Problem: I am trying to reproduce a round filled 2d contour plot in R using plotly (have tried ggplot2 also but plotly seemed to be easier).
Data: Sample data download link -
https://drive.google.com/file/d/10Mr5yWVReQckPI6TKLY_vzPT8zWiijKl/view?usp=sharing
The data to be plotted for contour is in a column format and typically called z variable, there is x and y data also available for all values of z. A simple dataframe would look like this:
Please ignore the repeat common x and y as I have truncated decimals. The data has about 25000 rows.
Approach: I first use akima package to interpolate z variable values for given x and y to map z in 2d. This makes the z column data fit in a xy grid for 2d plotting and show contours.
Expected outcome:
Code used:
dens <- akima::interp(x = dt$`Xvalue(mm)`,
y = dt$`Yvalue(mm)`,
z = dt$Values,
duplicate = "mean",
xo=seq(min(dt$`Xvalue(mm)`), max(dt$`Xvalue(mm)`), length = 10),
yo=seq(min(dt$`Yvalue(mm)`), max(dt$`Yvalue(mm)`), length = 10))
plot_ly(x = dens$x,
y = dens$y,
z = dens$z,
colors = c("blue","grey","red"),
type = "contour")
Actual outcome:
Help Needed:
To refine edges of the actual outcome plot to something of a close match to the expected outcome image.
Many thanks in advance for your comments and help.
I found that I could increase the grid output z matrix from akima::interp() from default 40x40 to custom using nx and ny input in function.
And then in plot_ly() add contours = list(coloring = 'fill', showlines = FALSE) to hide contour lines to get output close to my expected outcome.
So working code is like this:
dens <- akima::interp(x = dt$`Xvalue(mm)`,
y = dt$`Yvalue(mm)`,
z = dt$Values,
nx = 50,
ny = 50,
duplicate = "mean",
xo=seq(min(dt$`Xvalue(mm)`), max(dt$`Xvalue(mm)`), length = 50),
yo=seq(min(dt$`Yvalue(mm)`), max(dt$`Yvalue(mm)`), length = 50))
plot_ly(x = dens$x,
y = dens$y,
z = dens$z,
colors = c("blue","grey","red"),
type = "contour",
contours = list(coloring = 'fill', showlines = FALSE))
Plotly contour plot reference was very helpful in this case:
https://plot.ly/r/reference/#contour
Related
I plotted a 3d scatter plot in R using the scatter3d function.
Now, I want to plot the labels on every dot in the 3d scatter, such as every point has its ID next to it i.e., "1", "2" etc..
Here is what I tried:
library("car")
library("rgl")
scatter3d(geometry[,1],geometry[,2],geometry[,3] , surface=FALSE, labels = rownames(geometry), id.n=nrow(geometry))
This tutorial says that adding arguments labels=rownames(geometry), id.n=nrow(geometry) should display the labels on every dot but that did not work.
EDIT:
I uploaded the coordinate file here, you can read it like this
geometry = read.csv("geometry.txt",sep = " ")
colnames(geometry) = c("x","y","z")
EDIT:
Actually, even the example from the tutorial does not label the points and does not produce the plot displayed. There is probably something wrong with the package.
scatter3d(x = sep.l, y = pet.l, z = sep.w,
surface=FALSE, labels = rownames(iris), id.n=nrow(iris))
I can give you a quick fix if you want to use any other function other than scatter3d. This can be achieved using plot3d and text3d function. I have provided the basic code block of how it can be implemented. You can customize it to your needs.
plot3d(geometry[,1],geometry[,2],geometry[,3])
text3d(geometry[,1],geometry[,2],geometry[,3],rownames(geometry))
points3d(geometry[,1],geometry[,2],geometry[,3], size = 5)
After much messing around I got it (I also have the method for plot_ly if you,re interested)
test2 <- cbind(dataSet[,paste(d)],set.final$Groups,test)
X <- test2[,1]
Y <- test2[,2]
Z <- test2[,3]
# 3D plot with the regression plane
scatter3d(x = X, y = Y, z = Z, groups = test2$`set.final$Groups`,
grid = FALSE, fit = "linear",ellipsoid = FALSE, surface=FALSE,
surface.col = c("green", "blue", "red"),
#showLabels(x = x, y = y, z = z, labels=test2$test, method="identify",n = nrow(test2), cex=1, col=carPalette()[1], location=c("lr"))
#labels = test2$test,
id=list(method = "mahal", n = length(test2$test), labels = test2$test)
#id.n=nrow(test2$test)
)
#identify3d(x = X, y = Y, z = Z, labels = test2$test, n = length(test2$test), plot = TRUE, adj = c(-0.1, 0.5), tolerance = 20, buttons = c("right"))
rglwidget()
I want to plot a matrix of z values with x rows and y columns as a surface similar to this graph from MATLAB.
Surface plot:
Code to generate matrix:
# Parameters
shape<-1.849241
scale<-38.87986
x<-seq(from = -241.440, to = 241.440, by = 0.240)# 2013 length
y<-seq(from = -241.440, to = 241.440, by = 0.240)
matrix_fun<-matrix(data = 0, nrow = length(x), ncol = length(y))
# Generate two dimensional travel distance probability density function
for (i in 1:length(x)) {
for (j in 1:length(y)){
dxy<-sqrt(x[i]^2+y[j]^2)
prob<-1/(scale^(shape)*gamma(shape))*dxy^(shape-1)*exp(-(dxy/scale))
matrix_fun[i,j]<-prob
}}
# Rescale 2-d pdf to sum to 1
a<-sum(matrix_fun)
matrix_scale<-matrix_fun/a
I am able to generate surface plots using a couple methods (persp(), persp3d(), surface3d()) but the colors aren't displaying the z values (the probabilities held within the matrix). The z values only seem to display as heights not as differentiated colors as in the MATLAB figure.
Example of graph code and graphs:
library(rgl)
persp3d(x=x, y=y, z=matrix_scale, color=rainbow(25, start=min(matrix_scale), end=max(matrix_scale)))
surface3d(x=x, y=y, z=matrix_scale, color=rainbow(25, start=min(matrix_scale), end=max(matrix_scale)))
persp(x=x, y=y, z=matrix_scale, theta=30, phi=30, col=rainbow(25, start=min(matrix_scale), end=max(matrix_scale)), border=NA)
Image of the last graph
Any other tips to recreate the image in R would be most appreciated (i.e. legend bar, axis tick marks, etc.)
So here's a ggplot solution which seems to come a little bit closer to the MATLAB plot
# Parameters
shape<-1.849241
scale<-38.87986
x<-seq(from = -241.440, to = 241.440, by = 2.40)
y<-seq(from = -241.440, to = 241.440, by = 2.40)
df <- expand.grid(x=x,y=y)
df$dxy <- with(df,sqrt(x^2+y^2))
df$prob <- dgamma(df$dxy,shape=shape,scale=scale)
df$prob <- df$prob/sum(df$prob)
library(ggplot2)
library(colorRamps) # for matlab.like(...)
library(scales) # for labels=scientific
ggplot(df, aes(x,y))+
geom_tile(aes(fill=prob))+
scale_fill_gradientn(colours=matlab.like(10), labels=scientific)
BTW: You can generate your data frame of probabilities much more efficiently using the built-in dgamma(...) function, rather than calculating it yourself.
In line with alexis_laz's comment, here is an example using filled.contour. You might want to increase your by to 2.40 since the finer granularity increases the time it takes to generate the plot by a lot but doesn't improve quality.
filled.contour(x = x, y = y, z = matrix_scale, color = terrain.colors)
# terrain.colors is in the base grDevices package
If you want something closer to your color scheme above, you can fiddle with the rainbow function:
filled.contour(x = x, y = y, z = matrix_scale,
color = (function(n, ...) rep(rev(rainbow(n/2, ...)[1:9]), each = 3)))
Finer granularity:
filled.contour(x = x, y = y, z = matrix_scale, nlevels = 150,
color = (function(n, ...)
rev(rep(rainbow(50, start = 0, end = 0.75, ...), each = 3))[5:150]))
I have a following data:
library(rgl)
x <- c(rep(1,6),
rep(3,6),
rep(6,6),
rep(9,6),
rep(12,6))
y <- c(1.35,1.39,1.48,1.29,1.35,1.32,
NA,1.5,1.44,1.6,1.5,1.41,
NA,NA,1.72,1.56,1.6,1.55,
NA,NA,NA,1.95,1.9,1.75,
NA,NA,NA,NA,2.05,1.95)
z <- rep(1:6,5)
open3d()
plot3d(x,y,z, type = 'n')
lines3d(x,y,z)
Which is plotting lines in 3d as I expect.
But I cannot get it to plot a surface3d.
As I already read some threads I might need to interpolate my data. RGL docs has not cover this subject well. I tried akima but it doesn't accept NA's.
I would like to link lines to create a surface in linear way. I aware of the NA, so I expect that surface will be decreasing in the area for bigger x (more NA's).
Do I need to perform interpolation? If yes, how to do that on my sample data?
If no, how to achieve the surface3d on my sample data?
Thanks
the solution comes to me from this thread:
Making a wireframe plot from an x,y,z data.frame
below code will work for the sample data provided above (just switch x->y,y->z,z->x)
zmat <- matrix(data = z, nrow = 6, ncol = 5, byrow = FALSE)
surface3d(x = 1:6, y = c(1,3,6,9,12), z = zmat, alpha = 0.4, colour = 'blue')
I have 4 columns of data in R that looks like this
x y z group
the group columns has categorical values, so it is a discrete set of values, whereas the other three columns are continuous.
I want to make a 3d plot in R with x, y, and z, and where the color of the dot is given by "group". I also want to have a legend to this plot. How can I do this? I don't have a particular preference on the actual colors. I suppose rainbow(length(unique(group)) should do fine.
Here is an example using scatterplot3d and based on the example in the vignette
library(scatterplot3d)
# some basic dummy data
DF <- data.frame(x = runif(10),
y = runif(10),
z = runif(10),
group = sample(letters[1:3],10, replace = TRUE))
# create the plot, you can be more adventurous with colour if you wish
s3d <- with(DF, scatterplot3d(x, y, z, color = as.numeric(group), pch = 19))
# add the legend using `xyz.convert` to locate it
# juggle the coordinates to get something that works.
legend(s3d$xyz.convert(0.5, 0.7, 0.5), pch = 19, yjust=0,
legend = levels(DF$group), col = seq_along(levels(DF$group)))
Or, you could use lattice and cloud, in which case you can construct the key using key
cloud(z~x+y, data = DF, pch= 19, col.point = DF$group,
key = list(points = list(pch = 19, col = seq_along(levels(DF$group))),
text = list(levels(DF$group)), space = 'top', columns = nlevels(DF$group)))
If my X axis is time, and my Y is numeric data, how can I add a point at an arbitrary Y value (Say 500) whenever a point exists?
I am overlaying using lines on top of other plots.
Just add more points, with the same x values as your previous data, and a fixed y value.
So you have something like:
dfr <- data.frame(x = sample(100, 10, replace = TRUE), y = runif(10))
with(dfr, plot(x, y))
and you want to add
points(dfr$x, rep.int(0.5, 10), col = "blue")
Having time for x values shouldn't affect anything.