I'm hoping to keep in the image below the ticks on the vertical z axis, but remove ticks and numbers from the x and y axes. I would like to be able to label my x and y axes with a label for each condition in my matrix, but have not figured out how to do this with text3D. For some reason (because I'm on a mac?) I can't download axes3D, which is one potential solution I've seen in other responses.
Here is my code:
x = c(0,1)
y = c(0,1)
zval = c(104.1861, 108.529, 110.3675, 110.4112)
z = matrix (zval, nrow=2, ncol=2, byrow=TRUE)
hist3D(x,y,z, zlim=c(101,111), colvar = NULL, d=2, col = "lightblue", NAcol = "white", breaks = NULL, colkey = NULL, theta=-60, phi=20, nticks=10, axes=TRUE, ticktype="detailed", space=0.5, lighting=TRUE, light="diffuse", shade=.5, ltheta = 50, bty = "g")
My output

Ultimately, I'd like something more along the lines of this:
I'm very new to R.
stackoverflow.com/questions/26794236/ggplot2-3d-bar-plot
^ this seems like it might be what I need, but I couldn't replicate the code without an error. When I tried to run this piece I got an error because my x and z (in this case) axes aren't numerical:
cloud(y~x+z, d, panel.3d.cloud=panel.3dbars, col.facet='grey', xbase=0.4, ybase=0.4, scales=list(arrows=FALSE, col=1), par.settings = list(axis.line = list(col = "transparent")))
Maybe this might be helpful (with the caveat that 3D plots can sometimes make interpretation more challenging).
First, I recreated a data frame d based on something similar to what you started with:
x = c(0, 0, 1, 1)
y = c(0, 1, 0, 1)
z = c(104.1861, 108.529, 110.3675, 110.4112)
d <- data.frame(
x = factor(as.logical(x)),
y = factor(as.logical(y)),
z = z
)
Note that for x and y I converted the 0 and 1 to FALSE and TRUE with as.logical, then made them factors.
Then for the plot:
library(latticeExtra)
cloud(z ~ x + y, data = d, panel.3d.cloud=panel.3dbars, col.facet='grey',
xbase=0.4, ybase=0.4, scales=list(arrows=FALSE, col=1),
par.settings = list(axis.line = list(col = "transparent")))
You will want the formula as z ~ x + y where z is a numeric response.
Edit: If you wish to customize the axis labels, you can set the factor labels as follows (for example):
d <- data.frame(
x = factor(as.logical(x), labels = c("Hi", "Lo")),
y = factor(as.logical(y), labels = c("Label1", "Label2")),
z = z
)
Plot
Related
I have a problem where I have data with (x,y) coordinates that I want to plot in the x-y plane. Furthermore, I have some box constraints such that -7 < x < 7 and -5 < y < 5 need to be drawn and checked. All points that fall outside of this box constraint I would like to color red. To do this I have the following code in R:
library(rgl)
x <- 7
y <- 5
data.x <- rnorm(10,0,5)
data.y <- rnorm(10,0,5)
plot(data.x, data.y, xlim = c(min(-x,data.x),max(x,data.x)),
ylim = c(min(-y,data.y),max(y,data.y)), pch = 19)
rect(-x, -y, x, y, col = "lightgrey")
idx <- abs(data.x) > x | abs(data.y) > y
points(data.x[idx], data.y[idx], col = "red", pch = 19)
points(data.x[!idx], data.y[!idx], col = "deepskyblue", pch = 19)
Now, where I am stuck, is on how to plot this type of data picture when I have a third group of data and a third constraint. I.e.,
### How to generalize when I have a third axis and constraint, i.e., a 3D cube
z <- 4
data.z <- rnorm(10, 0, 5)
So essentially I want to plot a box constraint as a cube in the x-y-z plane, and to color the points that fall outside the box constraint red again.
Also, I should say I understand there are functions for plottig 3d scatter plots in R, however, what I am struggling with is how to draw the 3D cube that defines the constraints.
The difficulty with a 3D plot such as this is being able to interpret the "depth" of the points in the image. An animated 3D image might be helpful here:
library(plot3D)
x <- 7
y <- 5
z <- 6
set.seed(123)
data.x <- rnorm(10, 0, 5)
data.y <- rnorm(10, 0, 5)
data.z <- rnorm(10, 0, 5)
in_out <- abs(data.x) > x | abs(data.y) > y | abs(data.z) > z
for(i in seq(0, 358, 2)) {
png(paste0("box", sprintf("%03d", i), ".png"))
box3D(-x, -y, -z, x, y, z, col = NA, border = "gray50", theta = i, phi = 15,
xlim = c(-10, 10), ylim = c(-10, 10), zlim = c(-10, 10),
axes = TRUE, ticktype = "detailed")
points3D(data.x, data.y, data.z, colvar = in_out, pch = 16, cex = 3,
add = TRUE, colkey = FALSE, col = c("lightblue", "red"))
dev.off()
}
library(magick)
list.files(pattern = 'box\\d+\\.png', full.names = TRUE) %>%
image_read() %>%
image_join() %>%
image_animate(fps=50) %>%
image_write("box.gif")
box.gif
Does any one know how do you apply this
set.seed(101)
x <- 1:10
y <- rnorm(10)
## second data set on a very different scale
z <- runif(10, min=1000, max=10000)
par(mar = c(5, 4, 4, 4) + 0.3) # Leave space for z axis
plot(x, y) # first plot
par(new = TRUE)
plot(x, z, type = "l", axes = FALSE, bty = "n", xlab = "", ylab = "")
axis(side=4, at = pretty(range(z)))
mtext("z", side=4, line=3)
but using ggplot.
In ggplot you can only create sec.axis() or dup.axis() using a transformation of y axis. What about a whole new independent y axis which will be applied only for z variable and the simple y axis to be applied for the y variable.
ggplot2::sec_axis provides only one mechanism for providing a second axis, and it took a lot of convincing to get that into the codebase. You are responsible for coming up with the transformation. This transform must be linear in some way, so if either axis needs to be non-linear (e.g., exponential, logarithmic, etc), then your expert math skills will be put to the test.
If you can use scales, then this process becomes trivial:
dat <- data.frame(x, y, z)
ggplot(dat, aes(x, y)) +
geom_point() +
geom_line(
aes(y = zmod),
data = ~ transform(., zmod = scales::rescale(z, range(y), range(z)))
) +
scale_y_continuous(
sec.axis = sec_axis(~ scales::rescale(., range(dat$z), range(dat$y)),
breaks = c(2000,4000,6000,8000))
)
Unless I've missed something (I just checked ggplot2-3.3.5's NEWS.md file), this has not changed.
Recently I've been playing with an R package named latticeExtra and created the following chart
with this code
library(latticeExtra)
heisenberg <- read.csv(file="ABSOLUTE_FILE_LOCATION")
x <- heisenberg[[1]]
y <- heisenberg[[2]]
z <- heisenberg[[3]]
cloud(z~x+y, heisenberg, panel.3d.cloud=panel.3dbars, col.facet='green',
xbase=0.4, ybase=0.4,scales=list(arrows=FALSE, col=1),
par.settings = list(axis.line = list(col = "transparent")))
I'm still not happy just by looking at the cube containing the chart from this angle / perspective.
Do you know how I can change the view angle to like the opposite side?
In order to change the perspective, we use
screen = list(z = 0, x = 0, y = 0)
and change the values of z, x and y in accordance to what we want.
This code where z = 90, x = -45, y=0
library(latticeExtra)
heisenberg <- read.csv(file="C:/Users/tiago/Desktop/VIS/DATA/4/sample_1000.csv")
x <- heisenberg[[1]]
y <- heisenberg[[2]]
z <- heisenberg[[3]]
cloud(z~x+y, heisenberg, panel.3d.cloud=panel.3dbars, col.facet='green',
xbase=0.4, ybase=0.4,scales=list(arrows=FALSE, col=1),
screen = list(z = 90, x = -45, y=0),
par.settings = list(axis.line = list(col = "transparent")))
Gives this chart
I'm trying to create a very simple 3D plot using the rgl package: I have a function that just maps x values into y values. For a given z (in my example: z = 1), I can plot this function in a 3D plot:
library(rgl)
mycurve <- function(x) { return (1/x)}
myx <- seq(1, 10, by = 0.1)
plot3d(x = NA, xlim = c(0, 10), ylim = c(0, 10), zlim = c(0, 5),
xlab = "x", ylab = "y", zlab = "height")
lines3d(x = myx, y = mycurve(myx), z = 1)
However, even after hours of trying to understand the documentation of ?persp3d and ?surface3d, I still have no idea how to add a surface to my plot that "connects" my line to the x-y plane – like this:
(To generate this image, I cheated by plotting many lines: for (i in seq(0, 1, by = 0.01)) { lines3d(x = myx, y = mycurve(myx), z = i) }.)
I suppose that I need to supply the correct values to surface3d somehow. From ?surface3d:
The surface is defined by the matrix of height values in z, with rows corresponding to the values in x and columns corresponding to the values in y.
Given that my space curve is "vertical", each value of x corresponds to only 1 value of y. Still, I need to specify two z values for each xy pair, which is why I do not know how to proceed.
How can I plot a space curve as shown in the second image?
In persp3d, all 3 arguments can be matrices, so you can plot arbitrary surfaces. For your needs, this works:
mycurve <- function(x) { return (1/x)}
myx <- seq(1, 10, by = 0.1)
xmat <- matrix(NA, 2, length(myx))
ymat <- matrix(NA, 2, length(myx))
zmat <- matrix(NA, 2, length(myx))
for (i in 0:1) {
xmat[i+1,] <- myx
ymat[i+1,] <- mycurve(myx)
zmat[i+1,] <- i
}
library(rgl)
persp3d(x = xmat, y = ymat, z = zmat, xlim = c(0, 10), ylim = c(0, 10), zlim = c(0, 5),
xlab = "x", ylab = "y", zlab = "height", col = "gray")
The image produced looks like this:
If you want z to depend on x or y, you'll likely want a smaller step size, but this works for the surface you're after.
To use the persp3d function one needs to create a matrix for z to correspond to all of the x and y values in the desired range.
I revised your function to take both the x and y parameters and return the desired z value. The outer function will call the function repeatedly to fill the matrix. Then plot, with the defined x and y axis and z (from the outer function)
library(rgl)
mycurve <- function(x, y) { return (1/x)}
myx <- seq(1, 10, by = 0.4)
myy <-seq(1, 10, by =0.4)
#create matrix
data<-outer(myx, myy, mycurve)
#plot points
persp3d(x=myx, y=myy, z=data,
xlab = "x", ylab = "y", zlab = "height")
I am looking to plot the following:
L<-((2*pi*h*c^2)/l^5)*((1/(exp((h*c)/(l*k*T)-1))))
all variables except l are constant:
T<-6000
h<-6.626070040*10^-34
c<-2.99792458*10^8
k<-1.38064852*10^-23
l has a range of 20*10^-9 to 2000*10^-9.
I have tried l<-seq(20*10^-9,2000*10^-9,by=1*10^-9), however this does not give me the results I expect.
Is there a simple solution for this in R, or do I have to try in another language?
Thank you.
Looking at the spectral radiance equation wikipedia page, it seems that your formula is a bit off. Your formula multiplies an additional pi (not sure if intended) and the -1 is inside the exp instead of outside:
L <- ((2*pi*h*c^2)/l^5)*((1/(exp((h*c)/(l*k*T)-1))))
Below is the corrected formula. Also notice I have converted it into a function with parameter l since this is a variable:
T <- 6000 # Absolute temperature
h <- 6.626070040*10^-34 # Plank's constant
c <- 2.99792458*10^8 # Speed of light in the medium
k <- 1.38064852*10^-23 # Boltzmann constant
L <- function(l){((2*h*c^2)/l^5)*((1/(exp((h*c)/(l*k*T))-1)))}
# Plotting
plot(L, xlim = c(20*10^-9,2000*10^-9),
xlab = "Wavelength (nm)",
ylab = bquote("Spectral Radiance" ~(KW*sr^-1*m^-2*nm^-1)),
main = "Plank's Law",
xaxt = "n", yaxt = "n")
xtick <- seq(20*10^-9, 2000*10^-9,by=220*10^-9)
ytick <- seq(0, 4*10^13,by=5*10^12)
axis(side=1, at=xtick, labels = (1*10^9)*seq(20*10^-9,2000*10^-9,by=220*10^-9))
axis(side=2, at=ytick, labels = (1*10^-12)*seq(0, 4*10^13,by=5*10^12))
The plot above is not bad, but I think we can do better with ggplot2:
h <- 6.626070040*10^-34 # Plank's constant
c <- 2.99792458*10^8 # Speed of light in the medium
k <- 1.38064852*10^-23 # Boltzmann constant
L2 <- function(l, T){((2*h*c^2)/l^5)*((1/(exp((h*c)/(l*k*T))-1)))} # Plank's Law
classical_L <- function(l, T){(2*c*k*T)/l^4} # Rayleigh-Jeans Law
library(ggplot2)
ggplot(data.frame(l = c(20*10^-9,2000*10^-9)), aes(l)) +
geom_rect(aes(xmin=390*10^-9, xmax=700*10^-9, ymin=0, ymax=Inf),
alpha = 0.3, fill = "lightblue") +
stat_function(fun=L2, color = "red", size = 1, args = list(T = 3000)) +
stat_function(fun=L2, color = "green", size = 1, args = list(T = 4000)) +
stat_function(fun=L2, color = "blue", size = 1, args = list(T = 5000)) +
stat_function(fun=L2, color = "purple", size = 1, args = list(T = 6000)) +
stat_function(fun=classical_L, color = "black", size = 1, args = list(T = 5000)) +
theme_bw() +
scale_x_continuous(breaks = seq(20*10^-9, 2000*10^-9,by=220*10^-9),
labels = (1*10^9)*seq(20*10^-9,2000*10^-9,by=220*10^-9),
sec.axis = dup_axis(labels = (1*10^6)*seq(20*10^-9,2000*10^-9,by=220*10^-9),
name = "Wavelength (\U003BCm)")) +
scale_y_continuous(breaks = seq(0, 4*10^13,by=5*10^12),
labels = (1*10^-12)*seq(0, 4*10^13,by=5*10^12),
limits = c(0, 3.5*10^13)) +
labs(title = "Black Body Radiation described by Plank's Law",
x = "Wavelength (nm)",
y = expression("Spectral Radiance" ~(kWsr^-1*m^-2*nm^-1)),
caption = expression(''^'\U02020' ~'Spectral Radiance described by Rayleigh-Jeans Law, which demonstrates the ultraviolet catastrophe.')) +
annotate("text",
x = c(640*10^-9, 640*10^-9, 640*10^-9, 640*10^-9,
150*10^-9, (((700-390)/2)+390)*10^-9, 1340*10^-9),
y = c(2*10^12, 5*10^12, 14*10^12, 31*10^12,
35*10^12, 35*10^12, 35*10^12),
label = c("3000 K", "4000 K", "5000 K", "6000 K",
"UV", "VISIBLE", "INFRARED"),
color = c(rep("black", 4), "purple", "blue", "red"),
alpha = c(rep(1, 4), rep(0.6, 3)),
size = 4.5) +
annotate("text", x = 1350*10^-9, y = 23*10^12,
label = deparse(bquote("Classical theory (5000 K)"^"\U02020")),
color = "black", parse = TRUE)
Notes:
I created L2 by also making absolute temperature T a variable
For each T, I plot the function L2 using different colors for representation. I've also added a classical_L function to demonstrate classical theory of spectral radiance
geom_rect creates the light blue shaded area for "VISIBLE" light wavelength range
scale_x_continuous sets the breaks of the x axis, while labels sets the axis tick labels. Notice I have multiplied the seq by (1*10^9) to convert the units to nanometer (nm). A second x-axis is added to display the micrometer scale
Analogously, scale_y_continuous sets the breaks and tick labels for y axis. Here I multiplied by (1*10^-12) or (1*10^(-3-9)) to convert from watts (W) to kilowatts (kW), and from inverse meter (m^-1) to inverse nanometer (nm^-1)
bquote displays superscripts correctly in the y axis label
annotate sets the coordinates and text for curve labels. I've also added the labels for "UV", "VISIBLE" and "INFRARED" light wavelengths
ggplot2
Plot from wikipedia:
Image source: https://upload.wikimedia.org/wikipedia/commons/thumb/1/19/Black_body.svg/600px-Black_body.svg.png