I think your way is probably the easiest (shockingly). For instance, here are
some alternatives - I think in both cases you have to calculate the
coefficient of determination (R^2) manually. My understanding is that
multiple R^2 in your case is the usual R^2 because you only have one
predictor variable, and the adjusted R^2 considers the degrees of freedom and
penalizes for additional predictors. Which is better... depends? (Perhaps
more stats-savvy people can help you on that one. I'm a chemical engineer so
I unjustifiably claim ignorance).
## Data input
input <-
"Year Count
1999 3
2000 5
2001 9
2002 30
2003 62
2004 154
2005 245
2006 321"
dat <- read.table(textConnection(input),header=TRUE)
dat[,] <- lapply(dat,function(x) x-x[1])
# shifting in origin; will need to add back in later
## Nonlinear least squares
plot(dat)
out <- nls(Count~b0*exp(b1*Year),data=dat,
start=list(b0=1,b1=1))
lines(dat[,1],fitted(out),col=2)
out <- nls(Count~b0+b1*Year+b2*Year^2,data=dat, #polynomial
start=list(b0=0,b1=1,b2=1))
lines(dat[,1],fitted(out),col=3)
## Optim
f <- function(.pars,.dat,.fun) sum((.dat[,2]-.fun(.pars,.dat[,1]))^2)
fitFun <- function(b,x) cbind(1,x,x^2)%*%b
expFun <- function(b,x) b[1]*exp(b[2]*x)
plot(dat)
out <- optim(c(0,1,1),f,.dat=dat,.fun=fitFun)
lines(dat[,1],fitFun(out$par,dat[,1]),col=2)
out <- optim(c(1,1),f,.dat=dat,.fun=expFun)
lines(dat[,1],expFun(out$par,dat[,1]),col=3)
--- Andrew Clegg <[EMAIL PROTECTED]> wrote:
> Hi folks,
>
> I've looked through the list archives and online resources, but I
> haven't really found an answer to this -- it's pretty basic, but I'm
> (very much) not a statistician, and I just want to check that my
> solution is statistically sound.
>
> Basically, I have a data file containing two columns of data, call it
> data.tsv:
>
> year count
> 1999 3
> 2000 5
> 2001 9
> 2002 30
> 2003 62
> 2004 154
> 2005 245
> 2006 321
>
> These look exponential to me, so what I want to do is plot these
> points on a graph with linear axes, and add an exponential curve over
> the top. I also want to give an R-squared for the fit.
>
> The way I did it was like so:
>
>
> # Read in the data, make a copy of it, and take logs
> data = read.table("data.tsv", header=TRUE)
> log.data = data
> log.data$count = log(log.data$count)
>
> # Fit a model to the logs of the data
> model = lm(log.data$count ~ year, data = log.data)
>
> # Plot the original data points on a graph
> plot(data)
>
> # Draw in the exponents of the model's output
> lines(data$year, exp(fitted(model)))
>
>
> Is this the right way to do it? log-ing the data and then exp-ing the
> results seems like a bit of a long-winded way to achieve the desired
> effect. Is the R-squared given by summary(model) a valid measurement
> of the fit of the points to an exponential curve, and should I use
> multiple R-squared or adjusted R-squared?
>
> The R-squared I get from this method (0.98 multiple) seems a little
> high going by the deviation of the last data point from the curve --
> you'll see what I mean if you try it.
>
> Thanks in advance for any help!
>
> Yours gratefully,
>
> Andrew.
>
> ______________________________________________
> R-help@stat.math.ethz.ch mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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