How can we plot a 3D plot for the in-out degree distribution.
I found the the plot here: http://en.wikipedia.org/wiki/Degree_distribution really interesting but I do not know if we can draw it using r ?
any ideas ?
update:
the data that I am using is the degree distribution of the in-degree and out-degree:
> head(dDistribution_in)
[1] 0.30117450 0.19379195 0.10654362 0.06291946 0.03775168 0.03313758
> head(dDistribution_out)
[1] 0.36115772 0.17072148 0.09228188 0.05369128 0.04572148 0.02055369
You can do this with perspective plots, here's an example that should help:
require(MASS)
set.seed(42)
persp(kde2d(rnorm(100), rnorm(100)), col="grey90", shade=1, theta=120, xlab="X")
If you're asking specifically about replicating this degree distribution, you can use the same idea on your data. Also, while 3D styled plots can look good, a clearer representation of that data may be something like a hexagonally-binned 2d histogram.
Update
Using the small sample of your data works fine:
d_in <- c(0.30117450, 0.19379195, 0.10654362, 0.06291946, 0.03775168, 0.03313758)
d_out <- c(0.36115772, 0.17072148, 0.09228188, 0.05369128, 0.04572148, 0.02055369)
persp(kde2d(d_in, d_out), col="grey90", shade=1, theta=120, xlab="X")
Not sure what you're doing wrong, but see this post on how to make a reproducible example.
Related
I'm trying to plot 3-dimensional vectors (x, y, z coordinates) onto a 3D coordinate system in R like in the picture below. Ideally, I would then like to construct 3d kernel density plots, also like in the image below.
Ideal result of vector plot and 3d kernel density plot
I have a matrix containing ~100 rows and one column for each coordinate (x, y , z). Initially, I tried arrow3D() from the plot3D package but I find the perspective to be sub-par, it's rather difficult to discern directions of the arrows from one perspective in the final plot. Next I tried the rgl package which gives me interactivity - great. Minimal working example:
library(rgl)
library(matlib)
data2 <- data.frame(replicate(6,rnorm(100))) #sample data set for minimum working example
colnames(data2) <- c("x_target", "y_target", "z_target", "x_start", "y_start", "z_start")
x1 <- data2$x_target - data2$x_start
y1 <- data2$y_target - data2$y_start
z1 <- data2$z_target - data2$z_start
vec <- (diag(6,3)) # coordinates for x, y and z axis
rownames(vec) <- c("X", "Y", "Z") # labels for x, y and z axis
z <- as.matrix((data.frame(x=x1, y=y1, z=z1)))
open3d()
vectors3d(vec, color=c(rep("black",3)), lwd=2, radius=1/25)
vectors3d(X=z, headlength=1/25)
(due to the random numbers generator the strange looking rods appear at different coordinates, not exactly like in the image i link to below)
The result of the code above is a version of the image link below. One set of coordinates produces a very strange looking more like rod object which is far longer then the coordinates would produce. If I plot the vectors individually, no such object is created. Anyone have any ideas why this happens? Also, if anyone has a tool (doesn't have to be R), that can create a 3D vector plot like in the first image, I'd be grateful. I find it to be very complicated in R, but I'm definitely a beginner.
Strange object to the right (long red rod that doesn't look like an arrow at all)
Thank you!
This is due to a bug in the matlib package, fixed in verson 0.9.2 of that package. I think you need to install it from Github instead of CRAN to get the bug fix:
devtools::install_github("friendly/matlib")
BTW, if you are using random numbers in a reproducible example, you can make it perfectly reproducible by something like
set.seed(123)
at the start (or some number other than 123). I saw reproducible problems with your example for set.seed(4).
There are a number of questions in this forum on locating intersections between a fitted model and some raw data. However, in my case, I am in an early stage project where I am still evaluating data.
To begin with, I have created a data frame that contains a ratio value whose ideal value should be 1.0. I have plotted the data frame and also used abline() function to plot a horizontal line at y=1.0. This horizontal line and the plot of ratios intersect at some point.
plot(a$TIME.STAMP, a$PROCESS.RATIO,
xlab='Time (5s)',
ylab='Process ratio',
col='darkolivegreen',
type='l')
abline(h=1.0,col='red')
My aim is to locate the intersection point, say x and draw two vertical lines at x±k, as abline(v=x-k) and abline(v=x+k) where, k is certain band of tolerance.
Applying a grid on the plot is not really an option because this plot will be a part of a multi-panel plot. And, because ratio data is very tightly laid out, the plot will not be too readable. Finally, the x±k will be quite valuable in my discussions with the domain experts.
Can you please guide me how to achieve this?
Here are two solutions. The first one uses locator() and will be useful if you do not have too many charts to produce:
x <- 1:5
y <- log(1:5)
df1 <-data.frame(x= 1:5,y=log(1:5))
k <-0.5
plot(df1,type="o",lwd=2)
abline(h=1, col="red")
locator()
By clicking on the intersection (and stopping the locator top left of the chart), you will get the intersection:
> locator()
$x
[1] 2.765327
$y
[1] 1.002495
You would then add abline(v=2.765327).
If you need a more programmable way of finding the intersection, we will have to estimate the function of your data. Unfortunately, you haven’t provided us with PROCESS.RATIO, so we can only guess what your data looks like. Hopefully, the data is smooth. Here’s a solution that should work with nonlinear data. As you can see in the previous chart, all R does is draw a line between the dots. So, we have to fit a curve in there. Here I’m fitting the data with a polynomial of order 2. If your data is less linear, you can try increasing the order (2 here). If your data is linear, use a simple lm.
fit <-lm(y~poly(x,2))
newx <-data.frame(x=seq(0,5,0.01))
fitline = predict(fit, newdata=newx)
est <-data.frame(newx,fitline)
plot(df1,type="o",lwd=2)
abline(h=1, col="red")
lines(est, col="blue",lwd=2)
Using this fitted curve, we can then find the closest point to y=1. Once we have that point, we can draw vertical lines at the intersection and at +/-k.
cross <-est[which.min(abs(1-est$fitline)),] #find closest to 1
plot(df1,type="o",lwd=2)
abline(h=1)
abline(v=cross[1], col="green")
abline(v=cross[1]-k, col="purple")
abline(v=cross[1]+k, col="purple")
I'm trying to draw a graph where the distance between vertices correspond to the edge weights* and I've founde that in graphviz there is a way to draw such graph. Is there a way to do this in R with the igraph package (specfically with graph.adkacency)?
Thanks,
Noam
(as once have been asked: draw a graph where the distance between vertices correspond to the edge weights)
This is not possible as you need triangle equality for every triangle to be able to plot such an object. So you can only approximate it. For this you can use "force embedded" algorithms. There are a few in igraph. The one I often use is the Fruchterman-Reingold algorithm.
See for details:
library("igraph")
?layout.fruchterman.reingold
Edit:
Note that the distance between nodes will correspond somewhat with the inverse of the absolute edge weight.
Like Sacha Epskamp mentioned, unless your data is perfect, you cannot draw a graph that would not violate some triangular inequalities. However, there are techniques named Multidimensional scaling (MDS) targeted at minimizing such violations.
One implementation in R is cmdscale from the stats package. I'd recommend the example at the bottom of ?cmdscale:
> require(graphics)
>
> loc <- cmdscale(eurodist)
> x <- loc[,1]
> y <- -loc[,2]
> plot(x, y, type="n", xlab="", ylab="", main="cmdscale(eurodist)")
> text(x, y, rownames(loc), cex=0.8)
Of course, you can plot x and y using any graphics packages (you were inquiring about igraph specifically).
Finally, I'm sure you'll find plenty of other implementations if you search for "multidimensional scaling" or "MDS". Good luck.
I have a set of 3D coordinates (below - just for a single point, in 3D space):
x <- c(-521.531433, -521.511658, -521.515259, -521.518127, -521.563416, -521.558044, -521.571228, -521.607178, -521.631165, -521.659973)
y <- c(154.499557, 154.479568, 154.438705, 154.398682, 154.580688, 154.365189, 154.3564, 154.559189, 154.341309, 154.344223)
z <- c(864.379272, 864.354675, 864.365479, 864.363831, 864.495667, 864.35498, 864.358582, 864.50415, 864.35553, 864.359863)
xyz <- data.frame(x,y,z)
I need to make a time-series plot of this point with a 3D rendering (so I can rotate the plot, etc.). The plot will visualize a trajectory of the point above in time (for example in the form of solid line). I used 'rgl' package with plot3d method, but I can't make it to plot time-series (below, just plot a single point from first frame in time-series):
require(rgl)
plot3d(xyz[1,1],xyz[1,2],xyz[1,3],axes=F,xlab="",ylab="",zlab="")
I found this post, but it doesn't really deal with a real-time rendered 3D plots. I would appreciate any suggestions. Thank you.
If you read help(plot3d) you can see how to draw lines:
require(rgl)
plot3d(xyz$x,xyz$y,xyz$z,type="l")
Is that what you want?
How about this? It uses rgl.pop() to remove a point and a line and draw them as a trail - change the sleep argument to control the speed:
ts <- function(xyz,sleep=0.3){
plot3d(xyz,type="n")
n = nrow(xyz)
p = points3d(xyz[1,])
l = lines3d(xyz[1,])
for(i in 2:n){
Sys.sleep(sleep)
rgl.pop("shapes",p)
rgl.pop("shapes",l)
p=points3d(xyz[i,])
l=lines3d(xyz[1:i,])
}
}
The solution was simpler than I thought and the problem was that I didn't use as.matrix on my data. I was getting error (list) object cannot be coerced to type 'double' when I was simply trying to plot my entire dataset using plot3d (found a solution for this here). So, if you need to plot time-series of set of coordinates (in my case motion capture data of two actors) here is my complete solution (only works with the data set below!):
download example data set
read the above data into a table:
data <- read.table("Bob12.txt",sep="\t")
extract XYZ coordinates into a separate matrixes:
x <- as.matrix(subset(data,select=seq(1,88,3)))
y <- as.matrix(subset(data,select=seq(2,89,3)))
z <- as.matrix(subset(data,select=seq(3,90,3)))
plot the coordinates on a nice, 3D rendered plot using 'rgl' package:
require(rgl)
plot3d(x[1:nrow(x),],y[1:nrow(y),],z[1:nrow(z),],axes=F,xlab="",ylab="",zlab="")
You should get something like on the image below (but you can rotate it etc.) - hope you can recognise there are joint centers for people there. I still need to tweak it to make it visually better - to have first frame as a points (to clearly see actor's joints), then a visible break, and then the rest of frames as a lines.
I need to make a topographic map of a terrain for which I have only fairly sparse samples of (x, y, altitude) data. Obviously I can't make a completely accurate map, but I would like one that is in some sense "smooth". I need to quantify "smoothness" (probably the reciprocal the average of the square of the surface curvature) and I want to minimize an objective function that is the sum of two quantities:
The roughness of the surface
The mean square distance between the altitude of the surface at the sample point and the actual measured altitude at that point
Since what I actually want is a topographic map, I am really looking for a way to construct contour lines of constant altitude, and there may be some clever geometric way to do that without ever having to talk about surfaces. Of course I want contour lines also to be smooth.
Any and all suggestions welcome. I'm hoping this is a well-known numerical problem. I am quite comfortable in C and have a working knowledge of FORTRAN. About Matlab and R I'm fairly clueless.
Regarding where our samples are located: we're planning on roughly even spacing, but we'll take more samples where the topography is more interesting. So for example we'll sample mountainous regions more densely than a plain. But we definitely have some choices about sampling, and could take even samples if that simplifies matters. The only issues are
We don't know how much terrain we'll need to map in order to find features that we are looking for.
Taking a sample is moderately expensive, on the order of 10 minutes. So sampling a 100x100 grid could take a long time.
Kriging interpolation may be of some use for smoothly interpolating your sparse samples.
R has many different relevant tools. In particular, have a look at the spatial view. A similar question was asked in R-Help before, so you may want to look at that.
Look at the contour functions. Here's some data:
x <- seq(-3,3)
y <- seq(-3,3)
z <- outer(x,y, function(x,y,...) x^2 + y^2 )
An initial plot is somewhat rough:
contour(x,y,z, lty=1)
Bill Dunlap suggested an improvement: "It often works better to fit a smooth surface to the data, evaluate that surface on a finer grid, and pass the result to contour. This ensures that contour lines don't cross one another and tends to avoid the spurious loops that you might get from smoothing the contour lines themselves. Thin plate splines (Tps from library("fields")) and loess (among others) can fit the surface."
library("fields")
contour(predict.surface(Tps(as.matrix(expand.grid(x=x,y=y)),as.vector(z))))
This results in a very smooth plot, because it uses Tps() to fit the data first, then calls contour. It ends up looking like this (you can also use filled.contour if you want it to be shaded):
For the plot, you can use either lattice (as in the above example) or the ggplot2 package. Use the geom_contour() function in that case. An example can be found here (ht Thierry):
ds <- matrix(rnorm(100), nrow = 10)
library(reshape)
molten <- melt(data = ds)
library(ggplot2)
ggplot(molten, aes(x = X1, y = X2, z = value)) + geom_contour()
Excellent review of contouring algorithm, you might need to mesh the surface first to interpolate onto a grid.
maybe you can use:
GEOMap
geomapdata
gtm
with
Matrix
SparseM
slam
in R