On Nov 13, 2010, at 10:12 PM, cran.30.miller_2...@spamgourmet.com wrote: > On Fri, Nov 12, 2010 at 5:28 PM, David Winsemius - > cran.30.miller_2...@spamgourmet.com > <+cran > +miller_2555+c0e7477398.dwinsemius#comcast....@spamgourmet.com> wrote: > > On Nov 12, 2010, at 5:07 PM, David Winsemius wrote: > > > On Nov 12, 2010, at 4:22 PM, cran.30.miller_2...@spamgourmet.com > wrote: > > Hi - > > I have a dataframe of (x,y) values. I'd like to fit an exponential > curve to the data for further statistical analysis (pretty much the > same > functionality provided by Excel's LOGEST worksheet array function). > Can > someone point me to the (set of) functions/ package that is best > suited to > provide this functionality? Admittedly, I am a novice in the use of R > statistical functions, so a brief example of how to compute a > correlation > coefficient off a fitted exponential curve would be greatly > appreciated > (though I could probably work through it over time if I knew the > proper R > tools). > > > Probably (not seeing a clear description of the LOGEST function): > > ?exp > ?log > ?lm > ?cor > > > I set up a OO.org Calc spreadsheet which has a lot of Excel work- > alike functions and does have a LOGEST. Giving an argument of x=1:26 > and y=exp(x) to the first two arguments of LOGEST, I get 1 and e. > The OO.org help page says > "FunctionType (optional). If Function_Type = 0, functions in the > form y = m^x will be calculated. Otherwise, y = b*m^x functions will > be calculated." > > This might be the equivalent R operation: > > > x<-1:26 > > y<-exp(x) > > lm(log(y) ~ x) > > Call: > lm(formula = log(y) ~ x) > > Coefficients: > (Intercept) x > 0 1 > > > exp(coef(lm(log(y) ~ x))) > (Intercept) x > 1.000000 2.718282 > > Note this is not a correlation coefficient but rather an > (exponentiated) regression coefficient. > > -- > David Winsemius, MD > West Hartford, CT > > > > Thanks, but I'm looking to fit an exponential curve (not a linear > model).
Then WHY did you ask for a function that would duplicate a particular Excel function????? ... especially an Excel function which DOES a linear fit on the log of the Y argument? http://support.microsoft.com/kb/828528 > However, I was able to identify the `nls()` function that works well > (adapted from John Fox's contribution "Nonlinear Regression and > Nonlinear Least Squares" [Jan 2002] ref: > http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-nonlinear-regression.pdf) > > . For those interested, the following short script highlights my > simple test case (though a little sloppy): > > mydf <- as.data.frame(cbind(1:6,rev(c(48.0000, 24.0000, 12.0000, > 6.0000, 3.0000, 1.5000)),rev(c( 51.4943, 12.4048, 12.9587, 3.7707, > 2.4253, 2.0400)))); > colnames(mydf) <- c("X","Y","Y2"); > > my.mod <- nls(Y2 ~ a*exp(b*X), data=mydf, start=list(a=3.00,b=2.00), > trace=T) > > plot(mydf[,"X"],residuals(my.mod)) > plot(mydf[,"X"],mydf[,"Y2"], lwd=1) > lines(mydf[,"X"],fitted.values(my.mod), lwd=2) > David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
