I am using ´persp´ to generate 3D-Plot. it is the result:
persp(TestMatrix ,theta = 30, phi = 30, expand = 0.19,scale=FALSE,shade=0.4,border=NA,box=FALSE)
in my diagram. I would like to change the color continuous from blue to red.
what should I do?
UPDATE
I would like generate a diagram like this:
If you just need a shaded red curve, try to add a colparameter in your persp function :
persp(TestMatrix,
theta = 30, phi = 30, expand = 0.19, scale=FALSE,
shade=0.4, col="red", border="blue",
box=FALSE)
Edit :
Thanks for your edit, I wasn't sure you needed a two-colors plot. There is an question on this problem here : Create 3D Plot Colored According to the Z-axis. You need a color matrix to specify the colors of each facet of the surface. The subtil point is to calculate the height (z) in the middle of each facet, ie the mean of the four summit of the grid (which correspond to z values).
Adapting the answer for your graph, the solution can be something like that :
# Color palette (100 colors)
col.pal<-colorRampPalette(c("blue", "red"))
colors<-col.pal(100)
# height of facets
z.facet.center <- (z[-1, -1] + z[-1, -ncol(z)] + z[-nrow(z), -1] + z[-nrow(z), -ncol(z)])/4
# Range of the facet center on a 100-scale (number of colors)
z.facet.range<-cut(z.facet.center, 100)
persp(x, y, z,
theta = 30, phi = 30, expand = 0.19, scale=FALSE,
shade=NA, col=colors[z.facet.range], border="grey80",
box=FALSE)
Related
I am a beginning R student and would appreciate any help that anyone here could offer!
I have plotted a bivariate distribution function using the following code:
shape <- function(x,y){
(6/5)*(x+y^2)
}
x <- y <- seq(0, 1, length = 500)
z <- outer(x, y, shape)
persp(x, y, z, theta = 30, phi = 15, col = "red", shade = 0.5)
Since I included the bit about col = "red" and shade = 0.5, I thought my shape would appear red and beautifully shaded in "an approximation to daylight illumination" as promised in the help section.
However, my shape is uniformly black with no shading at all. Could anyone tell me what I am missing?
I've been trying to color specific bins above a defined threshold in the following data frame (df)
df <- read.table("https://pastebin.com/raw/3En2GWG6", header=T)
I've been following this example (Change colour of specific histogram bins in R), but I cannot seem to get this to adapt their suggestions to my data, so I wanted to ask you here at stackoverflow
I would like all bins with values above 0.100 to be "red", and the rest all to be either no color, or just black (I defined black, but I would prefer no color)
Here is what I tried:
col<-(df$consumption>=0.100)
table(col) # I can see 40 points above 100, the rest below
col[which(col=="TRUE")] <- "firebrick1"
col[which(col=="FALSE")] <- "black"
hist(df$consumption, breaks = 1000, xlim = c(0,0.2), col=col,xlab= "Consumption [MG]")
However, the whole graph is red, and that doesn't make sense..?
In other words, I would like anything to the right side of the line below to be red
hist(df$consumption, breaks = 1000, xlim = c(0,0.2),xlab= "Consumption [MG]")
abline(v=c(.100), col=c("red"),lty=c(1), lwd=c(5))
Simply plot two histograms on top of each other using add=TRUE and sub-setting the second.
hist(df$consumption, breaks=1000, xlim=c(0,.2),xlab= "Consumption [MG]")
hist(df$consumption[df$consumption > .100], breaks=1000, xlim=c(0,.2), col=2, add=TRUE)
abline(v=.100, col=2, lty=3)
Here is along the lines of what you were doing. You do not want to count the points above your cutoff, but rather the number of histogram bins above your cutoff.
# store the histogram as an object
h <- hist(df$consumption, breaks = 1000)
# extract out the breaks, and assign a color vector accordingly
cols <- ifelse(h$breaks > 0.1, "firebrick1", "black")
# use the color vector
plot(h, col = cols, xlim=c(0,.2),xlab= "Consumption [MG]")
abline(v=c(.100), col=c("red"),lty=c(1), lwd=c(5))
Hy there,
I use persp for a 3D-Plot and i am try to find out how persp define the ticks when the parameter ticktype="detailed" is set.
I want to draw lines into the box around a surface corresponding to the ticks. Up till now, frist I draw the surface without any labels and axes and add all lines and axes afterwords. To make it clear what I have done -> example code:
z <- matrix(rep(1:10, each=10), nrow=10, ncol=10)
x.axis <- 1:nrow(z)
y.axis <- 1:ncol(z)
max.y <- max(y.axis)
# Drawing the surface without the axes and no lines on the surface
pmat <- persp(z = z, x = x.axis, y = y.axis ,
lphi = 100, phi = 25, theta = -30,
axes=F,
border = NA, # no lines on the surface
col="deepskyblue",
expand = 0.5,
shade = 0.65)
Now I add the the lines on the surface with different color and the axes with ticks and labels:
par(new=T)
pmat <- persp(z = z, x = x.axis, y = y.axis ,
lphi = 100, phi = 25, theta = -30,
ticktype = "detailed",
expand = 0.5,
cex.lab=0.75,
col=NA,
border="grey80")
par(new=F)
To get lines on the box around the surface I use the following:
for (z_high in c(2,4,6,8)) {
lines(trans3d(x.axis, max.y, z_high, pmat) , col="black", lty=3)
}
As you can see, I use a own defined vector c(2,4,6,8) which represents the z-values for the box lines in the back. If the input surface now changes, I have to adapted this vector by my own. Is there a way to get the ticks for all axes in the persp plot? Did anyone know how persp define the ticks?
Here is the problem:
x<-seq(0,10,length.out = 1000)
y1<-dnorm(x,mean = 2,sd=1)
y2<-dnorm(x,mean = 6,sd=1)
plot(x,y1,type="l")
lines(x,y2)
abline(v=x[380])
The graph is shown below. How can I fill 2 different colors, say red and blue, on the each side of vertical line but still below two normal density functions. I thought I can use polygon, but failed.
This is the graph without filling colors:
Here's one way:
First, we'll get the parallel minimum of your densities - this is a vector of the top y coordinates for our polygons.
y = pmin(y1, y2)
# set up your plot as in the question
plot(x, y1, type="l")
lines(x, y2)
# define a re-usable variable for the vertical line placement
x_vert = 380
abline(v = x[x_vert])
# Now we'll draw 2 polygons, one for the left side, one for the right.
# The first (x,y) pairs of the polygon are just the (x,y) coords of the
# density we're filling to, until the vertical line
# Then we need to connect the "bottom" points, which have coordinates
# (x[x_vert], 0) and (x[1], 0)
polygon(x = c(x[1:x_vert], x[x_vert], x[1]),
y = c(y[1:x_vert], 0, 0),
col = "blue")
# similar for the right hand polygon, but now going from x_vert to length(x)
polygon(x = c(x[x_vert:length(x)], x[length(x)], x[x_vert]),
y = c(y[x_vert:length(x)], 0, 0),
col = "red")
Voila!
Warning.... very novice question follows:
I am trying to plot a fairly regular distribution of several thousand (X,Y) points each associated with a value, let's call Z, which varies very irregularly between, say, -20 to +20. I am not interested in smoothing; I want the point Z values to plot according to a smoothly varying color palette much like Gnuplot can do with the proper smooth color palette. I've tried base R, ggplot2, and latticeExtra, and as best I can, I can come up with the following which does almost what I want:
library(lattice)
library(latticeExtra)
library(colorRamps)
df = read.table(file"whatever", header=T)
levelplot(Z~X*Y, df, panel=panel.levelplot.points, cex=0.2,
col.regions=colorRampPalette(c("red","white","blue"))(50))
One data point looks like: 1302525 225167 -3.5
When I plot my dataframe with the "50" in the last code line as 3, I get the predictable R recycle behavior of the red, white, and blue colors repeating five times with the 16th color bar segment white. Changing the 3 to a 7 causes more shades of red and blue creating 2 repeat color range segments with two reddish colors left over as the color range tries to recycle. This suggests making this number larger causes a finer graduation of colors. But if I put in a number greater than 16, that's all I get, 16 colored segments, evenly changing from red, to white, to blue. But I'd like the color scale even finer, and in a perfect world, force a Z of zero to be the white color.
My experience so far with R is when I can't do something as simple as this, I'm missing a very fundamental concept. What is it?
As far as lattice is concerned, you can set up your colors palette with RColorBrewer (or even colorspace). Using the example provided by #Chase, but with positive value for z:
dat <- data.frame(x = rnorm(1000), y = rnorm(1000), z = sample(0:40, 1000, TRUE))
library(RColorBrewer)
# see, e.g.
# display.brewer.all(9, type="seq")
# display.brewer.pal(11, "RdBu")
my.col <- colorRampPalette(brewer.pal(11, "RdBu"))(diff(range(dat$z)))
xyplot(y ~ x, data=dat, col=my.col[dat$z], pch=19, alpha=.5)
Note that it is also necessary here to increase the range of available colors by interpolation. Also, with levelplot(), you might want to play with cut= and pretty=.
Have you looked at scale_gradient in ggplot? Or scale_brewer for discrete colours? Here's an example of scale_gradient
dat <- data.frame(x = rnorm(1000), y = rnorm(1000), z = sample(-20:20, 1000, TRUE))
p <- ggplot(dat, aes(x, y, colour = z)) + geom_point()
p + scale_colour_gradient()
p + scale_colour_gradient(low = "red", high = "blue")
p + scale_colour_gradient2(low = "red", mid = "white", high = "blue")
The "concept" you are missing is the at argument to levelplot() which defines the breakpoints between colour levels and/or contour lines. The default is pretty(z) which results in only a few levels. You can set at to be a sequence covering the range of values you want.
library(latticeExtra)
dat <- data.frame(x = rnorm(1000), y = rnorm(1000), z = rnorm(1000, mean = 1))
## for centering the colour key around zero
maxz <- max(abs(dat$z))
levelplot(z ~ x * y, dat, at = seq(-maxz, maxz, length = 100),
panel = panel.levelplot.points, par.settings = custom.theme.2())