Fitting logarithmic curve in R

Question

If I have a set of points in R that are linear I can do the following to plot the points, fit a line to them, then display the line:

x=c(61,610,1037,2074,3050,4087,5002,6100,7015)
y=c(0.401244, 0.844381, 1.18922, 1.93864, 2.76673, 3.52449, 4.21855, 5.04368, 5.80071)

plot(x,y)    
Estimate = lm(y ~ x)    
abline(Estimate)

Now, if I have a set of points that looks like a logarithmic curve fit is more appropriate such as the following:

x=c(61,610,1037,2074,3050,4087,5002,6100,7015)        
y=c(0.974206,1.16716,1.19879,1.28192,1.30739,1.32019,1.35494,1.36941,1.37505)

I know I can get the standard regression fit against the log of the x values with the following:

logEstimate = lm(y ~ log(x))

But then how do I transform that logEstimate back to normal scaling and plot the curve against my linear curve from earlier?

Answer 1

Hmmm, I'm not quite sure what you mean by "plot the curve against my linear curve from earlier".

d <- data.frame(x,y)  ## need to use data in a data.frame for predict()
logEstimate <- lm(y~log(x),data=d)

Here are three ways to get predicted values:

(1) Use predict:

plot(x,y)
xvec <- seq(0,7000,length=101)
logpred <- predict(logEstimate,newdata=data.frame(x=xvec))
lines(xvec,logpred)

(2) Extract the numeric coefficient values:

coef(logEstimate)
## (Intercept)      log(x) 
##  0.6183839   0.0856404 
curve(0.61838+0.08564*log(x),add=TRUE,col=2)

(3) Use with() magic (you need back-quotes around the parameter estimate names because they contain parentheses)

with(as.list(coef(logEstimate)),
      curve(`(Intercept)`+`log(x)`*log(x),add=TRUE,col=4))

Maybe what you want is

est1 <- predict(lm(y~x,data=d),newdata=data.frame(x=xvec))
plot(est1,logpred)

... although I'm not sure why ...