Creating shaded areas in R
Sometimes we want to shade areas under a density on a graphic, for instance to illustrate a p-value or a region under the normal curve.
Using the polygon() function from base R we can make graphics like the following one:
Standard normal density. Shaded area shows P(1 < X < 2)
How to do it
First we have to plot the graphic of the function whose areas we are going to shade. We can usually do that using the function curve(). When we have a standard normal density we can use:
The command above will create the graphic over an arbitrary range on the x values though. Since we want to shade a specific area under the curve, it is advisable to specify which interval on the x-axis we want the function plotted on. We can do that using the xlim parameter:
Now that we have our curve plotted, we are going to use the polygon(x,y) function. Its most important parameters are x and y, which define what kind of polygon R is going to draw. If x and y are vectors with i=1,…n elements, polygon(x,y) draws a polygon with vertices (xi,yi), i=1,…n. The trick now consists of expressing our region of interest as a polygon. The idea might seem stupid at first, as the area of interest is usually over a continuous curve, but if we take a large enough number of vertices the approximation will be very reasonable.
Let’s say we want to shade the region represented by P(-3 < X < -2). The first vertex we want for our polygon is (-3,0). We have then:
The second vertex will be (-3,f(-3)), where f(-3) is the normal density evaluated at -3. We can readily obtain this value on R using the dnorm() function:
As a third and fourth vertices we can consider (-2,f(-2)) and (-2,0). We have now:
We then issue:
Obtaining (if you followed our instructions correctly ):
First attempt at creating a shaded area in R
Not that bad, but we can make it better than that. We just have to create a sequence of vertices between the two extrema of the density (f(-3) and f(-2) here), so that the larger number of vertices gives us a better approximation. In order to accomplish that we are going to use the seq() function, to create a sequence between -3 and 2, with steps of 0.01:
1 2 3 4 | cord.x <- c(-3,seq(-3,-2,0.01),-2) cord.y <- c(0,dnorm(seq(-3,-2,0.01)),0) curve(dnorm(x,0,1),xlim=c(-3,3),main='Standard Normal') polygon(cord.x,cord.y,col='skyblue') |
Which gives us the result:
Final graphic.
Commentaries
Note that we used the standard normal density. If we wanted to use another probability distribution its parameters would have to be properly specified. Try playing with areas under the exponential distribution, for example. To learn more about the uses and misuses of the polygon function, take a look at its documentation (?polygon).
This is a great explaination! It worked like a charm.
Hi Fernando,
I recently came across your website as i was trying to figure out how to do a polygon in a plot for a specific range. Your information was very helpful. Could you tell me how do i apply this to a graph where i have to shade say from -1 and 1 only, and leave rest of it blank when given an assignment to graph P(-1.5 ≤ Z ≤ 1) ?
This is as far as I got:
x<-seq(-4,4,.2)
y=-1],-1.5),c(y[x>=-1],y[x==4]),col=”blue”)
but i need it to only go till 1 and not all the way to 4.
Please let me know at your earliest convenience.
I would greatly appreciate it.Thank you.
Tijana
Hi Tijana, you’d just need to update my last code snippet in the appropriate places, that is -3 by -1 and -2 by 1:
cord.x <- c(-1,seq(-1,1,0.01),1)
cord.y <- c(0,dnorm(seq(-1,1,0.01)),0)
curve(dnorm(x,0,1),xlim=c(-3,3),main='Standard Normal')
polygon(cord.x,cord.y,col='blue')
That should generate what you’re expecting.
This really is a great explanation. I do wonder, however, how this works for custom distributions, like the distribution of results. I tried it and it says “Error in xy.coords(x, y) : ‘x’ and ‘y’ lengths differ”. Do you have an idea why this might be?
Hi Paul,
Well, in this case I guess you would have to use the empirical distribution function (?ecdf). As an example with results coming from an Exponential distribution:
x <- rexp(100)
fnx <- ecdf(x)
cord.x <- c(0,seq(0,2,0.01),2)
cord.y <- c(0,fnx(seq(0,2,0.01)),0)
curve(fnx(x),xlim=c(0,5),main='Estimated Density of Results')
polygon(cord.x,cord.y,col='skyblue')
Is that what you were trying to achieve?
Brilliant!! Thank you.
First result on google for “shaded curve R” and it works great. Thanks!
Hi
I have a small question and I will highly appreciate your insight
my code will plot two normal curves (layout) I need to shade the area between in common under these two curves
the code is in R
x=seq(-7,10,length=200)
y1=dnorm(x,mean=0,sd=1)
plot(x,y1,type=”l”,lwd=2,col=”red”)
y2=dnorm(x,mean=3,sd=2)
lines(x,y2,type=”l”,lwd=2,col=”blue”)
I will highly appreciate any help
Hi Muna,
Very interesting question. The first thing you’ll need to find is where the two curves intersect. This is quite simple to do, just write out the two densities (X ~ N(0,1) and Y ~ N(3,2)) and make them equal. Solve for x, and you’ll end up with a quadratic equation (x^2 + 2x -3 + 8/3 ln(1/2) = 0), where the solutions are x = 1.418345 and x = -3.418345. These are the two points where the two normal curves intersect.
Now the catch is to actually create two polygons instead of one: you’ll need one to shade from the two points where the two curves intersect (shading it over the ‘fatter’ curve, and then from that second intersection point until the thinner curve goes to zero (I set it arbitrarily at 4). You have to add lty=0 to the polygon to avoid a line dividing them. Here’s the final code:
x=seq(-7,10,length=200)
y1=dnorm(x,mean=0,sd=1)
plot(x,y1,type='l',lwd=2,col='red')
y2=dnorm(x,mean=3,sd=2)
lines(x,y2,type='l',lwd=2,col='blue')
# First block (we shade under the blue curve)
cord.x <- c(-3.418345,seq(-3.418345,1.418345,0.01),1.418345)
cord.y <- c(0,dnorm(seq(-3.418345,1.418345,0.01),3,2),0)
polygon(cord.x,cord.y,col='blue',lty=0)
# Second block (we shade under the red curve)
cord.x <- c(1.418345,seq(1.418345,4,0.01),4)
cord.y <- c(0,dnorm(seq(1.418345,4,0.01)),0)
polygon(cord.x,cord.y,col='red',lty=0)
Thank you so much fernandohrosa
That was great:)
Dear Fernando
I have kind of a complicated problem, hope you can be patient with me
My code simulate one data set, then use this data set to get estimates for my parameters namely beta.t and theta.t.
What I need help with (if this is possible) is that I need my code to repeat this process as many as 10000 times… in each time it should generate a data set, find theta.t and beta.t and save these values in a vector so I can do all kind of work on them
like average variance..etc etc
following is my code
Thank you in advance
Muna
## Use EM algorithm to estimate parameters of pareto based
# on progressive censored data
n=20;m=5;R<-c(3,3,3,3,3)
theta=0.5; beta=3
W<-matrix((runif(m,0,1)),m,1)
E<-matrix(NA,m,1);V<-matrix(NA,m,1);U<-matrix(NA,m,1)
i<-1
while (i<=m) {
E[i]<- 1/( i+ sum(R[(m+1-i):m]))
i<-i+1
}
V<-W^E
i<-1
while (i<=m) {
U[i]<- (1- prod( V[(m+1-i):m]))
i<-i+1
}
x<-( (1-U)^(-1/theta)-1 )/beta
Yobs<-x
em.fct <- function(Y){
r <- length(Yobs)
##initial value
theta.t=theta
beta.t=beta
# Define log-likelihood function
ll <- function(y, k, c){
n*log(c*k)-(k+1)*( sum(log(1+c*Y))+sum( R*log(1+c*Y)+1/k ) )
}
# Compute the log-likelihood for the initial values
lltm1 <- ll(Yobs, theta.t, beta.t)
repeat{
# E-step
Bbar<- function(theta.t,beta.t){
sum(R*(1+beta.t*(theta.t+1)*Yobs)/(beta.t*(theta.t+1)*(1+beta.t*Yobs)))
}
Abar<- function(theta.t,beta.t){
sum( R* ( log(1+beta.t*Yobs)+ 1/theta.t ))
}
theta.t1<- n/( sum(log(1+beta.t*Yobs)) + Abar(theta.t,beta.t))
# M-step
beta.t1<-((theta.t1+ 1)/n *(sum(Yobs/(1+beta.t*Yobs))+Bbar(theta.t,beta.t)))^(-1)
theta.t1<- n/( sum(log(1+beta.t*Yobs)) + Abar(theta.t,beta.t))
beta.t <-beta.t1
theta.t<-theta.t1
# compute log-likelihood using current estimates
llt <- ll(Yobs, theta.t, beta.t)
# Print current parameter values and likelihood
cat(theta.t, beta.t, llt, "\n")
# Stop if converged
if ( abs(lltm1 – llt) < 0.0001) break
lltm1 <- llt
}
return(theta.t,beta.t)
}
em.fct(Yobs)