I've plotted up a series of points using the rgl package in R. I've plotted them in two dimensions for simplicity, but the issue still exists in three dimensions. The code snippet and plot below show a basic line of points plotted in the xy-plane:
library(rgl)
seq <- seq(1, 10, by = 0.1)
df <- data.frame(x = seq, y = seq / 10)
clear3d("all")
bg3d(color = "white")
points3d(x = df$x, y = df$y, z = 0)
axes3d()
rgl.viewpoint(theta = 0, phi = 0)
The points plot as expected. However, if I take these same points and translate them by a significant amount, the graphics device does not seem to be able to handle the points:
library(rgl)
seq <- seq(1, 10, by = 0.1)
df <- data.frame(x = seq, y = seq / 10)
# Translate points
df <- df + 1000000
clear3d("all")
bg3d(color = "white")
points3d(x = df$x, y = df$y, z = 0)
axes3d()
rgl.viewpoint(theta = 0, phi = 0)
Is this a known limitation? Is the problem with OpenGL, or with the package? I'm working with some points and surfaces that have an associated coordinate system, so I'd prefer not to translate my data back to the origin.
#derhass had the right idea. From the rgl manual:
Note that many of these calculations are done on the graphics card using single precision; you will likely see signs of rounding error if your scene requires more than 4 or 5 digit precision to distinguish values in any coordinate.
Related
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
Problem: I am trying to reproduce a round filled 2d contour plot in R using plotly (have tried ggplot2 also but plotly seemed to be easier).
Data: Sample data download link -
https://drive.google.com/file/d/10Mr5yWVReQckPI6TKLY_vzPT8zWiijKl/view?usp=sharing
The data to be plotted for contour is in a column format and typically called z variable, there is x and y data also available for all values of z. A simple dataframe would look like this:
Please ignore the repeat common x and y as I have truncated decimals. The data has about 25000 rows.
Approach: I first use akima package to interpolate z variable values for given x and y to map z in 2d. This makes the z column data fit in a xy grid for 2d plotting and show contours.
Expected outcome:
Code used:
dens <- akima::interp(x = dt$`Xvalue(mm)`,
y = dt$`Yvalue(mm)`,
z = dt$Values,
duplicate = "mean",
xo=seq(min(dt$`Xvalue(mm)`), max(dt$`Xvalue(mm)`), length = 10),
yo=seq(min(dt$`Yvalue(mm)`), max(dt$`Yvalue(mm)`), length = 10))
plot_ly(x = dens$x,
y = dens$y,
z = dens$z,
colors = c("blue","grey","red"),
type = "contour")
Actual outcome:
Help Needed:
To refine edges of the actual outcome plot to something of a close match to the expected outcome image.
Many thanks in advance for your comments and help.
I found that I could increase the grid output z matrix from akima::interp() from default 40x40 to custom using nx and ny input in function.
And then in plot_ly() add contours = list(coloring = 'fill', showlines = FALSE) to hide contour lines to get output close to my expected outcome.
So working code is like this:
dens <- akima::interp(x = dt$`Xvalue(mm)`,
y = dt$`Yvalue(mm)`,
z = dt$Values,
nx = 50,
ny = 50,
duplicate = "mean",
xo=seq(min(dt$`Xvalue(mm)`), max(dt$`Xvalue(mm)`), length = 50),
yo=seq(min(dt$`Yvalue(mm)`), max(dt$`Yvalue(mm)`), length = 50))
plot_ly(x = dens$x,
y = dens$y,
z = dens$z,
colors = c("blue","grey","red"),
type = "contour",
contours = list(coloring = 'fill', showlines = FALSE))
Plotly contour plot reference was very helpful in this case:
https://plot.ly/r/reference/#contour
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')