I am using filled.contour() to plot data stored in a matrix. The data is generated by a (highly) non-linear function, hence its distribution is not uniform at all and the range is very large.
Consequently, I have to use the option "levels" to fine tune the plot. However, filled.contour() does not use these custom levels to make an appropriate color key for the heat map, which I find quite surprising.
Here is a simple example of what I mean:
x = c(20:200/100)
y = c(20:200/100)
z = as.matrix(exp(x^2)) %*% exp(y^2)
filled.contour(x=x,y=y,z=z,color.palette=colorRampPalette(c('green','yellow','red')),levels=c(1:60/3,30,50,150,250,1000,3000))
As you can see, the color key produced with the code above is pretty much useless. I would like to use some sort of projection (perhaps sin(x) or tanh(x)?), so that the upper range is not over-represented in the key (in a linear way).
At this point, I would like to:
1) know if there is something very simple/obvious I am missing, e.g.: an option to make this "key range adapting" automagically;
2) seek suggestions/help on how to do it myself, should the answer to 1) be negative.
Thanks a lot!
PS: I apologize for my English, which is far from perfect. Please let me know if you need me to clarify anything.
I feel your frustration. I never found a way to do this with filled contour, so have usually reverted to using image and then adding my own scale as a separate plot. I wrote the function image.scale to help out with this (link). Below is an example of how you can supply a log-transform to your scale in order to stretch out the small values - then label the scale with the non-log-transformed values as labels:
Example:
source("image.scale.R") # http://menugget.blogspot.de/2011/08/adding-scale-to-image-plot.html
x = c(20:200/100)
y = c(20:200/100)
z = as.matrix(exp(x^2)) %*% exp(y^2)
pal <- colorRampPalette(c('green','yellow','red'))
breaks <- c(1:60/3,30,50,150,250,1000,3000)
ncolors <- length(breaks)-1
labs <- c(0.5, 1, 3,30,50,150,250,1000,3000)
#x11(width=6, height=6)
layout(matrix(1:2, nrow=1, ncol=2), widths=c(5,1), heights=c(6))
layout.show(2)
par(mar=c(5,5,1,1))
image(x=x,y=y,z=log(z), col=pal(ncolors), breaks=log(breaks))
box()
par(mar=c(5,0,1,4))
image.scale(log(z), col=pal(ncolors), breaks=log(breaks), horiz=FALSE, xlab="", ylab="", xaxt="n", yaxt="n")
axis(4, at=log(labs), labels=labs)
box()
Result:
Related
I'm doing stochastic dominance analysis with diferent income distributions using Pen's Parade. I can plot a single Pen's Parade using Pen function from ineq package, but I need a visual comparison and I want multiple lines in the same image. I don't know how extract values from the function, so I can't do this.
I have the following reproducible example:
set.seed(123)
x <- rnorm(100)
y <- rnorm(100, mean = 0.2)
library(ineq)
Pen(x)
Pen(y)
I obtain the following plots:
I want obtain sometime as the following:
You can use add = TRUE:
set.seed(123)
x <- rnorm(100)
y <- rnorm(100, mean = 0.2)
library(ineq)
Pen(x); Pen(y, add = TRUE)
From help("Pen"):
add logical. Should the plot be added to an existing plot?
While the solution mentioned by M-M in the comments is a more general solution, in this specific case it produces a busy Y axis:
Pen(x)
par(new = TRUE)
Pen(y)
I would generalize the advice for plotting functions in this way:
Check the plotting function's help file. If it has an add argument, use that.
Otherwise, use the par(new = TRUE) technique
Update
As M-M helpfully mentions in the comments, their more general solution will not produce a busy Y axis if you manually suppress the Y axis on the second plot:
Pen(x)
par(new = TRUE)
Pen(y, yaxt = "n")
Looking at ?ineq::Pen() it seems to work like plot(); therefore, followings work for you.
Pen(x)
Pen(y, add=T)
Note: However, add=T cuts out part of your data since second plot has points which fall out of the limit of the first.
Update on using par(new=T):
Using par(new=T) basically means overlaying two plots on top of each other; hence, it is important to make them with the same scale. We can achieve that by setting the same axis limits. That said, while using add=T argument it is desired to set limits of the axis to not loose any part of data. This is the best practice for overlaying two plots.
Pen(x, ylim=c(0,38), xlim=c(0,1))
par(new=T)
Pen(y, col="red", ylim=c(0,38), xlim=c(0,1), yaxt='n', xaxt='n')
Essentially, you can do the same with add=T.
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 am creating a plot in R and I dont like the x axis values being plotted by R.
For example:
x <- seq(10,200,10)
y <- runif(x)
plot(x,y)
This plots a graph with the following values on the X axis:
50, 100, 150, 200
However, I want to plot the 20 values 10,20, 30 ... 200 stored in variable x, as the X axis values. I have scoured through countless blogs and the terse manual - after hours of searching, the closest I've come to finding anything useful is the following (summarized) instructions:
call plot() or par(), specifying argument xaxt='n'
call axis() e.g. axis(side = 1, at = seq(0, 10, by = 0.1), labels = FALSE, tcl = -0.2)
I tried it and the resulting plot had no x axis values at all. Is it possible that someone out there knows how to do this? I can't believe that no one has ever tried to do this before.
You'll find the answer to your question in the help page for ?axis.
Here is one of the help page examples, modified with your data:
Option 1: use xaxp to define the axis labels
plot(x,y, xaxt="n")
axis(1, xaxp=c(10, 200, 19), las=2)
Option 2: Use at and seq() to define the labels:
plot(x,y, xaxt="n")
axis(1, at = seq(10, 200, by = 10), las=2)
Both these options yield the same graphic:
PS. Since you have a large number of labels, you'll have to use additional arguments to get the text to fit in the plot. I use las to rotate the labels.
Take a closer look at the ?axis documentation. If you look at the description of the labels argument, you'll see that it is:
"a logical value specifying whether (numerical) annotations are
to be made at the tickmarks,"
So, just change it to true, and you'll get your tick labels.
x <- seq(10,200,10)
y <- runif(x)
plot(x,y,xaxt='n')
axis(side = 1, at = x,labels = T)
# Since TRUE is the default for labels, you can just use axis(side=1,at=x)
Be careful that if you don't stretch your window width, then R might not be able to write all your labels in. Play with the window width and you'll see what I mean.
It's too bad that you had such trouble finding documentation! What were your search terms? Try typing r axis into Google, and the first link you will get is that Quick R page that I mentioned earlier. Scroll down to "Axes", and you'll get a very nice little guide on how to do it. You should probably check there first for any plotting questions, it will be faster than waiting for a SO reply.
Hope this coding will helps you :)
plot(x,y,xaxt = 'n')
axis(side=1,at=c(1,20,30,50),labels=c("1975","1980","1985","1990"))
In case of plotting time series, the command ts.plot requires a different argument than xaxt="n"
require(graphics)
ts.plot(ldeaths, mdeaths, xlab="year", ylab="deaths", lty=c(1:2), gpars=list(xaxt="n"))
axis(1, at = seq(1974, 1980, by = 2))
I am creating a plot in R and I dont like the x axis values being plotted by R.
For example:
x <- seq(10,200,10)
y <- runif(x)
plot(x,y)
This plots a graph with the following values on the X axis:
50, 100, 150, 200
However, I want to plot the 20 values 10,20, 30 ... 200 stored in variable x, as the X axis values. I have scoured through countless blogs and the terse manual - after hours of searching, the closest I've come to finding anything useful is the following (summarized) instructions:
call plot() or par(), specifying argument xaxt='n'
call axis() e.g. axis(side = 1, at = seq(0, 10, by = 0.1), labels = FALSE, tcl = -0.2)
I tried it and the resulting plot had no x axis values at all. Is it possible that someone out there knows how to do this? I can't believe that no one has ever tried to do this before.
You'll find the answer to your question in the help page for ?axis.
Here is one of the help page examples, modified with your data:
Option 1: use xaxp to define the axis labels
plot(x,y, xaxt="n")
axis(1, xaxp=c(10, 200, 19), las=2)
Option 2: Use at and seq() to define the labels:
plot(x,y, xaxt="n")
axis(1, at = seq(10, 200, by = 10), las=2)
Both these options yield the same graphic:
PS. Since you have a large number of labels, you'll have to use additional arguments to get the text to fit in the plot. I use las to rotate the labels.
Take a closer look at the ?axis documentation. If you look at the description of the labels argument, you'll see that it is:
"a logical value specifying whether (numerical) annotations are
to be made at the tickmarks,"
So, just change it to true, and you'll get your tick labels.
x <- seq(10,200,10)
y <- runif(x)
plot(x,y,xaxt='n')
axis(side = 1, at = x,labels = T)
# Since TRUE is the default for labels, you can just use axis(side=1,at=x)
Be careful that if you don't stretch your window width, then R might not be able to write all your labels in. Play with the window width and you'll see what I mean.
It's too bad that you had such trouble finding documentation! What were your search terms? Try typing r axis into Google, and the first link you will get is that Quick R page that I mentioned earlier. Scroll down to "Axes", and you'll get a very nice little guide on how to do it. You should probably check there first for any plotting questions, it will be faster than waiting for a SO reply.
Hope this coding will helps you :)
plot(x,y,xaxt = 'n')
axis(side=1,at=c(1,20,30,50),labels=c("1975","1980","1985","1990"))
In case of plotting time series, the command ts.plot requires a different argument than xaxt="n"
require(graphics)
ts.plot(ldeaths, mdeaths, xlab="year", ylab="deaths", lty=c(1:2), gpars=list(xaxt="n"))
axis(1, at = seq(1974, 1980, by = 2))
I am having problems getting segments of small lengths to appear in my plot.
Assuming the following sample data:
x=c(11,22,33,44,55)
y=c(15,23,33,45,57)
z=strptime(20120101:20120105,'%Y%m%d')
If I were to create segments out of this data my segment for the third record does not show up if I want square or butt line ends. It does show up if I allow my line ends to be round lend=0.
plot(z,x,type='n')
segments(as.numeric(z),x,as.numeric(z),y,lwd=5,lend=2)
If I try this:
segments(as.numeric(z),x,as.numeric(z),y,lwd=5,lend=0)
It shows a circle at 33. Is there a way to get at the very least a flat line that will appear at 33 (hopefully in base)?
I would have used my actual data which is also doing this when the range is small for instance 33.0005 to 33.0010, but that data is huge and I was hoping solving for when they are identical would also solve for small ranges.
ETA: If lwd=15 the circle looks even more ridiculous.
Maybe segments are not the right way to approach this?
This is for a candlestick chart, so these numbers would represent open and close. I also have high and low numbers which extend beyond this range and are drawn using lwd=1 under these segments.
As #Joran points out, this may well be the "correct" behaviour.
But a kludgy workaround is to simply add an arbitrary small number to the values. This value should be small enough to not "distort" the data, but large enough to show up in your plot, given your plot device resolution.
delta <- pmax(0.2, y - x)
plot(z,x,type='n')
segments(as.numeric(z),x ,y1 = y + delta, lwd=10, lend=1)
PS. I advise against this. You have been warned.
Base graphics does supply rect. And in fact, it does what you want. Using your definitions above.
xdiff <- max(as.numeric(z)) - min(as.numeric(z))
segwidth <- xdiff/50
plot(z,x,type='n')
rect(z-segwidth/2, x, z+segwidth/2, y, col="black")
Given the edits to your question, I suspect the way to go about this is to plot points to indicate your open and close, and a segment to indicate the range.
In this way, if your open and close points are identical (or close), you get a symbol at the correct point.
x <- strptime(20120101:20120105,'%Y%m%d')
y1 <- c(11,22,33,44,55)
y2 <- c(15,23,33,45,57)
r <- range(c(y1, y2))
plot(c(x, x), c(y1, y2), type="n", xlab="Date", ylab="y")
points(x, y1, pch=18)
points(x, y2, pch=18)
segments(as.numeric(x), y0=y1, y1=y2)
There's something a little odd about "square" lineend
library(grid)
epsilon <- 1e-4
grid.newpage()
grid.points(x=c(0.5-epsilon,0.5+epsilon), y=c(0.5,0.5), pch="+", gp=gpar(cex=2), def="npc")
grid.segments(0.5-epsilon, 0.5, 0.5+epsilon, 0.5, gp=gpar(lineend="square",lwd=50, alpha=0.2))
grid.segments(0.5-epsilon, 0.5, 0.5+epsilon, 0.5, gp=gpar(lineend="round",lwd=50, alpha=0.2))
grid.segments(0.5-epsilon, 0.5, 0.5+epsilon, 0.5, gp=gpar(lineend="butt",lwd=50, alpha=0.2))
the behavior has a jump at epsilon = 0,
for epsilon=1e-4 vs
for epsilon=0
As a workaround, I would draw rectangles instead of lines; they always have at least one linewidth.
grid.newpage()
grid.rect(x=0.5, y=0.5, width=0.01, height=0, gp=gpar(fill="black", col="red", lwd=10, linejoin="mitre"))