I have come across a number of situations where I want to plot more points than I really ought to be -- the main holdup is that when I share my plots with people or embed them in papers, they occupy too much space. It's very straightforward to randomly sample rows in a dataframe.
if I want a truly random sample for a point plot, it's easy to say:
ggplot(x,y,data=myDf[sample(1:nrow(myDf),1000),])
However, I was wondering if there were more effective (ideally canned) ways to specify the number of plot points such that your actual data is accurately reflected in the plot. So here is an example.
Suppose I am plotting something like the CCDF of a heavy tailed distribution, e.g.
ccdf <- function(myList,density=FALSE)
{
# generates the CCDF of a list or vector
freqs = table(myList)
X = rev(as.numeric(names(freqs)))
Y =cumsum(rev(as.list(freqs)));
data.frame(x=X,count=Y)
}
qplot(x,count,data=ccdf(rlnorm(10000,3,2.4)),log='xy')
This will produce a plot where the x & y axis become increasingly dense. Here it would be ideal to have fewer samples plotted for large x or y values.
Does anybody have any tips or suggestions for dealing with similar issues?
Thanks,
-e
I tend to use png files rather than vector based graphics such as pdf or eps for this situation. The files are much smaller, although you lose resolution.
If it's a more conventional scatterplot, then using semi-transparent colours also helps, as well as solving the over-plotting problem. For example,
x <- rnorm(10000); y <- rnorm(10000)
qplot(x, y, colour=I(alpha("blue",1/25)))
Beyond Rob's suggestions, one plot function I like as it does the 'thinning' for you is hexbin; an example is at the R Graph Gallery.
Here is one possible solution for downsampling plot with respect to the x-axis, if it is log transformed. It log transforms the x-axis, rounds that quantity, and picks the median x value in that bin:
downsampled_qplot <- function(x,y,data,rounding=0, ...) {
# assumes we are doing log=xy or log=x
group = factor(round(log(data$x),rounding))
d <- do.call(rbind, by(data, group,
function(X) X[order(X$x)[floor(length(X)/2)],]))
qplot(x,count,data=d, ...)
}
Using the definition of ccdf() from above, we can then compare the original plot of the CCDF of the distribution with the downsampled version:
myccdf=ccdf(rlnorm(10000,3,2.4))
qplot(x,count,data=myccdf,log='xy',main='original')
downsampled_qplot(x,count,data=myccdf,log='xy',rounding=1,main='rounding = 1')
downsampled_qplot(x,count,data=myccdf,log='xy',rounding=0,main='rounding = 0')
In PDF format, the original plot takes up 640K, and the downsampled versions occupy 20K and 8K, respectively.
I'd either make image files (png or jpeg devices) as Rob already mentioned, or I'd make a 2D histogram. An alternative to the 2D histogram is a smoothed scatterplot, it makes a similar graphic but has a more smooth cutoff from dense to sparse regions of space.
If you've never seen addictedtor before, it's worth a look. It has some very nice graphics generated in R with images and sample code.
Here's the sample code from the addictedtor site:
2-d histogram:
require(gplots)
# example data, bivariate normal, no correlation
x <- rnorm(2000, sd=4)
y <- rnorm(2000, sd=1)
# separate scales for each axis, this looks circular
hist2d(x,y, nbins=50, col = c("white",heat.colors(16)))
rug(x,side=1)
rug(y,side=2)
box()
smoothscatter:
library("geneplotter") ## from BioConductor
require("RColorBrewer") ## from CRAN
x1 <- matrix(rnorm(1e4), ncol=2)
x2 <- matrix(rnorm(1e4, mean=3, sd=1.5), ncol=2)
x <- rbind(x1,x2)
layout(matrix(1:4, ncol=2, byrow=TRUE))
op <- par(mar=rep(2,4))
smoothScatter(x, nrpoints=0)
smoothScatter(x)
smoothScatter(x, nrpoints=Inf,
colramp=colorRampPalette(brewer.pal(9,"YlOrRd")),
bandwidth=40)
colors <- densCols(x)
plot(x, col=colors, pch=20)
par(op)
Related
How can I get the area under overlapping density curves?
How can I solve the problem with R? (There is a solution for python here: Calculate overlap area of two functions )
set.seed(1234)
df <- data.frame(
sex=factor(rep(c("F", "M"), each=200)),
weight=round(c(rnorm(200, mean=55, sd=5),
rnorm(200, mean=65, sd=5)))
)
(Source: http://www.sthda.com/english/wiki/ggplot2-density-plot-quick-start-guide-r-software-and-data-visualization )
ggplot(df, aes(x=weight, color=sex, fill=sex)) +
geom_density(aes(y=..density..), alpha=0.5)
"The points used in the plot are returned by ggplot_build(), so you can access them." So now, I have the points, and I can feed them to approxfun, but my problem is that i don't know how to subtract the density functions.
Any help greatly appreciated! (And I believe in high demand, there is no solution for this readily available.)
I was looking for a way to do this for empirical data, and had the problem of multiple intersections as mentioned by user5878028. After some digging I found a very simple solution, even for a total R noob like me:
Install and load the libraries "overlapping" (which performs the calculation) and "lattice" (which displays the result):
library(overlapping)
library(lattice)
Then define a variable "x" as a list that contains the two density distributions you want to compare. For this example, the two datasets "data1" and "data2" are both columns in a text file called "yourfile":
x <- list(X1=yourfile$data1, X2=yourfile$data2)
Then just tell it to display the output as a plot which will also display the estimated % overlap:
out <- overlap(x, plot=TRUE)
I hope this helps someone like it helped me! Here's an example overlap plot
I will make a few base R plots, but the plots are not actually part of
the solution. They are just there to confirm that I am getting the right
answer.
You can get each of the density functions and solve for where they intersect.
## Create the two density functions and display
FDensity = approxfun(density(df$weight[df$sex=="F"], from=40, to=80))
MDensity = approxfun(density(df$weight[df$sex=="M"], from=40, to=80))
plot(FDensity, xlim=c(40,80), ylab="Density")
curve(MDensity, add=TRUE)
Now solve for the intersection
## Solve for the intersection and plot to confirm
FminusM = function(x) { FDensity(x) - MDensity(x) }
Intersect = uniroot(FminusM, c(40, 80))$root
points(Intersect, FDensity(Intersect), pch=20, col="red")
Now we can just integrate to get the area of the overlap.
integrate(MDensity, 40,Intersect)$value +
integrate(FDensity, Intersect, 80)$value
[1] 0.2952838
The above two proposed methods give different results.
If the data in the first answer is given to the overlap function it will result in overlap% of 0.18, while the first one results in overlap% of 0.29.
X1 = df$weight[df$sex=="F"]
X2 = df$weight[df$sex=="M"]
x=list(X1=X1, X2=X2)
out <- overlap(x, plot=TRUE)
out$OV
X1-X2
0.1754
I've followed the instructions on this website from STHDA to plot correlation matrices and correlograms in R. The website and examples are really good. However, I'd like to plot the upper part of the correlogram over the upper part of the correlation matrix.
Here's the code:
library(PerformanceAnalytics)
chart.Correlation(mtcars, histogram=TRUE, pch=19)
This should give me the correlation matrix using scatter plots, together with the histogram, which I'd like to maintain. But for the upper part of the plot, I'd like to have the correlogram obtained from this code:
library(corrplot)
corrplot(cor(mtcars), type="upper", order="hclust", tl.col="black", tl.srt=45)
The obvious way of doing it is exporting all graphs in pdf and then work with Inkscape, but it would be nicer if I could get this directly from R. Is there any possible way for doing this?
Thanks.
The trick to using the panel functions within pairs is found in help(pairs):
A panel function should not attempt to start a new plot, but just plot within a given coordinate system: thus 'plot' and 'boxplot' are not panel functions.
So, you should use graphic-adding functions, such as points, lines, polygon, or perhaps (when available) plot(..., add=TRUE), but not a straight-up plot. What you were suggesting in your comment (with SpatialPolygons) might have worked with some prodding if you actually tried to plot it on a device vice just returning it from your plotting function.
In my example below, I actually do "create a new plot", but I cheat (based on this SO post) by adding a second plot on top of the one already there. I do this to shortcut an otherwise necessary scale/shift, which would still not be perfect since you appear to want a "perfect circle", something that can really only be guaranteed with asp=1 (aspect ratio fixed at 1:1).
colorRange <- c('#69091e', '#e37f65', 'white', '#aed2e6', '#042f60')
## colorRamp() returns a function which takes as an argument a number
## on [0,1] and returns a color in the gradient in colorRange
myColorRampFunc <- colorRamp(colorRange)
panel.cor <- function(w, z, ...) {
correlation <- cor(w, z)
## because the func needs [0,1] and cor gives [-1,1], we need to
## shift and scale it
col <- rgb( myColorRampFunc( (1+correlation)/2 )/255 )
## square it to avoid visual bias due to "area vs diameter"
radius <- sqrt(abs(correlation))
radians <- seq(0, 2*pi, len=50) # 50 is arbitrary
x <- radius * cos(radians)
y <- radius * sin(radians)
## make them full loops
x <- c(x, tail(x,n=1))
y <- c(y, tail(y,n=1))
## I trick the "don't create a new plot" thing by following the
## advice here: http://www.r-bloggers.com/multiple-y-axis-in-a-r-plot/
## This allows
par(new=TRUE)
plot(0, type='n', xlim=c(-1,1), ylim=c(-1,1), axes=FALSE, asp=1)
polygon(x, y, border=col, col=col)
}
pairs(mtcars, upper.panel=panel.cor)
You can manipulate the size of the circles -- at the expense of unbiased visualization -- by playing with the radius. The colors I took directly from the page you linked to originally.
Similar functions can be used for your lower and diagonal panels.
I've run a 2d simulation in some modelling software from which i've got an export of x,y point locations with a set of 6 attributes. I wish to recreate a figure that combines the data, like this:
The ellipses and the background are shaded according to attribute 1 (and the borders of these are of course representing the model geometry, but I don't think I can replicate that), the isolines are contours of attribute 2, and the arrow glyphs are from attributes 3 (x magnitude) and 4 (y magnitude).
The x,y points are centres of the triangulated mesh I think, and look like this:
I want to know how I can recreate a plot like this with R. To start with I have irregularly-spaced data due to it being exported from an irregular mesh. That's immediately where I get stuck with R, having only ever used it for producing box-and-whisper plots and the like.
Here's the data:
https://dl.dropbox.com/u/22417033/Ellipses_noheader.txt
Edit: fields: x, y, heat flux (x), heat flux (y), thermal conductivity, Temperature, gradT (x), gradT (y).
names(Ellipses) <- c('x','y','dfluxx','dfluxy','kxx','Temps','gradTx','gradTy')
It's quite easy to make the lower plot (making the assumption that there is a dataframe named 'edat' read in with:
edat <- read.table(file=file.choose())
with(edat, plot(V1,V2), cex=0.2)
Things get a bit more beautiful with:
with(edat, plot(V1,V2, cex=0.2, col=V5))
So I do not think your original is being faithfully represented by the data. The contour lines are NOT straight across the "conductors". I call them "conductors" because this looks somewhat like iso-potential lines in electrostatics. I'm adding some text here to serve as a search handle for others who might be searching for plotting problems in real world physics: vector-field (the arrows) , heat equations, gradient, potential lines.
You can then overlay the vector field with:
with(edat, arrows(V1,V2, V1-20*V6*V7, V2-20*V6*V8, length=0.04, col="orange") )
You could"zoom in" with xlim and ylim:
with(edat, plot(V1,V2, cex=0.3, col=V5, xlim=c(0, 10000), ylim=c(-8000, -2000) ))
with(edat, arrows(V1,V2, V1-20*V6*V7, V2-20*V6*V8, length=0.04, col="orange") )
Guessing that the contour requested if for the Temps variable. Take your pick of contourplots.
require(akima)
intflow<- with(edat, interp(x=x, y=y, z=Temps, xo=seq(min(x), max(x), length = 410),
yo=seq(min(y), max(y), length = 410), duplicate="mean", linear=FALSE) )
require(lattice)
contourplot(intflow$z)
filled.contour(intflow)
with( intflow, contour(x=x, y=y, z=z) )
The last one will mix with the other plotting examples since those were using base plotting functions. You may need to switch to points instead of plot.
There are several parts to your plot so you will probably need several tools to make the different parts.
The background and ellipses can be created with polygon (once you figure where they should be).
The contourLines function can calculate the contour lines for you which you can add with the lines function (or contour has and add argument and could probably be used to add the lines directly).
The akima package has a function interp which can estimate values on a grid given the values ungridded.
The my.symbols function along with ms.arrows, both from the TeachingDemos package, can be used to draw the vector field.
#DWin is right to say that your graph don't represent faithfully your data, so I would advice to follow his answer. However here is how to reproduce (the closest I could) your graph:
Ellipses <- read.table(file.choose())
names(Ellipses) <- c('x','y','dfluxx','dfluxy','kxx','Temps','gradTx','gradTy')
require(splancs)
require(akima)
First preparing the data:
#First the background layer (the 'kxx' layer):
# Here the regular grid on which we're gonna do the interpolation
E.grid <- with(Ellipses,
expand.grid(seq(min(x),max(x),length=200),
seq(min(y),max(y),length=200)))
names(E.grid) <- c("x","y") # Without this step, function inout throws an error
E.grid$Value <- rep(0,nrow(E.grid))
#Split the dataset according to unique values of kxx
E.k <- split(Ellipses,Ellipses$kxx)
# Find the convex hull delimiting each of those values domain
E.k.ch <- lapply(E.k,function(X){X[chull(X$x,X$y),]})
for(i in unique(Ellipses$kxx)){ # Pick the value for each coordinate in our regular grid
E.grid$Value[inout(E.grid[,1:2],E.k.ch[names(E.k.ch)==i][[1]],bound=TRUE)]<-i
}
# Then the regular grid for the second layer (Temp)
T.grid <- with(Ellipses,
interp(x,y,Temps, xo=seq(min(x),max(x),length=200),
yo=seq(min(y),max(y),length=200),
duplicate="mean", linear=FALSE))
# The regular grids for the arrow layer (gradT)
dx <- with(Ellipses,
interp(x,y,gradTx,xo=seq(min(x),max(x),length=15),
yo=seq(min(y),max(y),length=10),
duplicate="mean", linear=FALSE))
dy <- with(Ellipses,
interp(x,y,gradTy,xo=seq(min(x),max(x),length=15),
yo=seq(min(y),max(y),length=10),
duplicate="mean", linear=FALSE))
T.grid2 <- with(Ellipses,
interp(x,y,Temps, xo=seq(min(x),max(x),length=15),
yo=seq(min(y),max(y),length=10),
duplicate="mean", linear=FALSE))
gradTgrid<-expand.grid(dx$x,dx$y)
And then the plotting:
palette(grey(seq(0.5,0.9,length=5)))
par(mar=rep(0,4))
plot(E.grid$x, E.grid$y, col=E.grid$Value,
axes=F, xaxs="i", yaxs="i", pch=19)
contour(T.grid, add=TRUE, col=colorRampPalette(c("blue","red"))(15), drawlabels=FALSE)
arrows(gradTgrid[,1], gradTgrid[,2], # Here I multiply the values so you can see them
gradTgrid[,1]-dx$z*40*T.grid2$z, gradTgrid[,2]-dy$z*40*T.grid2$z,
col="yellow", length=0.05)
To understand in details how this code works, I advise you to read the following help pages: ?inout, ?chull, ?interp, ?expand.grid and ?contour.
Suppose I have a number of replications of bivariate experiments which I wish to display simultaneously in hexagonally binned plots, with common cell counts. Is there existing code to do this? Is there an easy way to modify the hexbin package to do this for me?
For example:
library(hexbin)
x <- replicate(9, rnorm(10000), simplify=FALSE)
y <- replicate(9, rnorm(10000), simplify=FALSE)
h <- mapply(hexbin, x, y)
par(mfrow=c(3,3))
lapply(h, plot)
This code doesn't display a grid of hexbin plots with common cell counts, but I'd like it to.
hexbin objects are plotted using grid graphics so your par(mfrow=c(3,3)) does not do anything. Each graph is plotted on a separate page. To get the details of the plot options:
?gplot.hexbin
In this case, we want to set maxcnt to the largest cell count:
lapply(h, plot, maxcnt=max(unlist(lapply(h, function(x) max(x#count)))))
This will apply the same legend to each graph.
I am using the sm package in R to draw a density plot of several variables with different sample sizes, like this:
var1 <- density(vars1[,1])
var2 <- density(vars2[,1])
var3 <- density(vars3[,1])
pdf(file="density.pdf",width=8.5,height=8)
plot(var1,col="BLUE")
par(new=T)
plot(var2,axes=FALSE,col="RED")
par(new=T)
plot(var3,axes=FALSE,col="GREEN")
dev.off()
The problem I'm having, is that I want the y-axis to show the proportions so I can compare the different variables with each other in a more meaningful way. The maxima of all three density plots are now exactly the same, and I'm pretty sure that they wouldn't be if the y-axis showed proportions. Any suggestions? Many thanks!
Edit:
I just learned that I should not plot on top of an existing plot, so now the plotting part of the code looks like this:
pdf(file="density.pdf",width=8.5,height=8)
plot(var1,col="BLUE")
lines(var2,col="RED")
lines(var3,col="GREEN")
dev.off()
The maxima of those lines however are now very much in line with the sample size differences. Is there a way to put the proportions on the y-axis for all three variables, so the area under the curve is equal for all three variables? Many thanks!
Don't plot on top of an existing plot, because they axes may be different. Instead, use lines() to plot the second and third densities after plotting the first. If necessary, adjust the ylim parameter in plot() so that they all fit.
An example for how sample size ought not matter:
set.seed(1)
D1 <- density(rnorm(1000))
D2 <- density(rnorm(10000))
D3 <- density(rnorm(100000))
plot(D1$x,D1$y,type='l',col="red",ylim=c(0,.45))
lines(D2$x,D2$y,lty=2,col="blue")
lines(D3$x,D3$y,lty=3,col="green")
You could make tim's solution a little more flexible by not hard-coding in the limits.
plot(D1$x,D1$y,type='l',col="red",ylim=c(0, max(sapply(list(D1, D2, D3),
function(x) {max(x$y)}))))
This would also cater for Vincent's point that the density functions are not necessarily constrained in their range.