I'm having trouble displaying multiple 3d plots with rgl's mfrow3d and misc3d's contour3d.
In particular, plotting a new subplot will result in all previous subplots being deleted. Here is a simple example:
library(rgl)
library(misc3d)
# setup rgl subplots
mfrow3d(1,2)
# step into first subplot
next3d()
# Draw a ball
f <- function(x, y, z)x^2+y^2+z^2
x <- seq(-2,2,len=20)
contour3d(f,4,x,x,x)
# advance to next subplot
next3d()
# Ball with one corner removed.
contour3d(f,4,x,x,x,
mask = function(x,y,z) x > 0 | y > 0 | z > 0,
screen = list(x = 290, y = -20),
color = "red", color2 = "white")
# the first subplot is removed
In the first call to contour3d, the first ball draws fine on the left. However, after the second call to contour3d, the second plot is drawn on the right, but the first plot is deleted.
What am I missing here? My hunch is that I'm missing an argument to contour3d, as mfrow3d works fine with other *3d plotting functions, but not with with contour3d.
Like base graphics, rgl graphics come in two types: low level (things like drawing points, lines, etc.) and high level (like plot3d or persp3d). By default high level plots first advance to the next frame (by calling next3d()), while low level plots add to the current one.
The misc3d::contour3d function draws everything using low-level commands, but it assumes it has control of the full window, so instead of calling next3d() to advance to the next frame, it calls clear3d() which clears the whole window.
To work around this, you can call next3d() yourself (only after the first plot, you don't need it before the first one), and then tell contour3d() to add to the scene. That is, change your code like this:
library(rgl)
library(misc3d)
# setup rgl subplots
mfrow3d(1,2)
# Draw a ball
f <- function(x, y, z)x^2+y^2+z^2
x <- seq(-2,2,len=20)
contour3d(f,4,x,x,x)
# advance to next subplot
next3d()
# Ball with one corner removed.
contour3d(f,4,x,x,x,
mask = function(x,y,z) x > 0 | y > 0 | z > 0,
screen = list(x = 290, y = -20),
color = "red", color2 = "white", add = TRUE)
Related
Two questions in one: Given a line plotted in Julia, how can I
delete it from the plot and legend (without clearing the whole plot)
change its properties (such as color, thickness, opacity)
As a concrete example in the code below, how can I 1. delete previous regression lines OR 2. change their opacity to 0.1?
using Plots; gr()
f = x->.3x+.2
g = x->f(x)+.2*randn()
x = rand(2)
y = g.(x)
plt = scatter(x,y,c=:orange)
plot!(0:.1:1, f, ylim=(0,1), c=:green, alpha=.3, linewidth=10)
anim = Animation()
for i=1:200
r = rand()
x_new, y_new = r, g(r)
push!(plt, x_new, y_new)
push!(x, x_new)
push!(y, y_new)
A = hcat(fill(1., size(x)), x)
coefs = A\y
plot!(0:.1:1, x->coefs[2]*x+coefs[1], c=:blue) # plot new regression line
# 1. delete previous line
# 2. set alpha of previous line to .1
frame(anim)
end
gif(anim, "regression.gif", fps=5)
I tried combinations of delete, pop! and remove but without success.
A related question in Python can be found here: How to remove lines in a Matplotlib plot
Here is a fun and illustrative example of how you can use pop!() to undo plotting in Julia using Makie. Note that you will see this goes back in the reverse order that everything was plotted (think, like adding and removing from a stack), so deleteat!(scene.plots, ind) will still be necessary to remove a plot at a specific index.
using Makie
x = range(0, stop = 2pi, length = 80)
f1(x) = sin.(x)
f2(x) = exp.(-x) .* cos.(2pi*x)
y1 = f1(x)
y2 = f2(x)
scene = lines(x, y1, color = :blue)
scatter!(scene, x, y1, color = :red, markersize = 0.1)
lines!(scene, x, y2, color = :black)
scatter!(scene, x, y2, color = :green, marker = :utriangle, markersize = 0.1)
display(scene)
sleep(10)
pop!(scene.plots)
display(scene)
sleep(10)
pop!(scene.plots)
display(scene)
You can see the images above that show how the plot progressively gets undone using pop(). The key idea with respect to sleep() is that if we were not using it (and you can test this on your own by running the code with it removed), the fist and only image shown on the screen will be the final image above because of the render time.
You can see if you run this code that the window renders and then sleeps for 10 seconds (in order to give it time to render) and then uses pop!() to step back through the plot.
Docs for sleep()
I have to say that I don't know what the formal way is to accomplish them.
There is a cheating method.
plt.series_list stores all the plots (line, scatter...).
If you have 200 lines in the plot, then length(plt.series_list) will be 200.
plt.series_list[1].plotattributes returns a dictionary containing attributes for the first line(or scatter plot, depends on the order).
One of the attributes is :linealpha, and we can use it to modify the transparency of a line or let it disappear.
# your code ...
plot!(0:.1:1, x->coefs[2]*x+coefs[1], c=:blue) # plot new regression line
# modify the alpha value of the previous line
if i > 1
plt.series_list[end-1][:linealpha] = 0.1
end
# make the previous line invisible
if i > 2
plt.series_list[end-2][:linealpha] = 0.0
end
frame(anim)
# your code ...
You cannot do that with the Plots package. Even the "cheating" method in the answer by Pei Huang will end up with the whole frame getting redrawn.
You can do this with Makie, though - in fact the ability to interactively change plots was one of the reasons for creating that package (point 1 here http://makie.juliaplots.org/dev/why-makie.html)
Not sure about the other popular plotting packages for Julia.
I am quite new to programming/R and I'm having a very unusual problem. I've made a scatterplot and I would like to simply put the x y axis at 0 on the plot. However, when I use abline they are slightly off. I managed to get them to 0 using trial and error, but trying to plot other lines becomes impossible.
library('car')
scatterplot(cost~qaly, reg.line=FALSE, smooth=FALSE, spread=FALSE,
boxplots='xy', span=0.5, xlab="QALY", ylab="COST", main="Bootstrap",
cex=0.5, data=scat2, xlim=c(-.05,.05), grid=FALSE)
abline(v = 0, h = 0)
This gives lines which are slightly to the left and below 0.
here is an image of what this returns:
(I can't post an image since I'm new apparently)
I found that these values put the lines on 0:
abline(v=0.003)
abline(h=3000)
Thanks in advance for the help!
Using #Laterow's example, reproduce the issue
require(car)
set.seed(10)
x <- rnorm(1000); y <- rnorm(1000)
scatterplot(y ~ x)
abline(v=0, h=0)
scatterplot seems to be resetting the par settings on exit. You can sort of check this with locator(1) around some point, eg, for {-3,-3} I get
# $x
# [1] -2.469414
#
# $y
# [1] -2.223922
Option 1
As #joran points out, reset.par = FALSE is the easiest way
scatterplot(y ~ x, reset.par = FALSE)
abline(v=0, h=0)
Option 2
In ?scatterplot, it says that ... is passed to plot meaning you can use plot's very useful panel.first and panel.last arguments (among others).
scatterplot(y ~ x, panel.first = {grid(); abline(v = 0)}, grid = FALSE)
Note that if you were to do the basic
scatterplot(y ~ x, panel.first = abline(v = 0))
you would be unable to see the line because the default scatterplot grid covers it up, so you can turn that off, plot a grid first then do the abline.
You could also do the abline in panel.last, but this would be on top of your points, so maybe not as desirable.
Hopefully a straightforward question but I made a simple figure in R using filled.contour(). It looks fine, and what it should like given the data. However, I want to add a reference line along a contour for 0 (level = 0), and the plotted line doesn't match the colors on the filled.contour figure. The line is close, but not matching with the figure (and eventually crossing over another contour from the filled.contour plot). Any ideas why this is happening?
aa <- c(0.05843150, 0.11300040, 0.15280030, 0.183524400, 0.20772430, 0.228121000)
bb <- c(0.01561055, 0.06520635, 0.10196237, 0.130127650, 0.15314544, 0.172292410)
cc <- c(-0.02166599, 0.02306650, 0.05619421, 0.082193680, 0.10334837, 0.121156780)
dd <- c(-0.05356592, -0.01432910, 0.01546647, 0.039156660, 0.05858709, 0.074953650)
ee <- c(-0.08071987, -0.04654243, -0.02011676, 0.000977798, 0.01855881, 0.033651089)
ff <- c(-0.10343798, -0.07416114, -0.05111547, -0.032481132, -0.01683215, -0.003636035)
gg <- c(-0.12237798, -0.09753544, -0.07785126, -0.061607548, -0.04788856, -0.036169540)
hh <-rbind(aa,bb,cc,dd,ee,ff,gg)
z <- as.matrix(hh)
y <- seq(0.5,1.75,0.25)
x <- seq(1,2.5,0.25)
filled.contour(x,y,z,
key.title = title(main=expression("log"(lambda))),
color.palette = topo.colors) #This works
contour(x,y,z, level=0,add=T,lwd=3) #This line doesn't match plot
This is completely answered in the ?filled.contour help page. In the Notes section it states
The output produced by filled.contour is actually a combination of two plots; one is the filled contour and one is the legend. Two separate coordinate systems are set up for these two plots, but they are only used internally – once the function has returned these coordinate systems are lost. If you want to annotate the main contour plot, for example to add points, you can specify graphics commands in the plot.axes argument. See the examples.
And the examples given in that help page show how to annotate on top of the main plot. In this particular case, the correct way would be
filled.contour(x,y,z,
key.title = title(main=expression("log"(lambda))),
color.palette = topo.colors,
plot.axes = {
axis(1)
axis(2)
contour(x,y,z, level=0,add=T,lwd=3)
}
)
which produces
I am trying to plot a Rgraphviz object with two edge labels. Unfortunately the labels fall outside the plot. Here is my example:
require('Rgraphviz')
set.seed(123)
g1 <- randomGraph(letters[1:10], 1:4, 0.4)
eAttrs <- list()
eAttrs$label <- c("a~g" = "I have a very long label 1", "a~i" = "and a long label 2")
plot(g1, edgeAttrs = eAttrs)
Here is my plot:
I tried several things with no success:
1.
Set a larger bounding box
z <- agopen(g1, "foo")
z#boundBox#upRight#x <- z#boundBox#upRight#x + 300
z#boundBox#upRight#y <- z#boundBox#upRight#y + 300
plot(z, edgeAttrs = eAttrs)
2.
Decrease the label fontsize (not really what I want in my application, anyways)
eAttrs$labelfontsize=c("a~g"="3")
plot(g1, edgeAttrs = eAttrs)
3.
Change par attributes:
par(oma=c(10,10,10,10))
plot(g1, edgeAttrs = eAttrs)
4.
Change node, edge and general attributes from ?Rgraphviz::GraphvizAttributes
attrs <- list(graph=list(size=c(1, 1)))
attrs$edge$fontsize<-8
plot(g1, edgeAttrs = eAttrs, attrs=attrs)
None of my attempts seem to work. Does anyone have an idea?
The problem
Calling plot() on a graph object dispatches an S4 method (shown by getMethod("plot", "graph")), which in turn calls the function shown by typing getMethod("plot", "Ragraph"). That function contains the following rather unfortunate lines which, regardless of any related parameter settings you've made, will override them to reset the left and right margins to 0. Frustrating!
oldpars <- par(mai = c(sheight, 0, mheight, 0))
on.exit(par(oldpars), add = TRUE)
A workaround
One workaround is to construct a three panel layout in which the left and right panels are just there to provide a bit of buffering space. Turn off clipping, plot your graph object in the middle panel, and it then seems to work:
layout(matrix(1:3, nrow=1), widths=c(1,5,1))
par(xpd=NA) ## turn off all clipping
plot.new() ## blank plot in Panel 1
plot(g1, edgeAttrs = eAttrs) ## graph in Panel 2
plot.new() ## blank plot in Panel 3
I found another solution: In my original question I changed the size of the bounding box in a laid out graph I got with agopen. Plotting the laid out graph showed no edge labels at all. I found that it is possible to pass the edge attributes from the graph object to agopen and then change the bounding box:
require('Rgraphviz')
set.seed(123)
g1 <- randomGraph(letters[1:10], 1:4, 0.4)
eAttrs <- list()
eAttrs$label <- c("a~g" = "I have a very long label 1", "a~i" = "and a long label 2")
z <- agopen(g1, "foo", edgeAttr=eAttrs)
z#boundBox#botLeft#x <- z#boundBox#botLeft#x - 400 ##left
z#boundBox#upRight#x <- z#boundBox#upRight#x + 200 ##right
plot(z)
The plot:
Im drawing a knn-classification plot in R using plot to plot the samples and contour to plot the lines that classify the plane.
Here is my code:
k<-1
datax<-rbind(matrix(rnorm(30,-1,5.25),15,2),matrix(rnorm(36,1,5.25),18,2))
datay<-rbind(matrix(1,15,1),matrix(0,18,1))
plot(datax[,1], datax[,2],pch = datay+1,axes=FALSE,ann=FALSE)
box()
n <- 1000
xp <- seq(length=n, from = min(datax[,1]), to = max(datax[,1]))
yp <- seq(length=n,from = min(datax[,2]) ,to = max(datax[,2]))
gr <- expand.grid(xp, yp)
library(class)
z <- as.numeric(knn(datax, gr, datay,k))-1
zM <- matrix(z, n, n, byrow = FALSE)
contour(xp, yp, zM, xlab="x",ylab="",nlevels = 1 ,lwd=2, add=TRUE, drawlabels =FALSE)
My question is: How can i color the enclosed areas in the plot? I tried filled.contour but there is no add parameter. I simply want the area where the classifier is = 0 white and where it classifies = 1 in blue. How should i do this?
thanks
Instead of contour, you can use contourLines to keep the coordinates of the edges of the contour lines and plot them with polygon.
plot(datax[,1], datax[,2],axes=FALSE,ann=FALSE, type="n")
box()
cL <- contourLines(xp, yp, zM,nlevels = 1)
lapply(cL,function(x)polygon(x$x,x$y,col="red"))
points(datax[,1], datax[,2],pch = datay+1)
However it is not perfect with contour lines that reach the edges of the plot (see the left lower corner of the second plot), so it will need some hand-made tuning:
Edit: In the case of nested contour lines, I don't think there is an easy way to deal with it but here is one way:
library(splancs)
ord <- sapply(lapply(cL,function(x)datay[inout(datax,cbind(x$x,x$y))]),
median) #Check what values are present in the polygon and
#take the most common one
plot(datax[,1], datax[,2],axes=FALSE,ann=FALSE, type="n")
box()
lapply(cL[ord==1],function(x)polygon(x$x,x$y,col="blue"))
lapply(cL[ord==0],function(x)polygon(x$x,x$y,col="white"))
points(datax[,1], datax[,2],pch = datay+1)
2nd Edit: There is of course also the possibility of using function image in your case:
image(xp, yp, zM, col=c("transparent","blue"))
points(datax[,1], datax[,2],pch = datay+1)