From the parametric equations on wiki page ( http://en.wikipedia.org/wiki/Parametric_equation ), I can plot 2d equations as follows:
#for a circle:
x = seq(-pi, pi, length.out=30)
plot(sin(x),cos(x))
# for a star:
a=10; b=10/1.8
x=seq(-50,50,length.out=500)
plot((a-b)*cos(x)+b*cos(x*((a/b)-1)), (a-b)*sin(x)-b*sin(x*((a/b)-1)), ylim=range(-13,13))
How can I plot 3d equations on a 3d plot of a Helix given by equations:
x=a*cos(t)
y=a*sin(t)
z=b*t
From searching I found that the 3d plotting functions take either a matrix or x,y,z values but not math curve functions.
You can plot 3D equations like you did the 2D ones.
library(lattice)
t<-seq(-2*pi, 2*pi, length.out=200)
cloud(z~x+y,data.frame(x=3*cos(t),y=3*sin(t), z=2*t))
So yes, you can't supply a raw function directly, but you can easily calculate points to plot based on those functions. Let me know if you had something else in mind.
Here's a two-parameter torus
t <- seq(0, 2*pi, length.out=50);
u <- seq(0, 2*pi, length.out=50);
tu<-expand.grid(t=t,u=u)
R <- 6;
r <- 3;
tu <- transform(tu,
x = cos(t)*(R+r*cos(u)),
y = sin(t)*(R+r*cos(u)),
z = r*sin(u)
)
rr<-c(-10,10)
cloud(z~x+y, tu, xlim=rr, ylim=rr, zlim=rr, screen=list(y=20));
Actually, I just realized wireframe is better, just took me a bit longer to figure out the syntax.
xm<-outer(t,u,function(t, u)cos(t)*(R+r*cos(u)))
ym<-outer(t,u,function(t, u)sin(t)*(R+r*cos(u)))
zm<-outer(t,u,function(t, u) r*sin(u))
rr<-c(-10,10)
wireframe(zm~xm+ym, xlim=rr, ylim=rr, zlim=rr, screen=list(y=30))
More details found on the ?cloud help page
With plotly:
library(plotly)
t <- seq(-2*pi, 2*pi, length.out=200)
dat <- data.frame(x=3*cos(t), y=3*sin(t), z=2*t)
plot_ly(dat, x = ~x, y = ~y, z = ~z, type = 'scatter3d', mode = 'lines',
line = list(width = 4))
Related
I am analyzing difference scores with polynomial regression in R. Based on [Edwards and Parry's (1993)][1] recommendations I have been trying to combine a persp() plot with a contour() plot. I would also need to plot the first two principal axes on the contour plot. My attempts so far have only provided me with each individual plot, but I don't know how to combine them. An example for the end-result is :
Edwards & Parry (1993) example difference score visualisation
I manage to get the persp() plot just fine. I have also obtained the contour plot. I can't seem to find any way to combine the two. I have managed to make the plot in plotly using the add_surface() option in the pipeline. My problem with the output is that the surface is smooth, and the contourplot lacks the values in the plot. Basically: persp() and contour() are visualised in a way that is extremely similar to the look I'm aiming for, per the example in the source.
My current attempt (in minimalistic code) is as follows:
surface <- function(e, i){
y <- .2*e + .14*i + .08*e^2 + + .1*e*i + .2*i^2
}
e <- i <- seq(-3, 3, length= 20)
y <- outer(e, i, surface)
persp(e, i, y,
xlab = 'Explicit',
ylab = 'Implicit',
zlab = 'Depression',
theta = 45)
contour(e,i,y)
So basically my question is: how can I make a plot like Edwards and Parry (1993) make, with a similar visual style, in R. It does not have to be base-R, I'm happy with any method. I've been stuck on this problem for a week now.
My attempt in plotly (to compare it to my desired end-result) is:
if(!"plotly" %in% installed.packages){install.packages('plotly')}
library(plotly)
plot_ly(z = ~y) %>% add_surface(x = ~e, y= ~i, z= ~y,
contours = list(
z = list(
show=TRUE,
usecolormap=FALSE,
highlightcolor="#ff0000",
project=list(z=TRUE)
)
)
) %>%
layout(
scene=list(
xaxis = list(title = "Explicit"),
yaxis = list(title = "Implicit"),
zaxis = list(title = "Depression")
)
)
[1]: Edwards, J. R., & Parry, M. E. (1993). On the use of polynomial regression as an alternative to difference scores. Academy of Management Journal, 36(6), 1577–1613. https://doi.org/10.2307/256822
I have found an answer and I will share it here. It seems it cannot be done in base-R. But the RSM-package allows for the addition of contour lines to the base of the plot.
In this answer I will give a minimal example of:
the persp() plot
the contour lines in the base
addition of x=y and x=-y axis
calculation and addition of the first and second principal axis
The only thing I could not solve is that the lines now are drawn over the surface. I don't know how to solve it.
library(rsm)
x <- seq(-3,3,by=0.25)
y <- seq(-3,3,by=0.25)
d <- expand.grid(x=x,y=y)
z <- c(data=NA,1089)
b0 = .140; b1 = -.441; b2 = -.154; b3 = .161 ; b4 =-.106; b5 = .168
k=1
for (i in 1:25) {
for (j in 1:25) {
z[k]=b0+b1*x[i]+b2*y[j]+b3*x[i]*x[i]+b4*x[i]*y[j]+ b5*y[j]*y[j]
k=k+1
} }
data.lm <- lm(z~poly(x,y,degree=2),data=d)
res1 <- persp(data.lm,x~y,
zlim=c(-2,max(z)),
xlabs = c('X','Y'),
zlab = 'Z',
contour=list(z="bottom"),
theta=55,
phi=25)
# draw x=y line (lightly dotted)
xy_pos <- matrix(c(-3,-3,3,3),ncol=2,byrow = T)
lines(trans3d(xy_pos[,2], xy_pos[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 3,
col = 'darkgrey')
# draw x=-y line (lightly dotted)
xy_neg <- matrix(c(-3,3,3,-3),ncol=2,byrow = T)
lines(trans3d(xy_neg[,2], xy_neg[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 3,
col = 'darkgrey')
# Find stationary points:
X0 <- (b2*b4 - 2*b1*b5) / (4*b3*b5 - b4^2)
Y0 <- (b1*b4 - 2*b2*b3) / (4*b3*b5 - b4^2)
# First Principal Axis
p11 = (b5-b3+sqrt((b3-b5)^2+b4^2))/b4
p10 = Y0 - p11*X0
Ypaf1 = p10 + p11*x
# plot first principal axis (full line)
xypaf1 <- matrix(c(Ypaf1[1], -3, Ypaf1[25], 3),ncol=2, byrow=T)
lines(trans3d(xypaf1[,2], xypaf1[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 1,
col = 'black')
# Second Principal Axis
p21 = (b5-b3-sqrt((b3-b5)^2+b4^2))/b4
p20 = Y0 - p21*X0
Ypaf2 = p20 + p21*x
# plot second principal axis (dashed line)
xypaf2 <- matrix(c(Ypaf2[1], -3, Ypaf2[25], 3),ncol=2, byrow=T)
lines(trans3d(xypaf2[,2], xypaf2[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 2,
col = 'black')
So I have this code that produces the exact surface
f = function(x, y){
z = ((x^2)+(3*y^2))*exp(-(x^2)-(y^2))
}
plot3d(f, col = colorRampPalette(c("blue", "white")),
xlab = "X", ylab = "Y", zlab = "Z",
xlim = c(-3, 3), ylim = c(-3, 3),
aspect = c(1, 1, 0.5))
Giving the following plot:
Now I have some code that does a random walk metropolis algorithm to reproduce the above image. I think it works as if I do another plot of these calculated values I get the next image with 500 points. Here is the code
open3d()
plot3d(x0, y0, f(x0, y0), type = "p")
Which gives the following plot:
I know it's hard looking at this still image but being able to rotate the sampling is working.
Now here is my question: How can I use plot3d() so that I can have a surface that connects all these points and gives a more jagged representation of the exact plot? Or how can I have each point in the z axis as a bar from the xy plane? I just want something more 3 dimensional than points and I can't find how to do this.
Thanks for your help
You can do this by triangulating the surface. You don't give us your actual data, but I can create some similar data using
f = function(x, y){
z = ((x^2)+(3*y^2))*exp(-(x^2)-(y^2))
}
x <- runif(500, -3, 3)
y <- runif(500, -3, 3)
z <- f(x, y)
Then the plotting is done using the method in ?persp3d.deldir:
library(deldir)
library(rgl)
col <- colorRampPalette(c("blue", "white"))(20)[1 + round(19*(z - min(z))/diff(range(z)))]
dxyz <- deldir::deldir(x, y, z = z, suppressMsge = TRUE)
persp3d(dxyz, col = col, front = "lines", back = "lines")
This might need some cosmetic fixes, e.g.
aspect3d(2, 2, 1)
After some rotation, this gives me the following plot:
I'm not sure to understand what you want. If my understanding is correct, here is a solution. Define a parametric representation of your surface:
fx <- function(u,v) u
fy <- function(u,v) v
fz <- function(u,v){
((u^2)+(3*v^2))*exp(-(u^2)-(v^2))
}
Let's say you have these points:
x0 <- seq(-3, 3, length.out = 20)
y0 <- seq(-3, 3, length.out = 20)
Then you can use the function parametric3d of the misc3d package, with the option fill=FALSE to get a wireframe:
library(misc3d)
parametric3d(fx, fy, fz, u=x0, v=y0,
color="blue", fill = FALSE)
Is it what you want?
To get some vertical bars, use the function segments3d of rgl:
i <- 8
bar <- rbind(c(x0[i],y0[i],0),c(x0[i],y0[i],f(x0[i],y0[i])))
segments3d(bar, color="red")
Here is a plot with only 50 points using my original code.
When I then apply what was said by Stéphane Laurent I then get this plot which feels too accurate when given the actual points I have
Perhaps you need to explain to me what is actually happening in the function parametric3d
I am looking to use the R plotly library to create a 3d surface plot of x,y,z coordinate data, similar to what is shown at the link below:
https://plot.ly/r/3d-surface-plots/
It appears that the plot_ly function requires the z coordinates to be in a matrix of dimensions x * y, as seen in datasets::volcano, used in the linked example. I'd appreciate some guidance on how to construct this matrix. Here is my sample x,y coordinate data:
## x coordinates
xSeq = seq(0, 1, .01)
## y coordinates
ySeq = seq(0, 1, .01)
## list with x, y coordinates
exmplList = list(x = xSeq, y = ySeq)
The z coordinates would be calculated via a formula from the x,y pairs (example formula used here is x + y). I've played around with something like:
exmplList = within(exmplList, z <- matrix(x + y, nrow = length(xSeq), ncol = length(ySeq)))
But that doesn't accomplish the pair combinations that I am trying to achieve.
Plotly surface needs a matrix so you could simply use this bit directly:
z = matrix(xSeq + ySeq, nrow = length(xSeq), ncol = length(ySeq))
Instead of doing a list. So, by running the following code:
## x coordinates
xSeq = seq(0, 1, 0.01)
## y coordinates
ySeq = seq(0, 1, 0.01)
## list with x, y coordinates
z = matrix(xSeq + ySeq, nrow = length(xSeq), ncol = length(ySeq))
fig = plot_ly(z = ~z) %>% add_surface()
fig
One obtains the following plot:
You might need to click and rotate a bit to see the plane. Hope it helps, cheers.
I want to plot a matrix of z values with x rows and y columns as a surface similar to this graph from MATLAB.
Surface plot:
Code to generate matrix:
# Parameters
shape<-1.849241
scale<-38.87986
x<-seq(from = -241.440, to = 241.440, by = 0.240)# 2013 length
y<-seq(from = -241.440, to = 241.440, by = 0.240)
matrix_fun<-matrix(data = 0, nrow = length(x), ncol = length(y))
# Generate two dimensional travel distance probability density function
for (i in 1:length(x)) {
for (j in 1:length(y)){
dxy<-sqrt(x[i]^2+y[j]^2)
prob<-1/(scale^(shape)*gamma(shape))*dxy^(shape-1)*exp(-(dxy/scale))
matrix_fun[i,j]<-prob
}}
# Rescale 2-d pdf to sum to 1
a<-sum(matrix_fun)
matrix_scale<-matrix_fun/a
I am able to generate surface plots using a couple methods (persp(), persp3d(), surface3d()) but the colors aren't displaying the z values (the probabilities held within the matrix). The z values only seem to display as heights not as differentiated colors as in the MATLAB figure.
Example of graph code and graphs:
library(rgl)
persp3d(x=x, y=y, z=matrix_scale, color=rainbow(25, start=min(matrix_scale), end=max(matrix_scale)))
surface3d(x=x, y=y, z=matrix_scale, color=rainbow(25, start=min(matrix_scale), end=max(matrix_scale)))
persp(x=x, y=y, z=matrix_scale, theta=30, phi=30, col=rainbow(25, start=min(matrix_scale), end=max(matrix_scale)), border=NA)
Image of the last graph
Any other tips to recreate the image in R would be most appreciated (i.e. legend bar, axis tick marks, etc.)
So here's a ggplot solution which seems to come a little bit closer to the MATLAB plot
# Parameters
shape<-1.849241
scale<-38.87986
x<-seq(from = -241.440, to = 241.440, by = 2.40)
y<-seq(from = -241.440, to = 241.440, by = 2.40)
df <- expand.grid(x=x,y=y)
df$dxy <- with(df,sqrt(x^2+y^2))
df$prob <- dgamma(df$dxy,shape=shape,scale=scale)
df$prob <- df$prob/sum(df$prob)
library(ggplot2)
library(colorRamps) # for matlab.like(...)
library(scales) # for labels=scientific
ggplot(df, aes(x,y))+
geom_tile(aes(fill=prob))+
scale_fill_gradientn(colours=matlab.like(10), labels=scientific)
BTW: You can generate your data frame of probabilities much more efficiently using the built-in dgamma(...) function, rather than calculating it yourself.
In line with alexis_laz's comment, here is an example using filled.contour. You might want to increase your by to 2.40 since the finer granularity increases the time it takes to generate the plot by a lot but doesn't improve quality.
filled.contour(x = x, y = y, z = matrix_scale, color = terrain.colors)
# terrain.colors is in the base grDevices package
If you want something closer to your color scheme above, you can fiddle with the rainbow function:
filled.contour(x = x, y = y, z = matrix_scale,
color = (function(n, ...) rep(rev(rainbow(n/2, ...)[1:9]), each = 3)))
Finer granularity:
filled.contour(x = x, y = y, z = matrix_scale, nlevels = 150,
color = (function(n, ...)
rev(rep(rainbow(50, start = 0, end = 0.75, ...), each = 3))[5:150]))
I have a following data:
library(rgl)
x <- c(rep(1,6),
rep(3,6),
rep(6,6),
rep(9,6),
rep(12,6))
y <- c(1.35,1.39,1.48,1.29,1.35,1.32,
NA,1.5,1.44,1.6,1.5,1.41,
NA,NA,1.72,1.56,1.6,1.55,
NA,NA,NA,1.95,1.9,1.75,
NA,NA,NA,NA,2.05,1.95)
z <- rep(1:6,5)
open3d()
plot3d(x,y,z, type = 'n')
lines3d(x,y,z)
Which is plotting lines in 3d as I expect.
But I cannot get it to plot a surface3d.
As I already read some threads I might need to interpolate my data. RGL docs has not cover this subject well. I tried akima but it doesn't accept NA's.
I would like to link lines to create a surface in linear way. I aware of the NA, so I expect that surface will be decreasing in the area for bigger x (more NA's).
Do I need to perform interpolation? If yes, how to do that on my sample data?
If no, how to achieve the surface3d on my sample data?
Thanks
the solution comes to me from this thread:
Making a wireframe plot from an x,y,z data.frame
below code will work for the sample data provided above (just switch x->y,y->z,z->x)
zmat <- matrix(data = z, nrow = 6, ncol = 5, byrow = FALSE)
surface3d(x = 1:6, y = c(1,3,6,9,12), z = zmat, alpha = 0.4, colour = 'blue')