I have a plot with two logarithmic axes. I'd like to add a circle to a certain position of the plot. I tried to use plotrix, but this does not give options for "log-radius".
# data to plot
x = 10^(-1 * c(5:0))
y = x ^-1.5
#install.packages("plotrix", dependencies=T)
# use require() within functions
library("plotrix")
plot (x, y, log="xy", type="o")
draw.circle(x=1e-2, y=1e2, radius=1e1, col=2)
How can I add a circle to my log-log plot?
As krlmlr suggests, the easiest solution is to slightly modify plotrix::draw.circle(). The log-log coordinate system distorts coordinates of a circle given in the linear scale; to counteract that, you just need to exponentiate the calculated coordinates, as I've done in the lines marked with ## <- in the code below:
library("plotrix")
draw.circle.loglog <-
function (x, y, radius, nv = 100, border = NULL, col = NA, lty = 1,
lwd = 1)
{
xylim <- par("usr")
plotdim <- par("pin")
ymult <- (xylim[4] - xylim[3])/(xylim[2] - xylim[1]) * plotdim[1]/plotdim[2]
angle.inc <- 2 * pi/nv
angles <- seq(0, 2 * pi - angle.inc, by = angle.inc)
if (length(col) < length(radius))
col <- rep(col, length.out = length(radius))
for (circle in 1:length(radius)) {
xv <- exp(cos(angles) * log(radius[circle])) * x[circle] ## <-
yv <- exp(sin(angles) * ymult * log(radius[circle])) * y[circle] ## <-
polygon(xv, yv, border = border, col = col[circle], lty = lty,
lwd = lwd)
}
invisible(list(x = xv, y = yv))
}
# Try it out
x = 10^(-1 * c(5:0))
y = x ^-1.5
plot (x, y, log="xy", type="o")
draw.circle.loglog(x = c(1e-2, 1e-3, 1e-4), y = c(1e2, 1e6, 1e2),
radius = c(2,4,8), col = 1:3)
A work around would be to apply log10 explicitly.
plot (log10(x), log10(y), type="o")
draw.circle(x=log10(1e-2), y=log10(1e2), radius=log10(1e1), col=2)
Edit (using symbols):
plot (x, y, log="xy", type="o",xlim=c(1e-5,1), ylim=c(1,1e8))
par(new=T)
symbols(x=1e-2, y=1e2, circles=1e1, xlim=c(1e-5,1), ylim=c(1,1e8),
xaxt='n', yaxt='n', ann=F, log="xy")
The function draw.circle from the plotrix package looks like that on my system:
> draw.circle
function (x, y, radius, nv = 100, border = NULL, col = NA, lty = 1,
lwd = 1)
{
xylim <- par("usr")
plotdim <- par("pin")
ymult <- (xylim[4] - xylim[3])/(xylim[2] - xylim[1]) * plotdim[1]/plotdim[2]
angle.inc <- 2 * pi/nv
angles <- seq(0, 2 * pi - angle.inc, by = angle.inc)
if (length(col) < length(radius))
col <- rep(col, length.out = length(radius))
for (circle in 1:length(radius)) {
xv <- cos(angles) * radius[circle] + x
yv <- sin(angles) * radius[circle] * ymult + y
polygon(xv, yv, border = border, col = col[circle], lty = lty,
lwd = lwd)
}
invisible(list(x = xv, y = yv))
}
<environment: namespace:plotrix>
What happens here is essentially that the circle is approximated by a polygon of 100 vertices (parameter nv). You can do either of the following:
Create your own version of draw.circle that does the necessary coordinate transformation to "undo" the log transform of the axes.
The function invisibly returns the list of coordinates that are used for plotting.
(If you pass a vector as radius, then only the coordinates of the last circle are returned.) You might be able to apply a transform to those coordinates and call polygon on the result. Pass appropriate values for border, col, lty and/or lwd to hide the polygon drawn by the functions itself.
The first version sounds easier to me. Simply replace the + x by a * x, same for y, inside the for loop, and you're done. Equivalently, for the second version, you subtract x and then multiply by x each coordinate, same for y. EDIT: These transformations are slightly wrong, see Josh's answer for the correct ones.
Related
I've been trying to create a combination of radar/polar chart of a given vector of polygon vertices, without packages, but just with base R, which I really struggle with. So far, with some help, I have reached the following point:
a <- a <- abs(rnorm(5, mean = 4, sd = 2))
names(a) <- LETTERS[1:5]
stars(matrix(a,nrow=1),axes=TRUE, scale=FALSE,col.lines="blue",radius=FALSE)
center <- c(x=2.1, y=2.1) #the starchart for some reason chooses this as a center
half <- seq(0, pi, length.out = 51)
angle=45
for (D in a) {
Xs <- D * cos(half); Ys <- D * sin(half)
lines(center["x"] + Xs, center["y"] + Ys, col = "gray", xpd = NA, lty="dashed")
lines(center["x"] + Xs, center["y"] - Ys, col = "gray", xpd = NA, lty="dashed")
}
which gives me something this:
What I would need to take further is:
center this mixed radar/polar chart at (0,0) and mark the center
color the polygon area transparently
add radii starting from the outermost circle and reaching the center through the polygon vertices
put the vector name labels on the ends of the radii on the outermost circle
So, the final result should look something like this:
I have experimented with the polygon(), symbols() functions and par() graphic parametres, but I am really struggling to combine them...My problem is that I don't understand how the stars() function plot coordinates selection relates to my input.
Did not liked the stars functions... so I made a full rondabout with polygon:
polar_chart <- function(values){
k <- length(values)
m <- max(values)
# initialise plot
plot(1, type="n", xlab="", ylab="", xlim=1.2*m*c(-1,1), ylim=1.2*m*c(-1,1))
# radial lines & letters
sapply(k:1, function(x){
text(1.1*m*cos(-(x-1)*2*pi/k + 2*pi/3), 1.1*m*sin(-(x-1)*2*pi/k + 2*pi/3),
LETTERS[x], cex = 0.75)
lines(c(0, m*cos((x-1)*2*pi/k + 2*pi/3)), c(0, m*sin((x-1)*2*pi/k + 2*pi/3)),
col = "grey",lty="dashed")
})
# circles
aux <- seq(2*pi + 0.1, 0, -0.1)
sapply(values, function(x) lines(x*cos(aux), x*sin(aux), col = "grey",lty="dashed"))
# polygon
x <- values*cos(-(1:k-1)*2*pi/k + 2*pi/3)
y <- values*sin(-(1:k-1)*2*pi/k + 2*pi/3)
polygon(c(x, x[1]),c(y, y[1]), col = "red", border = "blue", density = 50)
}
values <- abs(rnorm(5, mean = 4, sd = 2))
polar_chart(values)
And returns a plot like the following:
I have a plot where I draw arrows from points to points. I would like to put this arrow heads not to the end of the line, but to middle. Is there a simple way to do it other than placing extra arrows with half length of the according line?
My code is this:
plot(x, y, xlim=range(x), ylim=range(y), xlab="x", ylab="y", pch=16,
main="Filled Plane")
for(i in 1:20){
arrows(x[i], y[i], x[i+1], y[i+1], length = 0.25, angle = 30, col = i)
}
Make a custom function myArrow() and add one new argument cut to control the proportion of the arrows
myArrow <- function(x0, y0, x1, y1, cut = 1, ...){
x.new <- (1 - cut) * x0 + cut * x1
y.new <- (1 - cut) * y0 + cut * y1
# segments(x0, y0, x1, y1, ...)
arrows(x0, y0, x.new, y.new, ...)
}
Note1 : The computation of x.new and y.new in this custom function uses a simple mathematical concept, i.e. the Section Formula. The value of cut must be between 0 to 1.
Note2 : The use of this function is equivalent to that of the original functionarrows() other than that it has one more new argument cut.
Note3 : If you want complete lines behind the arrows, just remove the hash(#) in the function.
Plot and try different cut value. For example, I use cut = 0.7. (If you want the arrowheads to the middle, use cut = 0.5.)
# Toy Data
x <- seq(1, 5.5, by = 0.5)
y <- rep(c(1, 5), 5)
plot(x, y, pch = 16)
for(i in 1:9){
myArrow(x[i], y[i], x[i+1], y[i+1], cut = 0.7, col = i, lwd = 2)
}
Since you do not provide your x and y, I made up some data. There is no need for the loop. arrows will handle a vector of coordinates. One way is to draw a full-length arrow with no arrowhead and another that just goes halfway but has the arrowhead.
## Some bogus data
set.seed(123)
x = runif(4)
y = runif(4)
## Compute the midpoints
midx = diff(x)/2 + x[-length(x)]
midy = diff(y)/2 + y[-length(y)]
## Draw it
plot(x,y,xlim=range(x), ylim=range(y), xlab="x", ylab="y",
main="Filled Plane",pch=16)
arrows(x[-length(x)], y[-length(y)],x[-1],y[-1],
angle = 0, col = 1:3)
arrows(x[-length(x)], y[-length(y)],midx,midy,
length = 0.25, angle = 30, col = 1:3)
I have been trying to generate random values so that I can construct a circle.
The values of x and y are expected to satisfy the following equation
x^2 + y^2 = 1
Here is the code that I used.
par(type = "s")
x <- runif(1000, min = -1, max = 1)
y <- sqrt(1 - x^2)
z <- NULL
z$x <- x
z$y <- y
z <- as.data.frame(z)
plot.new()
plot(z$x, z$y, type = "p")
plot.window(xlim = c(-10,10), ylim = c(-10,10), asp = 1)
But the graph I get is not quite what I expected it to be.
The graph resembles an upper half of an ellipse rather than a semicircle
Why are there no values for y where y < 0
Please find the plot here.
I am also interested in finding out, how to generate random values for x, y, z, a; where x^2 + y^2 + z^2 + a^2 = 10
Maybe you missed #thelatemail's comment:
png()
plot(z$x, z$y, type = "p", asp=1)
dev.off()
The reason passing asp=1 to plot.window would fail(if it were called first, and this is what you might have tried) is that plot itself calls plot.window again, and in the process reacquires the default values. You can see that in the code of plot.default:
> plot.default
function (x, y = NULL, type = "p", xlim = NULL, ylim = NULL,
log = "", main = NULL, sub = NULL, xlab = NULL, ylab = NULL,
ann = par("ann"), axes = TRUE, frame.plot = axes, panel.first = NULL,
panel.last = NULL, asp = NA, ...)
{
localAxis <- function(..., col, bg, pch, cex, lty, lwd) Axis(...)
localBox <- function(..., col, bg, pch, cex, lty, lwd) box(...)
localWindow <- function(..., col, bg, pch, cex, lty, lwd) plot.window(...)
#.... omitted the rest of the R code.
(Calling plot.window after that plot call should not be expected to have any favorable effect.)
The problem is within this part of your code:
x <- runif(1000, min = -1, max = 1)
y <- sqrt(1 - x^2)enter code here
This problem arises from interpreting two distinct mathematical entities as the same (functions and equations are two different things). A function f takes an input x, and returns a single output f(x). Equations don't have this limitation, so if you are encoding this equation as a function, you will lose half the points in the circle, you will generate all the points in the upper semicircle.
Since the circle equation has two y outputs for any x value you can just generate two pairs of coordinates for each point generated by your uniform distribution like this:
x1 = runif(1000, min = -1, max = 1)
x2 = x1
y1 = sqrt(1 - x1^2)
y2 = (-1)*y1
x = c(x1,x2)
y = c(y1,y2)
plot(x,y, asp=1)
As John Coleman recommended in his comment, i'd prefer using parametric/polar coordinates instead. Generate angles in radians between 0 and 2pi and then calculate the appropriate x and y positions using the generated angle and the radius you want.
radius = 1
theta = runif(1000, min = 0, max = 2*pi)
x = radius * cos(theta)
y = radius * sin(theta)
plot(x,y, asp=1)
For the last part of your question, for each value of a variable, you'd have to work out all the possible tuples that solve the equation, and if z and a are also variables, it may not be possible to represent it solely on a 2-dimensional graph.
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$z <- with(df, f(x, y))
wireframe(z ~ x * y, data = df)
Next the perspective plot:
dm <- outer(x, y, FUN=f)
persp(x, y, dm)
The contour plot:
contour(x, y, dm)
I've been working on a rather complicated chart in R. I have a wireframe with a surface and points distributed in X,Y,Z space all over (e.g. under the surface and over it).
The problem is that the points that plot don't "look" like they are underneath the surface.
I am trying to figure out how best to visualize this chart to make the points look under the surface. Some sample code for the wireframe & cloud come from here: R-List Posting
The code in an example:
library(lattice)
surf <-
expand.grid(x = seq(-pi, pi, length = 50),
y = seq(-pi, pi, length = 50))
surf$z <-
with(surf, {
d <- 3 * sqrt(x^2 + y^2)
exp(-0.02 * d^2) * sin(d)
})
g <- surf
pts <- data.frame(x =rbind(2,2,2), y=rbind(-2,-2,-2), z=rbind(.5,0,-.5))
wireframe(z ~ x * y, g, aspect = c(1, .5),
drape=TRUE,
scales = list(arrows = FALSE),
pts = pts,
panel.3d.wireframe =
function(x, y, z,
xlim, ylim, zlim,
xlim.scaled, ylim.scaled, zlim.scaled,
pts,
...) {
panel.3dwire(x = x, y = y, z = z,
xlim = xlim,
ylim = ylim,
zlim = zlim,
xlim.scaled = xlim.scaled,
ylim.scaled = ylim.scaled,
zlim.scaled = zlim.scaled,
...)
xx <-
xlim.scaled[1] + diff(xlim.scaled) *
(pts$x - xlim[1]) / diff(xlim)
yy <-
ylim.scaled[1] + diff(ylim.scaled) *
(pts$y - ylim[1]) / diff(ylim)
zz <-
zlim.scaled[1] + diff(zlim.scaled) *
(pts$z - zlim[1]) / diff(zlim)
panel.3dscatter(x = xx,
y = yy,
z = zz,
xlim = xlim,
ylim = ylim,
zlim = zlim,
xlim.scaled = xlim.scaled,
ylim.scaled = ylim.scaled,
zlim.scaled = zlim.scaled,
...)
})
Looking at my example, the points in pts are in actually in vertical line where X,Y =(2,-2) and the z goes from .5 to -.5.
However, to my eye the third point doesn't look like it is under the surface, to it looks like it is at coordinates(2,-3,0).
Is this just my eye mis-interpreting this ?
Does anyone have a suggestion on how to make my points look more "3D" ? Perhaps muting the color of the point to make it look "under the surface" by using some sort of transparency on the surface ?
I tried making the colors of the points different (red for over the surface, blue for under the surface) but that does not really help the graph much.
This might get you started:
library(emdbook)
sfun <- function(x,y) {
d <- 3 * sqrt(x^2 + y^2)
exp(-0.02 * d^2) * sin(d)
}
cc <- curve3d(sfun(x,y),xlim=c(-pi,pi),ylim=c(-pi,pi),n=c(50,50),
sys3d="rgl")
colvec <- colorRampPalette(c("pink","white","lightblue"))(100)
with(cc,persp3d(x,y,z,col=colvec[cut(z,100)],alpha=0.5))
pts <- data.frame(x=c(2,2,2), y=c(-2,-2,-2), z=c(.5,0,-.5))
with(pts,spheres3d(x,y,z,col="blue",radius=0.1))
rgl.snapshot("rgltmp1.png")