search for: modout

Hi: I use following code to smooth my data set, and use two command to calculate the std for predict error. For me, I think the two command used to calculate mod.std should be exactly the same. But the result is different. Can anyone give me some clue on this? mod.model = loess(modout ~ modin, span=0.1, degree=1, control=loess.control(surface='interpolate', statistics='approximate', trace.hat='approximate')); modpre = predict(mod.model); mod.std = sqrt(var(modout-modpre, na.rm=TRUE)); (#mod.std = sqrt(var(modout[1:length(modpre)] - modpre[1:length(mo...

R Users: How can I catch R errors and depend on this error call other R functions in Perl? The foolowing is the error message when I use perl call R. Error in nls(modout ~ exp(-b1 * modin)/(b2 + b3 * modin), trace = F, start = c(b1 = 0.005, : singular gradient Segmentation fault Thanks, Zhongming