I would like to superimpose, on each lattice histogram panel, an additional histogram (which will be the same in each panel). I want the overlayed histogram to have solid borders but empty fill (col), to allow comparison with the underlying histograms.
That is, the end result will be a series of panels, each with a different colored histogram, and each with the same extra outline histogram on top of the colored histogram.
Here's something that I tried, but it just produces empty panels:
foo.df <- data.frame(x=rnorm(40), categ=c(rep("A", 20), rep("B", 20)))
bar.df <- data.frame(x=rnorm(20))
histogram(~ x | categ, data=foo.df,
panel=function(...){histogram(...);
histogram(~ x, data=bar.df, col=NULL)})
(My guess is that I need to use panel.superpose, but this function is somewhat confusing. Sarkar's book doesn't explain how to use it, and the R help page has no examples. I'm finding it difficult to make sense of the panel.superpose help page without already having a basic understanding. There are a very small number of examples that I've found on the web, but I have been unable to figure out what aspects of those examples apply to my case. This answer is surely relevant, but I don't understand its use of panel.groups, and the example overlays three different groups from a single dataframe, whereas I want to repeatedly overlay the same data on multiple panels that also have different data .)
I continued working on this problem, and came up with an answer. I had been on the right track but got several crucial details wrong. Comments in the code below spell out important points.
# Main data, which will be displayed as solid histograms, different in each panel:
foo.df <- data.frame(y=rnorm(40), cat=c(rep("A", 20), rep("B", 20)))
# Comparison data: This will be displayed as an outline histogram in each panel:
bar.df <- data.frame(y=rnorm(30)-2)
# Define some vectors that we'll use in the histogram call.
# These have to be adjusted for the data by trial and error.
# Usually, panel.histogram will figure out reasonable default values for these.
# However, the two calls to panel.histogram below may figure out different values,
# producing pairs of histograms that aren't comparable.
bks <- seq(-5,3,0.5) # breaks that define the bar bins
yl <- c(0,50) # height of plot
# The key is to coordinate breaks in the two panel.histogram calls below.
# The first one inherits the breaks from the top-level call through '...' .
# Using "..." in the second call generates an error, so I specify parameters explicitly.
# It's not necessary to specify type="percent" at the top level, since that's the default,
# but it is necessary to specify it in the second panel.histogram call.
histogram(~ y | cat, data=foo.df, ylim=yl, breaks=bks, type="percent", border="cyan",
panel=function(...){panel.histogram(...)
panel.histogram(x=bar.df$y, col="transparent",
type="percent", breaks=bks)})
# col="transparent" is what makes the second set of bars into outlines.
# In the first set of bars, I set the border color to be the same as the value of col
# (cyan by default) rather than using border="transparent" because otherwise a filled
# bar with the same number of points as an outline bar will be slightly smaller.
Related
This answer shows how to use groups and panel.superpose to display overlapping histograms in the same panel, assigning different colors to each histogram. In addition, I want to give each histogram a different border color. (This will allow me to display one histogram as solid bars without a border, overlayed with a transparent, all-border histogram. The example below is a little different for the sake of clarity.)
Although it's possible to use border= to use different border colors in the plot, they are not assigned to groups as fill colors are with col=. If you give border= a sequence of colors, it seems to cycle through them one bar at at time. If the two histograms overlap, the effect is a bit silly (see below).
Is there a way to give each group a specific border color?
# This illustrates the problem: Assignment of border colors to bars ignores grouping:
# make some data
foo.df <- data.frame(x=c(rnorm(10),rnorm(10)+2), cat=c(rep("A", 10),rep("B", 10)))
# plot it
histogram(~ x, groups=cat, data=foo.df, ylim=c(0,75), breaks=seq(-3, 5, 0.5), lwd=2,
panel=function(...)panel.superpose(..., panel.groups=panel.histogram,
col=c("transparent", "cyan"),
border=c(rep("black", 3), rep("red", 3))))
Note that you can't just count how many bars there are in each group and provide those numbers to rep in the border setting. If the two histograms overlap, at least one of the histograms will use two border colors.
(It's the panel.superpose code that places the groups on the same panel and that assigns the colors. I don't have a deep understanding of it.)
panel.histogram() doesn't have a formal groups= argument, and if you examine its code, you'll see that it handles any supplied groups= argument differently and in a less standard way than panel.*() functions that do. The upshot of that design decision is that (as you've found) it's not in general easy to pass in to it vectors of graphical parameters specifying per-group appearance
As a workaround, I'd suggest using latticeExtra's +() and as.layer() functions to overlay a number of separate histogram() plots, one for each group. Here's how you might do that:
library(lattice)
library(latticeExtra)
## Split your data by group into separate data.frames
foo.df <- data.frame(x=c(rnorm(10),rnorm(10)+2), cat=c(rep("A", 10),rep("B", 10)))
foo.A <- subset(foo.df, cat=="A")
foo.B <- subset(foo.df, cat=="B")
## Use calls to `+ as.layer()` to layer each group's histogram onto previous ones
histogram(~ x, data=foo.A, ylim=c(0,75), breaks=seq(-3, 5, 0.5),
lwd=2, col="transparent", border="black") +
as.layer(
histogram(~ x, data=foo.B, ylim=c(0,75), breaks=seq(-3, 5, 0.5),
lwd=2, col="cyan", border="red")
)
I am trying to plot several histograms for the same data set, but with different numbers of bins. I am using Gadfly.
Suppose x is just an array of real values, plotting each histogram works:
plot(x=x, Geom.histogram(bincount=10))
plot(x=x, Geom.histogram(bincount=20))
But I'm trying to put all the histograms together. I've added the number of bins as another dimension to my data set:
x2 = vcat(hcat(10*ones(length(x)), x), hcat(20*ones(length(x)), x)
df = DataFrame(Bins=x2[:,1], X=x2[:,2])
Is there any way to send the number of bins (the value from the first column) to Geom.histogram when using Geom.subplot_grid? Something like this:
plot(df, x="X", ygroup="Bins", Geom.subplot_grid(Geom.histogram(?)))
I think you would be better off not using subplot grid at that point, and instead just combine them with vstack or hstack. From the docs
Plots can also be stacked horizontally with ``hstack`` or vertically with
``vstack``. This allows more customization in regards to tick marks, axis
labeling, and other plot details than is available with ``subplot_grid``.
I had some problems while trying to plot a histogram to show the frequency of every value while plotting the value as well. For example, suppose I use the following code:
x <- sample(1:10,1000,replace=T)
hist(x,label=TRUE)
The result is a plot with labels over the bar, but merging the frequencies of 1 and 2 in a single bar.
Apart from separate this bar in two others for 1 and 2, I also need to put the values under each bar.
For example, with the code above I would have the number 10 under the tick at the right margin of its bar, and I needed to plot the values right under the bars.
Is there any way to do both in a single histogram with hist function?
Thanks in advance!
Calling hist silently returns information you can use to modify the plot. You can pull out the midpoints and the heights and use that information to put the labels where you want them. You can use the pos argument in text to specify where the label should be in relation to the point (thanks #rawr)
x <- sample(1:10,1000,replace=T)
## Histogram
info <- hist(x, breaks = 0:10)
with(info, text(mids, counts, labels=counts, pos=1))
I'm looking to plot a set of sparklines in R with just a 0 and 1 state that looks like this:
Does anyone know how I might create something like that ideally with no extra libraries?
I don't know of any simple way to do this, so I'm going to build up this plot from scratch. This would probably be a lot easier to design in illustrator or something like that, but here's one way to do it in R (if you don't want to read the whole step-by-step, I provide my solution wrapped in a reusable function at the bottom of the post).
Step 1: Sparklines
You can use the pch argument of the points function to define the plotting symbol. ASCII symbols are supported, which means you can use the "pipe" symbol for vertical lines. The ASCII code for this symbol is 124, so to use it for our plotting symbol we could do something like:
plot(df, pch=124)
Step 2: labels and numbers
We can put text on the plot by using the text command:
text(x,y,char_vect)
Step 3: Alignment
This is basically just going to take a lot of trial and error to get right, but it'll help if we use values relative to our data.
Here's the sample data I'm working with:
df = data.frame(replicate(4, rbinom(50, 1, .7)))
colnames(df) = c('steps','atewell','code','listenedtoshell')
I'm going to start out by plotting an empty box to use as our canvas. To make my life a little easier, I'm going to set the coordinates of the box relative to values meaningful to my data. The Y positions of the 4 data series will be the same across all plotting elements, so I'm going to store that for convenience.
n=ncol(df)
m=nrow(df)
plot(1:m,
seq(1,n, length.out=m),
# The following arguments suppress plotting values and axis elements
type='n',
xaxt='n',
yaxt='n',
ann=F)
With this box in place, I can start adding elements. For each element, the X values will all be the same, so we can use rep to set that vector, and seq to set the Y vector relative to Y range of our plot (1:n). I'm going to shift the positions by percentages of the X and Y ranges to align my values, and modified the size of the text using the cex parameter. Ultimately, I found that this works out:
ypos = rev(seq(1+.1*n,n*.9, length.out=n))
text(rep(1,n),
ypos,
colnames(df), # These are our labels
pos=4, # This positions the text to the right of the coordinate
cex=2) # Increase the size of the text
I reversed the sequence of Y values because I built my sequence in ascending order, and the values on the Y axis in my plot increase from bottom to top. Reversing the Y values then makes it so the series in my dataframe will print from top to bottom.
I then repeated this process for the second label, shifting the X values over but keeping the Y values the same.
text(rep(.37*m,n), # Shifted towards the middle of the plot
ypos,
colSums(df), # new label
pos=4,
cex=2)
Finally, we shift X over one last time and use points to build the sparklines with the pipe symbol as described earlier. I'm going to do something sort of weird here: I'm actually going to tell points to plot at as many positions as I have data points, but I'm going to use ifelse to determine whether or not to actually plot a pipe symbol or not. This way everything will be properly spaced. When I don't want to plot a line, I'll use a 'space' as my plotting symbol (ascii code 32). I will repeat this procedure looping through all columns in my dataframe
for(i in 1:n){
points(seq(.5*m,m, length.out=m),
rep(ypos[i],m),
pch=ifelse(df[,i], 124, 32), # This determines whether to plot or not
cex=2,
col='gray')
}
So, piecing it all together and wrapping it in a function, we have:
df = data.frame(replicate(4, rbinom(50, 1, .7)))
colnames(df) = c('steps','atewell','code','listenedtoshell')
BinarySparklines = function(df,
L_adj=1,
mid_L_adj=0.37,
mid_R_adj=0.5,
R_adj=1,
bottom_adj=0.1,
top_adj=0.9,
spark_col='gray',
cex1=2,
cex2=2,
cex3=2
){
# 'adJ' parameters are scalar multipliers in [-1,1]. For most purposes, use [0,1].
# The exception is L_adj which is any value in the domain of the plot.
# L_adj < mid_L_adj < mid_R_adj < R_adj
# and
# bottom_adj < top_adj
n=ncol(df)
m=nrow(df)
plot(1:m,
seq(1,n, length.out=m),
# The following arguments suppress plotting values and axis elements
type='n',
xaxt='n',
yaxt='n',
ann=F)
ypos = rev(seq(1+.1*n,n*top_adj, length.out=n))
text(rep(L_adj,n),
ypos,
colnames(df), # These are our labels
pos=4, # This positions the text to the right of the coordinate
cex=cex1) # Increase the size of the text
text(rep(mid_L_adj*m,n), # Shifted towards the middle of the plot
ypos,
colSums(df), # new label
pos=4,
cex=cex2)
for(i in 1:n){
points(seq(mid_R_adj*m, R_adj*m, length.out=m),
rep(ypos[i],m),
pch=ifelse(df[,i], 124, 32), # This determines whether to plot or not
cex=cex3,
col=spark_col)
}
}
BinarySparklines(df)
Which gives us the following result:
Try playing with the alignment parameters and see what happens. For instance, to shrink the side margins, you could try decreasing the L_adj parameter and increasing the R_adj parameter like so:
BinarySparklines(df, L_adj=-1, R_adj=1.02)
It took a bit of trial and error to get the alignment right for the result I provided (which is what I used to inform the default values for BinarySparklines), but I hope I've given you some intuition about how I achieved it and how moving things using percentages of the plotting range made my life easier. In any event, I hope this serves as both a proof of concept and a template for your code. I'm sorry I don't have an easier solution for you, but I think this basically gets the job done.
I did my prototyping in Rstudio so I didn't have to specify the dimensions of my plot, but for posterity I had 832 x 456 with the aspect ratio maintained.
I try to overlay two histograms in the same plane but the option Probability=TRUE (relative frequencies) in hist() is not effective with the code below. It is a problem because the two samples have very different sizes (length(cl1)=9 and length(cl2)=339) and, with this script, I cannot vizualize differences between both histograms because each shows frequencies. How can I overlap two histograms with the same bin width, showing relative frequencies?
c1<-hist(dataList[["cl1"]],xlim=range(minx,maxx),breaks=seq(minx,maxx,pasx),col=rgb(1,0,0,1/4),main=paste(paramlab,"Group",groupnum,"cl1",sep=" "),xlab="",probability=TRUE)
c2<-hist(dataList[["cl2"]],xlim=range(minx,maxx),breaks=seq(minx,maxx,pasx),col=rgb(0,0,1,1/4),main=paste(paramlab,"Group",groupnum,"cl2",sep=" "),xlab="",probability=TRUE)
plot(c1, col=rgb(1,0,0,1/4), xlim=c(minx,maxx), main=paste(paramlab,"Group",groupnum,sep=" "),xlab="")# first histogram
plot(c2, col=rgb(0,0,1,1/4), xlim=c(minx,maxx), add=T)
cl1Col <- rgb(1,0,0,1/4)
cl2Col <- rgb(0,0,1,1/4)
legend('topright',c('Cl1','Cl2'),
fill = c(cl1Col , cl2Col ), bty = 'n',
border = NA)
Thanks in advance for your help!
When you call plot on an object of class histogram (like c1), it calls the S3 method for the histogram. Namely, plot.histogram. You can see the code for this function if you type graphics:::plot.histogram and you can see its help under ?plot.histogram. The help file for that function states:
freq logical; if TRUE, the histogram graphic is to present a
representation of frequencies, i.e, x$counts; if FALSE, relative
frequencies (probabilities), i.e., x$density, are plotted. The default
is true for equidistant breaks and false otherwise.
So, when plot renders a histogram it doesn't use the previously specified probability or freq arguments, it tries to figure it out for itself. The reason for this is obvious if you dig around inside c1, it contains all of the data necessarily for the plot, but does not specify how it should be rendered.
So, the solution is to reiterate the argument freq=FALSE when you run the plot functions. Notably, freq=FALSE works whereas probability=TRUE does not because plot.histogram does not have a probability option. So, your plot code will be:
plot(c1, col=rgb(1,0,0,1/4), xlim=c(minx,maxx), main=paste(paramlab,"Group",groupnum,sep=" "),xlab="",freq=FALSE)# first histogram
plot(c2, col=rgb(0,0,1,1/4), xlim=c(minx,maxx), add=T, freq=FALSE)
This all seems like a oversight/idiosyncratic decision (or lack thereof) on the part of the R devs. To their credit it is appropriately documented and is not "unexpected behavior" (although I certainly didn't expect it). I wonder where such oddness should be reported, if it should be reported at all.