I am trying to plot the plane determined by the equation y = x between 0 and 2 superimposed on a 3D surface described by z = 1/2 * x * y.
I "hacked" code from another question to get a minimally satisfactory representation of the surface (wouldn't mind a fresh start on that), but I have no idea how to superimposed the 2D plane.
Here is the idea:
And this is what I got so far:
with the borrow code:
my_surface <- function(f, n=10, ...) {
ranges <- rgl:::.getRanges()
x <- seq(ranges$xlim[1], ranges$xlim[2], length=n)
y <- seq(ranges$ylim[1], ranges$ylim[2], length=n)
z <- outer(x,y,f)
surface3d(x, y, z, ...)
}
library(rgl)
f <- function(x, y) 1/2 * x * y
g <- function(x, y) x = y
x <- seq(0, 2, by=0.001)
y <- seq(0,2, by=0.001)
z <- 1/2 * x * y
plot3d(x, y, z, xlab="X", ylab="y", zlab="z", site=5, type = "n")
my_surface(f, alpha=.5, col="red")
my_second_surface <- function(f, n=10, ...) {
ranges <- rgl:::.getRanges()
x <- seq(ranges$xlim[1], ranges$xlim[2], length=n)
y <- seq(ranges$ylim[1], ranges$ylim[2], length=n)
z <- outer(x,y,g)
surface3d(x, y, z, ...)
}
my_second_surface(g, alpha=.5, col="blue")
Notice that the main problem (or difference with the sketch on the first graph) is that I can't make the greenish triangle appear. Instead, I get an oblique plane in blue.
Related
I would like to draw a 3D plot of the function z=f(x,y)=x^2+y^2, y>x>0 see the picture
I know the way to do that without the constraint y>x>0 which is easy. could you please help me.
My code for the version without the constraint
fxy <- function(x,y){
return(x^2+y^2)}
x <- seq(-100 ,100, by=5)
y <- seq(-100 ,100, by=5)
z <- outer(x, y, fxy)
persp(x, y, z,
main="3D Plot of x2+y2",
zlab = "Function",
theta = 20, phi = 10,
col = "blue")
I don't know how to do this with the base persp function, but it's possible with rgl.
library(rgl)
fxy <- function(x,y){
return(x^2+y^2)}
x <- seq(-100 ,100, by=5)
y <- seq(-100 ,100, by=5)
z <- outer(x, y, fxy)
# First, draw the full surface:
full <- surface3d(x, y, z,
col = "blue")
# Convert it to a mesh object:
m1 <- as.mesh3d(full)
# Clip to x > 0
m2 <- clipMesh3d(m1, "x", bound = 0, greater = TRUE)
# Clip to y > x
m3 <- clipMesh3d(m2, function(x,y,z) y-x, bound = 0, greater = TRUE)
# Plot it
plot3d(m3)
You can now rotate it to whatever orientation you like.
I am trying to add a tangent to my plot at the point x = 30 and I want to calculate the x-intersection of the tangent at y = 0.08.
I already found a very useful example, which I tried to use, but for some reason it is not working for y = 0.08.
I don't understand the meaning of deriv in the predict() function nor the actual difference in between pred0 and pred1. Can someone please explain?
x <- seq(0,40)
y <- dnorm(seq(0,40), mean=25, sd=5)
plot(x, y)
spl <- smooth.spline(y ~ x)
lines(spl, col="green")
newx <- 30
pred0 <- predict(spl, x=newx, deriv=0)
pred1 <- predict(spl, x=newx, deriv=1)
yint <- pred0$y - (pred1$y*newx)
xint <- -yint/pred1$y
xint
plot(x, y)
abline(h=0, col="red")
lines(spl, col="red")
points(pred0,col="red", pch=19)
lines(x, yint + pred1$y*x)
points(xint, 0, col="red", pch=19)
It seems like you have no problem calculating the tangent line and the intersect, but rather need some help in finding the x value for a given y value. This method will work for any smooth curve, but mark Gregors warning. There might not be a corresponding x value, or there might be several.
x <- seq(0, 40, by=0.01)
y <- dnorm(x, mean=25, sd=5)
spl <- smooth.spline(y ~ x)
plot(spl, type="l")
yval <- 0.08
ad <- abs(yval - spl$y)
if (min(ad) > max(diff(spl$y))*10) {
warning("The supplied y value is out of bounds")
}
xval <- x[which(diff(sign(diff(ad))) > 1) + 1]
points(xval, rep(yval, length(xval)))
With that xval you can calculate the tangent as you've already done.
I am trying to visualize a curve for pollination distribution. I am very new to R so please don't be upset by my stupidity.
llim <- 0
ulim <- 6.29
f <- function(x,y) {(.156812/((2*pi)*(.000005^2)*(gamma(2/.156812)))*exp(-((sqrt(x^2+y^2))/.000005)^.156812))}
integrate(function(y) {
sapply(y, function(y) {
integrate(function(x) f(x,y), llim, ulim)$value
})
}, llim, ulim)
fv <- Vectorize(f)
curve(fv, from=0, to=1000)
And I get:
Error in y^2 : 'y' is missing
I'm not quite sure what you're asking to plot. But I know you want to visualise your scalar function of two arguments.
Here are some approaches. First we define your function.
llim <- 0
ulim <- 6.29
f <- function(x,y) {
(.156812/((2*pi)*(.000005^2)*(gamma(2/.156812)))*exp(-((sqrt(x^2+y^2))/.000005)^.156812))
}
From your title I thought of the following. The function defined below intf integrates your function over the square [0,ul] x [0,ul] and return the value. We then vectorise and plot the integral over the square as a function the length of the side of the square.
intf <- function(ul) {
integrate(function(y) {
sapply(y, function(y) {
integrate(function(x) f(x,y), 0, ul)$value
})
}, 0, ul)$value
}
fv <- Vectorize(intf)
curve(fv, from=0, to=1000)
If f is a distribution, I guess you can make your (somewhat) nice probability interpretation of this curve. (I.e. ~20 % probability of pollination(?) in the 200 by 200 meter square.)
However, you can also do a contour plot (of the log-transformed values) which illustrate the function we are integrating above:
logf <- function(x, y) log(f(x, y))
x <- y <- seq(llim, ulim, length.out = 100)
contour(x, y, outer(x, y, logf), lwd = 2, drawlabels = FALSE)
You can also plot some profiles of the surface:
plot(1, xlim = c(llim, ulim), ylim = c(0, 0.005), xlab = "x", ylab = "f")
y <- seq(llim, ulim, length.out = 6)
for (i in seq_along(y)) {
tmp <- function(x) f(x, y = y[i])
curve(tmp, llim, ulim, add = TRUE, col = i)
}
legend("topright", lty = 1, col = seq_along(y),
legend = as.expression(paste("y = ",y)))
They need to be modified a bit to make them publication worthy, but you get the idea. Lastly, you can do some 3d plots as others have suggested.
EDIT
As per your comments, you can also do something like this:
# Define the function times radius (this time with general a and b)
# The default of a and b is as before
g <- function(z, a = 5e-6, b = .156812) {
z * (b/(2*pi*a^2*gamma(2/b)))*exp(-(z/a)^b)
}
# A function that integrates g from 0 to Z and rotates
# As g is not dependent on the angle we just multiply by 2pi
intg <- function(Z, ...) {
2*pi*integrate(g, 0, Z, ...)$value
}
# Vectorize the Z argument of intg
gv <- Vectorize(intg, "Z")
# Plot
Z <- seq(0, 1000, length.out = 100)
plot(Z, gv(Z), type = "l", lwd = 2)
lines(Z, gv(Z, a = 5e-5), col = "blue", lwd = 2)
lines(Z, gv(Z, b = .150), col = "red", lwd = 2)
lines(Z, gv(Z, a = 1e-4, b = .2), col = "orange", lwd = 2)
You can then plot the curves for the a and b you want. If either is not specified, the default is used.
Disclaimer: my calculus is rusty and I just did off this top of my head. You should verify that I've done the rotation of the function around the axis properly.
The lattice package has several functions that can help you draw 3 dimensional plots, including wireframe() and persp(). If you prefer not to use a 3d-plot, you can create a contour plot using contour().
Note: I don't know if this is intentional, but your data produces a very large spike in one corner of the plot. This produces a plot that is for all intents flat, with a barely noticable spike in one corner. This is particularly problematic with the contour plot below.
library(lattice)
x <- seq(0, 1000, length.out = 50)
y <- seq(0, 1000, length.out = 50)
First the wire frame plot:
df <- expand.grid(x=x, y=y)
df$z <- with(df, f(x, y))
wireframe(z ~ x * y, data = df)
Next the perspective plot:
dm <- outer(x, y, FUN=f)
persp(x, y, dm)
The contour plot:
contour(x, y, dm)
I would like to create a waterfall plot in R (XYYY) from my data.
So far, I use this code:
load("myData.RData")
ls()
dim(data)
##matrix to xyz coords
library(reshape2)
newData <- melt(data, id="Group.1")
dim(newData)
head(newData)
tail(newData)
newDataO <- newData[c(2,1,3)]
head(newDataO)
##color scale for z axis
myColorRamp <- function(colors, values) {
v <- (values - min(values))/diff(range(values))
x <- colorRamp(colors)(v)
rgb(x[,1], x[,2], x[,3], maxColorValue = 255)
}
cols <- myColorRamp(c("darkblue","yellow","darkorange","red","darkred"),newDataO$value)
##3D scatter
library(rgl)
plot3d(newDataO$variable, newDataO$Group.1, newDataO$value, xlab="", ylab="", zlab="", type="p", col=cols, box=FALSE, axes=FALSE)
rgl.postscript("persptrial_060514.eps","eps")
to get this plot:
https://dl.dropboxusercontent.com/u/14906265/persptrial_060514.jpg
I have also use this option in 2d with polygon but the result does not properly show the differential effect between both plots (left vs right).
I do not know whether something like persp3d could do the job but I am not familiar enough with writing code to achieve it. Any help will be very much appreciated.
It seems to me that the simplest way of doing a waterfall plot in R is to add all the lines manually in a loop.
library(rgl)
# Function to plot
f <- function(x, y) sin(10 * x * y) * cos(4 * y^3) + x
nx <- 30
ny <- 100
x <- seq(0, 1, length = nx)
y <- seq(0, 1, length = ny)
z <- outer(x, y, FUN = f)
# Plot function and add lines manually
surface3d(x, y, z, alpha = 0.4)
axes3d()
for (i in 1:nx) lines3d(x[i], y, z[i, ], col = 'white', lwd = 2)
I'm trying to plot the two circumferences with dimensions xy together with the 3d plot and colour the intersection of the two circles, how can I do that?
# objective function
x <- seq(-1,1,.1)
y <- seq(-1,1,.1)
z <- x^2 + y^2
library(scatterplot3d)
library(plotrix)
scatterplot3d(x,y,z,pch=19,color="royalblue4")
draw.circle (1,1,1)
draw.circle (1,-1,1)
I'm not really into mathematic stuff, but I'll post as answer because it might be of use and, also, is too big for a comment. Excuse any ignorance of mine, though, if I post nonsense.
#your data
library(scatterplot3d)
x <- seq(-1,1,.1)
y <- seq(-1,1,.1)
z <- x^2 + y^2
ang = 60 #angle of the 3D plot. experiment with different values
#your 3D plot, with extended xx', yy' limits
sp3d <- scatterplot3d(x, y, z, pch=19, color="royalblue4",
xlim = c(-1, 3), ylim = c(-3, 3), angle = ang)
#to use parametric equations of circles
f <- seq(-2*pi, 2*pi, 0.1)
#circle1
sp3d$points(x = 1 + 1*cos(f), y = 1 + 1*sin(f), z = rep(0, length(f)), type = "l")
#circle2
sp3d$points(x = 1 + 1*cos(f), y = -1 + 1*sin(f), z = rep(0, length(f)), type = "l")
The plot is: