I need to create something like a spider chart in R without using any libraries. That’s my code for now. It creates a figure with points number equal to the length of vector ‘a’. However, I’d like each point to be at the distance from the coordinates center equal to a respective number in a vector, for example one point at a distance 1, another at 2, so on. Is it possible to do so?
a <- 1:6
angle <- seq(0, 2*pi, (2*pi)/length(a))
x <- cos(angle)
y <- sin(angle)
plot(x, y,
type = "l")
See ?stars:
a <- 1:6
stars(matrix(a, nrow=1), scale=FALSE)
For future reference, using R's built-in help search would have found this with ??spider
Related
I want to create 50 concentric circles. I did it with python but now I want to do this in R. I have tried the symbols function but with no result. I want my circles to start from x,y coordinates and the radius of each circle to be 3times bigger than the previous.
step=1
for(i in seq(1,50,1)){
symbols (x, y, circles=50, col="grey")
step=step+3
}
From this I get one circle as a result.
I am new in programming so it is probably very simple. Should I use a specific package?
The beauty of R is that many things can be vectorized, including the imput to the 'symbols' function. Here's an example for you:
#vector of radii
#written in a way that's easily changable
n_circles <- 50
my_circles <- seq(1,by=1,length.out = n_circles)
#generate x and y
x <- rep(1,n_circles)
y <- rep(1, n_circles)
#plot
symbols(x,y,1:n_circles)
I'm trying to make a 3D scatterplot with boudaries or zones based on combinations of 3 variables that return certain values. The variables each range between 0:1, and combine to make an index that ranges from -1:1 as follows:
f(x,y,z) = (x*y)-z
I'd like to create a visual representation that will highlight all combinations of variables that return a certain index value. As an example, I can easily show those variables where index > 0 using scatterplot3d (rgl would also work):
# Create imaginary dataset of 50 observations for each variable
x<-runif(50,0,1)
y<-runif(50,0,1)
z<-runif(50,0,1)
# Create subset where f(x,y,z) > 0
x1<-y1<-z1<-1
for (i in 1:length(x)){ if ((x[i]*y[i])-z[i] > 0) {
x1<-rbind(x1, x[i])
y1<-rbind(y1, y[i])
z1<-rbind(z1, z[i])}
}
s3d<-scatterplot3d(x,y,z) # Plot entire dataset
s3d$points3d(x1,y1,z1,pch=19, col="red") # Highlight subset where f(x,y,z) > 0
This gives me the following graph:
It seems fairly intuitive that there should be an easy way to plot either the surface (extending from top/right/back to bottom/left/front) separating the subset from the rest of the data, or else a volume/3D area within which these plots lie. E.g. my first instinct was to use something like surface3d, persp3d or planes3d. However, all attempts so far have only yielded error messages. Most solutions seem to use some form of z<-lm(y~x) but I obviously need something like q<-func((x*y)-z) for all values of x, y and z that yield q > 0.
I know I could calculate extreme points and use them as vertices for a 3D polygon, but that seems too "manual". It feels like I'm overlooking something fairly simple and obvious. I've looked at many similar questions on Stack but can't seem to find one that fits my particular problem. If I've missed any and this question has been answered already, please do point me in the right direction!
Here is a suggestion for an interactive 3D plot that is based on an example from the "R Graphics Cookbook" by Winston Chang.
set.seed(4321)
library(rgl)
interleave <- function(v1,v2) as.vector(rbind(v1,v2))
x <- runif(50)
y <- runif(50)
z <- runif(50)
plot3d(x, y, z, type="s", size=0.6, col=(2+(x*y<z)))
x0 <- y0 <- seq(0, 1, 0.1)
surface3d(x0, y0, outer(x0, y0), alpha=0.4) #plot the surface f(x,y)=x*y
x1 <- x[x * y > z] #select subset that is below the separating surface
y1 <- y[x * y > z]
z1 <- z[x * y > z]
segments3d(interleave(x1, x1), #highlight the distance of the points below the surface
interleave(y1, y1),
interleave(x1 * y1, z1), col="red", alpha=0.4)
If you don't like the red lines and only want the surface and the colored points, this will be sufficient:
plot3d(x,y,z,type="s",size=0.6,col=(2+(x*y<z)))
x0 <- y0 <- seq(0,1,0.1)
surface3d(x0,y0,outer(x0,y0),alpha=0.4)
Does this representation provide the information that you wanted to highlight?
The first thought was to see if the existing functions within scatterplot3d could handle the problem but I think not:
my.lm <- lm(z ~ I(x) * I(y)+0)
s3d$plane3d(my.lm, lty.box = "solid", col="red")
pkg:scatterplot3d doesn't really have a surface3d function so you will need to choose a package that provides that capability; say 'rgl', 'lattice', or 'plot3d'. Any of them should provide the needed facilities.
I have a regular grid in cylindrical co-ordinates (R, z, theta). At each grid point I have a value for the density at that point. I am looking for advice on how to interpolate the density values to get the value at a certain point within a 3D grid cell, using the values from the 8 grid points around it. Is there an R package that will make this simple?
I have tried to search for answers, but most seem to be related to the 'grid' plotting package.
e.g., some sample data:
R <- c(1,2,3)
z <- c(1,2,3)
th <- c(1,2,3)
dens <- array(rep(1, 3*3*3), dim=c(3, 3, 3))
dens[1,2,1] <- 2
How do I get the value of dens at (R,z,th) = (1.5,1.5,1.5) ?
I'm looking into plotting functions and I've run into persp and curve but I'm not able to follow them to plot a 2D function.
They are for surface plots, yes?
If I had a function like x^2 + y^2 [x,y] in [-3,3] how do I go about it?
Any links will be much appreciated and critique on existing packages (if multiple) ? gold.
Thanks.
To use persp, you need to supply values of x, values of y, and values of z for each combination of x and y. The easiest way to do this is to define x and y and then use outer to create a matrix that crosses x and y. You need to specify the way the two variables should be combined as the third argument to outer, in this case the function +:
x <- seq(-3,3,length.out=100)
y <- seq(-3,3,length.out=100)
z <- outer(x^2,y^2,`+`)
persp(x,y,z, col='blue')
You may also be interested in rotating the results. Here are some examples using the theta parameter:
par(mar=c(1,1,1,1))
layout(matrix(1:4, nrow=2))
s=lapply(c(0,30,60,90), function(t) persp(x,y,z, col='blue', theta=t))
EDIT: I understand from your comment you would like a 2D representation of this surface. The easiest way to get that in base R is with image of your z matrix:
image(z)
I'm wondering if it is possible to caclulate the area within a contour in R.
For example, the area of the contour that results from:
sw<-loess(m~l+d)
mypredict<-predict(sw, fitdata) # Where fitdata is a data.frame of an x and y matrix
contour(x=seq(from=-2, to=2, length=30), y=seq(from=0, to=5, length=30), z=mypredict)
Sorry, I know this code might be convoluted. If it's too tough to read. Any example where you can show me how to calculate the area of a simply generated contour would be helpful.
Thanks for any help.
I'm going to assume you are working with an object returned by contourLines. (An unnamed list with x and y components at each level.) I was expecting to find this in an easy to access location but instead found a pdf file that provided an algorithm which I vaguely remember seeing http://finzi.psych.upenn.edu/R/library/PBSmapping/doc/PBSmapping-UG.pdf (See pdf page 19, labeled "-11-") (Added note: The Wikipedia article on "polygon" cites this discussion of the Surveyors' Formula: http://www.maa.org/pubs/Calc_articles/ma063.pdf , which justifies my use of abs().)
Building an example:
x <- 10*1:nrow(volcano)
y <- 10*1:ncol(volcano)
contour(x, y, volcano);
clines <- contourLines(x, y, volcano)
x <- clines[[9]][["x"]]
y <- clines[[9]][["y"]]
level <- clines[[9]][["level"]]
level
#[1] 130
The area at level == 130 (chosen because there are not two 130 levels and it doesn't meet any of the plot boundaries) is then:
A = 0.5* abs( sum( x[1:(length(x)-1)]*y[2:length(x)] - y[1:(length(x)-1)]*x[2:length(x)] ) )
A
#[1] 233542.1
Thanks to #DWin for reproducible example, and to the authors of sos (my favourite R package!) and splancs ...
library(sos)
findFn("area polygon compute")
library(splancs)
with(clines[[9]],areapl(cbind(x,y)))
Gets the same answer as #DWin, which is comforting. (Presumably it's the same algorithm, but implemented within a Fortran routine in the splancs package ...)