levels indicates where the lines are drawn, with respect to the specific 'z' value of the bivariate normal density. Since max(z) is
0.09188815, levels of a <- c(.95,.90) can't be drawn.
To draw the line delimiting 95% of the mass I used the ellipse() function as suggested in this post (second answer from the top).
library(mixtools)
library(mnormt)
x <- seq(-5, 5, 0.25)
y <- seq(-5, 5, 0.25)
mu1 <- c(0, 0)
sigma1 <- matrix(c(2, -1, -1, 2), nrow = 2)
f <- function(x, y) dmnorm(cbind(x, y), mu1, sigma1)
z <- outer(x, y, f)
a <- c(.01,.05)
contour(x, y, z, levels = a)
ellipse(mu=mu1, sigma=sigma1, alpha = .05, npoints = 250, col="red")
I also found another solution in the book "Applied Multivariate Statistics with R" by Daniel Zelterman.
# Figure 6.5: Bivariate confidence ellipse
library(datasets)
library(MASS)
library(MVA)
#> Loading required package: HSAUR2
#> Loading required package: tools
biv <- swiss[, 2 : 3] # Extract bivariate data
bivCI <- function(s, xbar, n, alpha, m)
# returns m (x,y) coordinates of 1-alpha joint confidence ellipse of mean
{
x <- sin( 2* pi * (0 : (m - 1) )/ (m - 1)) # m points on a unit circle
y <- cos( 2* pi * (0 : (m - 1)) / (m - 1))
cv <- qchisq(1 - alpha, 2) # chisquared critical value
cv <- cv / n # value of quadratic form
for (i in 1 : m)
{
pair <- c(x[i], y[i]) # ith (x,y) pair
q <- pair %*% solve(s, pair) # quadratic form
x[i] <- x[i] * sqrt(cv / q) + xbar[1]
y[i] <- y[i] * sqrt(cv / q) + xbar[2]
}
return(cbind(x, y))
}
### pdf(file = "bivSwiss.pdf")
plot(biv, col = "red", pch = 16, cex.lab = 1.5)
lines(bivCI(var(biv), colMeans(biv), dim(biv)[1], .01, 1000), type = "l",
col = "blue")
lines(bivCI(var(biv), colMeans(biv), dim(biv)[1], .05, 1000),
type = "l", col = "green", lwd = 1)
lines(colMeans(biv)[1], colMeans(biv)[2], pch = 3, cex = .8, type = "p",
lwd = 1)
Created on 2021-03-15 by the reprex package (v0.3.0)

Generating the points of a circle

I'm just playing around with the image of circles under the complex map exp(z).
I couldn't find a built in R function to generate the points on a circle of given radius so I wrote one myself (integrating the equations of motion numerically):
# Integration points:
N <- 10000
e <- 0.001
dt <- seq(0, e*(N-1), by=e)
Rp = pi # radius of point circle
Rv = pi # radius of vector circle
# Initial conditions:
px <- c(Rp)
py <- c(0)
vx <- c(0)
vy <- c(Rv)
Rp <- c()
Rv <- c()
ax <- c()
ay <- c()
for (i in(2:N)) {
Rp[i-1] <- sqrt(px[i-1]^2 + py[i-1]^2)
Rv[i-1] <- sqrt(vx[i-1]^2 + vy[i-1]^2)
ax[i-1] <- -(Rv[i-1]^2/Rp[i-1]^2)*px[i-1] # acceleration toowards
ay[i-1] <- -(Rv[i-1]^2/Rp[i-1]^2)*py[i-1] # center of circle
px[i] <- px[i-1] + e*vx[i-1] # dp_x = epsilon * v_x
py[i] <- py[i-1] + e*vy[i-1] # dp_y = epsilon * v_y
vx[i] <- vx[i-1] + e*ax[i-1] # dv_x = epsilon * a_x
vy[i] <- vy[i-1] + e*ay[i-1] # dv_y = epslon * a_y
}
complex(real=px,imaginary=py)
This seems like a lot of work just to get a circle, and the program is slow. Is there a built in R function to do this for me?
par(mfrow=c(1,2))
plot(cbind(px,py))
plot(exp(zs))
Thanks!
Parameterize on angle:
circle_xy = function(n, r, close_loop = FALSE) {
theta = seq(0, 2 * pi, length.out = n + 1)
if(!close_loop) theta = theta[-(n + 1)]
cbind(x = r * cos(theta), y = r * sin(theta))
}
Gives x-y coords for n evenly spaced points on a circle of radius r. If close_loop = TRUE, the first point is repeated at the end. Takes about 0.2 seconds to generate 1MM points on my laptop.
And there is plot.formula function that would take that to an instantiation:
plot( y ~ x, data = xy<- circle_xy(100,1), type="l")

Fitting data points to an ellipse with its center at the origin using R

I have a question about fitting ellipses to data with the ellipse center at the origin. I have explored two methods that fit ellipses but generate an arbitrary center unless I manipulate the data with some imaginary mirror points.
Method#01
This portion of the script directly comes from this useful post. I'm copying the codes directly here for ease.
fit.ellipse <- function (x, y = NULL) {
# from:
# http://r.789695.n4.nabble.com/Fitting-a-half-ellipse-curve-tp2719037p2720560.html
#
# Least squares fitting of an ellipse to point data
# using the algorithm described in:
# Radim Halir & Jan Flusser. 1998.
# Numerically stable direct least squares fitting of ellipses.
# Proceedings of the 6th International Conference in Central Europe
# on Computer Graphics and Visualization. WSCG '98, p. 125-132
#
# Adapted from the original Matlab code by Michael Bedward (2010)
# michael.bedward#gmail.com
#
# Subsequently improved by John Minter (2012)
#
# Arguments:
# x, y - x and y coordinates of the data points.
# If a single arg is provided it is assumed to be a
# two column matrix.
#
# Returns a list with the following elements:
#
# coef - coefficients of the ellipse as described by the general
# quadratic: ax^2 + bxy + cy^2 + dx + ey + f = 0
#
# center - center x and y
#
# major - major semi-axis length
#
# minor - minor semi-axis length
#
EPS <- 1.0e-8
dat <- xy.coords(x, y)
D1 <- cbind(dat$x * dat$x, dat$x * dat$y, dat$y * dat$y)
D2 <- cbind(dat$x, dat$y, 1)
S1 <- t(D1) %*% D1
S2 <- t(D1) %*% D2
S3 <- t(D2) %*% D2
T <- -solve(S3) %*% t(S2)
M <- S1 + S2 %*% T
M <- rbind(M[3,] / 2, -M[2,], M[1,] / 2)
evec <- eigen(M)$vec
cond <- 4 * evec[1,] * evec[3,] - evec[2,]^2
a1 <- evec[, which(cond > 0)]
f <- c(a1, T %*% a1)
names(f) <- letters[1:6]
# calculate the center and lengths of the semi-axes
#
# see http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2288654/
# J. R. Minter
# for the center, linear algebra to the rescue
# center is the solution to the pair of equations
# 2ax + by + d = 0
# bx + 2cy + e = 0
# or
# | 2a b | |x| |-d|
# | b 2c | * |y| = |-e|
# or
# A x = b
# or
# x = Ainv b
# or
# x = solve(A) %*% b
A <- matrix(c(2*f[1], f[2], f[2], 2*f[3]), nrow=2, ncol=2, byrow=T )
b <- matrix(c(-f[4], -f[5]), nrow=2, ncol=1, byrow=T)
soln <- solve(A) %*% b
b2 <- f[2]^2 / 4
center <- c(soln[1], soln[2])
names(center) <- c("x", "y")
num <- 2 * (f[1] * f[5]^2 / 4 + f[3] * f[4]^2 / 4 + f[6] * b2 - f[2]*f[4]*f[5]/4 - f[1]*f[3]*f[6])
den1 <- (b2 - f[1]*f[3])
den2 <- sqrt((f[1] - f[3])^2 + 4*b2)
den3 <- f[1] + f[3]
semi.axes <- sqrt(c( num / (den1 * (den2 - den3)), num / (den1 * (-den2 - den3)) ))
# calculate the angle of rotation
term <- (f[1] - f[3]) / f[2]
angle <- atan(1 / term) / 2
list(coef=f, center = center, major = max(semi.axes), minor = min(semi.axes), angle = unname(angle))
}
Let's take a example distribution of polar points for illustration purpose
X<-structure(list(x_polar = c(0, 229.777200000011, 246.746099999989,
-10.8621999999741, -60.8808999999892, 75.8904999999795, -83.938199999975,
-62.9770000000135, 49.1650999999838, 52.3093000000226, 49.6891000000178,
-66.4248999999836, 34.3671999999788, 242.386400000018, 343.60619999998
), y_polar = c(0, 214.868299999973, 161.063599999994, -68.8972000000067,
-77.0230000000447, 93.2863000000361, -16.2356000000145, 27.7828000000445,
-17.8077000000048, 2.10540000000037, 25.6866000000155, -84.6034999999683,
-31.1800000000512, 192.010800000047, 222.003700000001)), .Names = c("x_polar",
"y_polar"), row.names = c(NA, -15L), class = "data.frame")
efit <- fit.ellipse(X)
e <- get.ellipse(efit)
#plot
par(bg=NA)
plot(X, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="", type='n',
ylim=c(min(X$y_polar)-150, max(X$y_polar)), xlim=c(min(X$x_polar)-150, max(X$x_polar))) #blank plot
points(X$x_polar, X$y_polar, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="") #observations
lines(e, col="red", lwd=3, lty=2) #plotting the ellipse
points(0,0,col=2, lwd=2, cex=2) #center/origin
To bring the origin of the ellipse at the center we could modify as follows (surely not the best way of doing it)
#generate mirror coordinates
X$x_polar_mirror<- -X$x_polar
X$y_polar_mirror<- -X$y_polar
mydata<-as.matrix(data.frame(c(X$x_polar, X$x_polar_mirror), c(X$y_polar, X$y_polar_mirror)))
#fit the data
efit <- fit.ellipse(mydata)
e <- get.ellipse(efit)
par(bg=NA)
plot(mydata, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="", type='n',
ylim=c(min(X$y_polar)-150, max(X$y_polar)), xlim=c(min(X$x_polar)-150, max(X$x_polar)))
points(X$x_polar, X$y_polar, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="")
lines(e, col="red", lwd=3, lty=2)
points(0,0,col=2, lwd=2, cex=2) #center
Well ... it sort of does the job but none would be happy with all those imaginary points considered in the calculation.
Method#02
This is another indirect way of fitting the data but again the ellipse center is not at the origin. Any workaround?
require(car)
dataEllipse(X$x_polar, X$y_polar, levels=c(0.15, 0.7),
xlim=c(-150, 400), ylim=c(-200,300))
My questions: (a) is there a robust alternative way of fitting these points with the ellipse center at the origin (0,0)? (b) is there a measure of the goodness of ellipse fit? Thank you in advance.
I'm not really happy with aproach I've concieved, there should be a closed form solution, but still:
# Ellipse equasion with center in (0, 0) with semiaxis pars[1] and pars[2] rotated by pars[3].
# t and pars[3] in radians
ellipsePoints <- function(t, pars) {
data.frame(x = cos(pars[3]) * pars[1] * cos(t) - sin(pars[3]) * pars[2] * sin(t),
y = sin(pars[3]) * pars[1] * cos(t) + cos(pars[3]) * pars[2] * sin(t))
}
# Way to fit an ellipse through minimising distance to data points.
# If weighted then points which are most remote from center will have bigger impact.
ellipseBrute <- function(x, y, pars, weighted = FALSE) {
d <- sqrt(x**2 + y**2)
t <- asin(y/d)
w <- (d/sum(d))**weighted
t[x == 0 & y == 0] <- 0
ep <- ellipsePoints(t, pars)
sum(w*(sqrt(ep$x**2 + ep$y**2) - d)**2)
}
# Fit through optim.
opt_res <- optim(c(diff(range(X$x_polar)),
diff(range(X$y_polar)),
2*pi)/2,
ellipseBrute,
x = X$x_polar, y = X$y_polar,
weighted = TRUE
)
# Check resulting ellipse throuh plot
df <- ellipsePoints(seq(0, 2*pi, length.out = 1e3), opt_res$par)
plot(y ~ x, df, col = 'blue', t = 'l',
xlim = range(c(X$x_polar, df$x)),
ylim = range(c(X$y_polar, df$y)))
points(0, 0, pch = 3, col = 'blue')
points(y_polar ~ x_polar, X)

Reproduce Fisher linear discriminant figure

Many books illustrate the idea of Fisher linear discriminant analysis using the following figure (this particular is from Pattern Recognition and Machine Learning, p. 188)
I wonder how to reproduce this figure in R (or in any other language). Pasted below is my initial effort in R. I simulate two groups of data and draw linear discriminant using abline() function. Any suggestions are welcome.
set.seed(2014)
library(MASS)
library(DiscriMiner) # For scatter matrices
# Simulate bivariate normal distribution with 2 classes
mu1 <- c(2, -4)
mu2 <- c(2, 6)
rho <- 0.8
s1 <- 1
s2 <- 3
Sigma <- matrix(c(s1^2, rho * s1 * s2, rho * s1 * s2, s2^2), byrow = TRUE, nrow = 2)
n <- 50
X1 <- mvrnorm(n, mu = mu1, Sigma = Sigma)
X2 <- mvrnorm(n, mu = mu2, Sigma = Sigma)
y <- rep(c(0, 1), each = n)
X <- rbind(x1 = X1, x2 = X2)
X <- scale(X)
# Scatter matrices
B <- betweenCov(variables = X, group = y)
W <- withinCov(variables = X, group = y)
# Eigenvectors
ev <- eigen(solve(W) %*% B)$vectors
slope <- - ev[1,1] / ev[2,1]
intercept <- ev[2,1]
par(pty = "s")
plot(X, col = y + 1, pch = 16)
abline(a = slope, b = intercept, lwd = 2, lty = 2)
MY (UNFINISHED) WORK
I pasted my current solution below. The main question is how to rotate (and move) the density plot according to decision boundary. Any suggestions are still welcome.
require(ggplot2)
library(grid)
library(MASS)
# Simulation parameters
mu1 <- c(5, -9)
mu2 <- c(4, 9)
rho <- 0.5
s1 <- 1
s2 <- 3
Sigma <- matrix(c(s1^2, rho * s1 * s2, rho * s1 * s2, s2^2), byrow = TRUE, nrow = 2)
n <- 50
# Multivariate normal sampling
X1 <- mvrnorm(n, mu = mu1, Sigma = Sigma)
X2 <- mvrnorm(n, mu = mu2, Sigma = Sigma)
# Combine into data frame
y <- rep(c(0, 1), each = n)
X <- rbind(x1 = X1, x2 = X2)
X <- scale(X)
X <- data.frame(X, class = y)
# Apply lda()
m1 <- lda(class ~ X1 + X2, data = X)
m1.pred <- predict(m1)
# Compute intercept and slope for abline
gmean <- m1$prior %*% m1$means
const <- as.numeric(gmean %*% m1$scaling)
z <- as.matrix(X[, 1:2]) %*% m1$scaling - const
slope <- - m1$scaling[1] / m1$scaling[2]
intercept <- const / m1$scaling[2]
# Projected values
LD <- data.frame(predict(m1)$x, class = y)
# Scatterplot
p1 <- ggplot(X, aes(X1, X2, color=as.factor(class))) +
geom_point() +
theme_bw() +
theme(legend.position = "none") +
scale_x_continuous(limits=c(-5, 5)) +
scale_y_continuous(limits=c(-5, 5)) +
geom_abline(intecept = intercept, slope = slope)
# Density plot
p2 <- ggplot(LD, aes(x = LD1)) +
geom_density(aes(fill = as.factor(class), y = ..scaled..)) +
theme_bw() +
theme(legend.position = "none")
grid.newpage()
print(p1)
vp <- viewport(width = .7, height = 0.6, x = 0.5, y = 0.3, just = c("centre"))
pushViewport(vp)
print(p2, vp = vp)
Basically you need to project the data along the direction of the classifier, plot a histogram for each class, and then rotate the histogram so its x axis is parallel to the classifier. Some trial-and-error with scaling the histogram is needed in order to get a nice result. Here's an example of how to do it in Matlab, for the naive classifier (difference of class' means). For the Fisher classifier it is of course similar, you just use a different classifier w. I changed the parameters from your code so the plot is more similar to the one you gave.
rng('default')
n = 1000;
mu1 = [1,3]';
mu2 = [4,1]';
rho = 0.3;
s1 = .8;
s2 = .5;
Sigma = [s1^2,rho*s1*s1;rho*s1*s1, s2^2];
X1 = mvnrnd(mu1,Sigma,n);
X2 = mvnrnd(mu2,Sigma,n);
X = [X1; X2];
Y = [zeros(n,1);ones(n,1)];
scatter(X1(:,1), X1(:,2), [], 'b' );
hold on
scatter(X2(:,1), X2(:,2), [], 'r' );
axis equal
m1 = mean(X(1:n,:))';
m2 = mean(X(n+1:end,:))';
plot(m1(1),m1(2),'bx','markersize',18)
plot(m2(1),m2(2),'rx','markersize',18)
plot([m1(1),m2(1)], [m1(2),m2(2)],'g')
%% classifier taking only means into account
w = m2 - m1;
w = w / norm(w);
% project data onto w
X1_projected = X1 * w;
X2_projected = X2 * w;
% plot histogram and rotate it
angle = 180/pi * atan(w(2)/w(1));
[hy1, hx1] = hist(X1_projected);
[hy2, hx2] = hist(X2_projected);
hy1 = hy1 / sum(hy1); % normalize
hy2 = hy2 / sum(hy2); % normalize
scale = 4; % set manually
h1 = bar(hx1, scale*hy1,'b');
h2 = bar(hx2, scale*hy2,'r');
set([h1, h2],'ShowBaseLine','off')
% rotate around the origin
rotate(get(h1,'children'),[0,0,1], angle, [0,0,0])
rotate(get(h2,'children'),[0,0,1], angle, [0,0,0])

R: Add points to surface plot with persp having the appropriate size

I would like to achieve that the points I add to the plot have their size adjusted to obtain a better 3D impression. I know that I somehow have to use the transformation matrix that is returned to compute the length of the vector orthogonal to the 2d plane to the respective point in 3d, but I don't know how to do that.
Here is an example:
x1 <- rnorm(100)
x2 <- 4 + rpois(100, 4)
y <- 0.1*x1 + 0.2*x2 + rnorm(100)
dat <- data.frame(x1, x2, y)
m1 <- lm(y ~ x1 + x2, data=dat)
x1r <- range(dat$x1)
x1seq <- seq(x1r[1], x1r[2], length=30)
x2r <- range(dat$x2)
x2seq <- seq(x2r[1], x2r[2], length=30)
z <- outer(x1seq, x2seq, function(a,b){
predict(m1, newdata=data.frame(x1=a, x2=b))
})
res <- persp(x1seq, x2seq, z)
mypoints <- trans3d(dat$x1, dat$x2, dat$y, pmat=res)
points(mypoints, pch=1, col="red")
You can use the function presented here to determine distance to the observer, then scale the pointsize (cex) to that distance:
# volcano data
z <- 2 * volcano # Exaggerate the relief
x <- 10 * (1:nrow(z)) # 10 meter spacing (S to N)
y <- 10 * (1:ncol(z)) # 10 meter spacing (E to W)
# draw volcano and store transformation matrix
pmat <- persp(x, y, z, theta = 35, phi = 40, col = 'green4', scale = FALSE,
ltheta = -120, shade = 0.75, border = NA, box = TRUE)
# take some xyz values from the matrix
s = sample(1:prod(dim(z)), size=500)
xx = x[row(z)[s] ]
yy = y[col(z)[s]]
zz = z[s] + 10
# depth calculation function (adapted from Duncan Murdoch at https://stat.ethz.ch/pipermail/r-help/2005-September/079241.html)
depth3d <- function(x,y,z, pmat, minsize=0.2, maxsize=2) {
# determine depth of each point from xyz and transformation matrix pmat
tr <- as.matrix(cbind(x, y, z, 1)) %*% pmat
tr <- tr[,3]/tr[,4]
# scale depth to point sizes between minsize and maxsize
psize <- ((tr-min(tr) ) * (maxsize-minsize)) / (max(tr)-min(tr)) + minsize
return(psize)
}
# determine distance to eye
psize = depth3d(xx,yy,zz,pmat,minsize=0.1, maxsize = 1)
# from 3D to 2D coordinates
mypoints <- trans3d(xx, yy, zz, pmat=pmat)
# plot in 2D space with pointsize related to distance
points(mypoints, pch=8, cex=psize, col=4)

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