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Graphing perpendicular offsets in a least squares regression plot in R

I'm interested in making a plot with a least squares regression line and line segments connecting the datapoints to the regression line as illustrated here in the graphic called perpendicular offsets: http://mathworld.wolfram.com/LeastSquaresFitting.html alt text
(from MathWorld - A Wolfram Web Resource: wolfram.com)

I have the plot and regression line done here:

## Dataset from http://www.apsnet.org/education/advancedplantpath/topics/RModules/doc1/04_Linear_regression.html

## Disease severity as a function of temperature

# Response variable, disease severity
diseasesev<-c(1.9,3.1,3.3,4.8,5.3,6.1,6.4,7.6,9.8,12.4)

# Predictor variable, (Centigrade)
temperature<-c(2,1,5,5,20,20,23,10,30,25)

## For convenience, the data may be formatted into a dataframe
severity <- as.data.frame(cbind(diseasesev,temperature))

## Fit a linear model for the data and summarize the output from function lm()
severity.lm <- lm(diseasesev~temperature,data=severity)

# Take a look at the data
plot(
 diseasesev~temperature,
        data=severity,
        xlab="Temperature",
        ylab="% Disease Severity",
        pch=16,
        pty="s",
        xlim=c(0,30),
        ylim=c(0,30)
)
abline(severity.lm,lty=1)
title(main="Graph of % Disease Severity vs Temperature")

Should I use some kind of for loop and segments http://www.iiap.res.in/astrostat/School07/R/html/graphics/html/segments.html to do the perpendicular offsets? Is there a more efficient way? Please provide an example if possible.

like image 960
D W Avatar asked Apr 14 '10 17:04

D W


1 Answers

You first need to figure out the coordinates for the base of the perpendicular segments, then call the segments function which can take vectors of coordinates as inputs (no need for a loop).

perp.segment.coord <- function(x0, y0, lm.mod){
 #finds endpoint for a perpendicular segment from the point (x0,y0) to the line
 # defined by lm.mod as y=a+b*x
  a <- coef(lm.mod)[1]  #intercept
  b <- coef(lm.mod)[2]  #slope
  x1 <- (x0+b*y0-a*b)/(1+b^2)
  y1 <- a + b*x1
  list(x0=x0, y0=y0, x1=x1, y1=y1)
}

Now just call segments:

ss <- perp.segment.coord(temperature, diseasesev, severity.lm)
do.call(segments, ss)
#which is the same as:
segments(x0=ss$x0, x1=ss$x1, y0=ss$y0, y1=ss$y1)

Note that the results will not look perpendicular unless you ensure that the x-unit and y-unit of your plot have the same apparent length (isometric scales). You can do that by using pty="s" to get a square plot and set xlim and ylim to the same range.

like image 52
Aniko Avatar answered Sep 25 '22 13:09

Aniko