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Setting an image as the colour of a polygon

I borrowed this code from statmethods[dot]net. The result is a coloured area under a normal distribution.

mean=100; sd=15
lb=80; ub=120

x <- seq(-4,4,length=100)*sd + mean
hx <- dnorm(x,mean,sd)

plot(x, hx, type="n", xlab="IQ Values", ylab="Density",
  main="Normal Distribution", axes=FALSE)

i <- x >= lb & x <= ub
lines(x, hx)
polygon(c(lb,x[i],ub), c(0,hx[i],0), col="red") 

area <- pnorm(ub, mean, sd) - pnorm(lb, mean, sd)
result <- paste("P(",lb,"< IQ <",ub,") =",
   signif(area, digits=3))
mtext(result,2)

enter image description here

I was wondering if it is possible to select an image as the colour of the red polygon?

Many thanks!

like image 254
Nikolay Nenov Avatar asked May 06 '13 16:05

Nikolay Nenov


1 Answers

Working with your code, I basically suggest plotting an image (I used a random image below) using the png package (based on advice from In R, how to plot with a png as background?) and then plot your lines over that.

mean=100; sd=15
lb=80; ub=120

x <- seq(-4,4,length=100)*sd + mean
hx <- dnorm(x,mean,sd)

# load package and an image
library(png)
ima <- readPNG("Red_Hot_Sun.PNG")

# plot an empty plot with your labels, etc.
plot(1,xlim=c(min(x),max(x)), type="n", xlab="IQ Values", ylab="Density",
  main="Normal Distribution", axes=FALSE)
# put in the image
lim <- par()
rasterImage(ima, lim$usr[1], lim$usr[3], lim$usr[2], lim$usr[4])

# add your plot
par(new=TRUE)
plot(x, hx, xlim=c(min(x),max(x)), type="l", xlab="", ylab="", axes=FALSE)
i <- x >= lb & x <= ub
lines(x, hx)
# add a polygon to cover the background above plot
polygon(c(x,180,180,20,20), c(hx,0,1,1,0), col="white")
# add polygons to cover the areas under the plot you don't want visible
polygon(c(-20,-20,x[x<=lb],lb), c(-10,min(hx),hx[x<=lb],-10), col="white")
polygon(c(ub,x[x>=ub],200,200), c(-1,hx[x>=ub],min(hx),-1), col="white")

# add your extra text
area <- pnorm(ub, mean, sd) - pnorm(lb, mean, sd)
result <- paste("P(",lb,"< IQ <",ub,") =",
   signif(area, digits=3))
mtext(result,2)

Gives you:

An image

like image 190
Thomas Avatar answered Oct 25 '22 16:10

Thomas