similar to: R-beta: ms, nls, etc?

Thanks very much Douglas for the pointer to nlm. Maybe the "Notes on R" maintainer can add at least a mention of nlm in the section on nonlinear fitting? I never did nonlinear fitting in S-Plus before, so I have nothing to unlearn, but I was hoping someone could show me how to do a least squares fit with nlm. example: x<-c(1,2,3,4,5,6) y<-.3*x^-.6 +.2 y<-y+rnorm(6,0,.01)

Thanks very much Douglas for the pointer to nlm. Maybe the "Notes on R" maintainer can add at least a mention of nlm in the section on nonlinear fitting? I never did nonlinear fitting in S-Plus before, so I have nothing to unlearn, but I was hoping someone could show me how to do a least squares fit with nlm. example: x<-c(1,2,3,4,5,6) y<-.3*x^-.6 +.2 y<-y+rnorm(6,0,.01)

I have a question on the use of return(). (Nothing on it in the docs I have) The test code below gives the error: Error: Object "x" not found when I do: thingy2(). How should it be fixed? Thanks very much for any help! (My original solution to this sort of problem was to use global variables x<<-... y<<-...) Bill Simpson ----------------------------- thingy<-function(k)

Simple-mindedly I tried getting MLE and SE for one-parameter model in the same way as for multi-param models. out<-nlm(fn,p=c(2),hessian=T) But sqrt(diag(solve(out$hessian))) gives the answer 1. The Hessian has only one entry, not really a matrix. diag(x) gives 1 if x is just a single number. Is this what I should be doing to get SE for MLE? sqrt(solve(out$hessian)) Thanks very much for

Simple-mindedly I tried getting MLE and SE for one-parameter model in the same way as for multi-param models. out<-nlm(fn,p=c(2),hessian=T) But sqrt(diag(solve(out$hessian))) gives the answer 1. The Hessian has only one entry, not really a matrix. diag(x) gives 1 if x is just a single number. Is this what I should be doing to get SE for MLE? sqrt(solve(out$hessian)) Thanks very much for

Questions on image: 1) How can I put labels on the x and y axes? 2) How to tell it to use e.g. 32 grey levels (not some colour map)? 3) How to know what the legend is (i.e. each grey level = what z value)? Thanks very much for any help. BTW I was wondering if persp was on the To Do list. That would be great! Bill Simpson

version .62: --------------------------------------------- > ?qpois The Poisson Distribution dpois(x, lambda) ppois(q, lambda) qpois(p, lambda) rpois(n, lambda) Arguments: x: vector of (positive) quantiles. p: vector of probabilities. n: number of random values to return. lambda: vector of positive means.

After some paper clipping problems I checked options() and saw that the default paper size was a4; I need US Letter. So as per instructions I uncommented the R_PAPERSIZE line in config.site, R_PAPERSIZE=letter Restarting R, everything is still the same though. I guess I have to reinstall? That seems awkward. Is there a way a prompt ("a4 or letter paper?") can be inserted in the

I would love to have bash-like command line editing in R. (Press up cursor and see previous command line; use left cursor to go back then edit it) Appendix B in Rnotes describes Splus I guess, not R. Starting R by R -e doesn't let any of the following described actions (B.3) work. I don't use vi or EMACs (I use Nedit), so I would prefer bash-like interface anyway. (I don't think I

I am running R-0.61.1 on linux under X. If I kill the R Graphics window (click on X box in upper right corner), then subsequently do x11(), R crashes. I find that things only work the right way if I close the graphics window using dev.off(). I was wondering if maybe it could be arranged that closing the graphics window via click would seem to R to be equivalent to typing dev.off(). BTW thanks

Can someone please tell me to make a density histogram? hist makes one with count or relative frequency on the y-axis. I want the density, which is (rel freq)/(bin width) In the help I see: intensities: values f^(x[i]), as estimated density values. If `all(diff(breaks) == 1)', they are the relative frequencies `counts/n' and in general satisfy

This surprised me. Is it the way it is supposed to be? > x<-c(1,2,3,4,5) > y<-c(1.2,1.3,1.4,1.5,NA) > x-y [1] -0.2 0.7 1.6 2.5 NA > (x-y)^2 [1] 0.04 0.49 2.56 6.25 NaN >>>so NA^2 = NaN? Why not still NA? > sum((x-y)^2) [1] NaN >>>yes that is reasonable So if you ever have a data set with missing observations (NAs), you can't do any nlm() least

If I do fit<-lm(y~x) Is it possible to access the SE of the slope? (Analogous to getting the slope like this: fit$coef[2]) If not, it would be handy. (I want SE of 1/slope, and an approx way is fit$se[2]/(fit$coef[2]^2)) Thanks for any help. Bill Simpson -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

I am using R Version 0.62.0 Unstable-snapshot (April 26, 1998) on Linux. I use the AfterStep window manager. Problem. Once upon a time, a workspace was saved. But at the end of this session, on quitting R I say "no", don't save workspace. On starting up, there are all the old objects again! I take it that the workspace is saved as ~/.RData. So is the only solution to manually rm

I include the argument pty="s" to image, but still when I make an image by the method > postscript("rstuff/test.ps") > tauseq<-seq(0,1,.5) > cif2d.image(x,,y,tauseq) > dev.off() the image plot created is NOT square. I thought pty="s" would make it square. Generally it would be nice if the image saved to disk were like the one we see while in R

I do: x<-rnorm(1000) quantile(x,c(.025,.975)) 2% 98% -1.844753 1.931762 Since I want to find a 95% confidence interval, I take the .025 and .975 quantiles. HOWEVER R says I have the 2% (not 2.5%) and 98% (not 97.5%) points. Is it just rounding the printed 2% and 98%, or is it REALLY finding .02 and .98 points instead of .025 and .975? Thanks for any help. Bill Simpson

I do: x<-rnorm(1000) quantile(x,c(.025,.975)) 2% 98% -1.844753 1.931762 Since I want to find a 95% confidence interval, I take the .025 and .975 quantiles. HOWEVER R says I have the 2% (not 2.5%) and 98% (not 97.5%) points. Is it just rounding the printed 2% and 98%, or is it REALLY finding .02 and .98 points instead of .025 and .975? Thanks for any help. Bill Simpson

When I look at ppoints I see: ppoints<-function (x) { n <- length(x) if (n == 1) n <- x (1:n - 0.5)/n } However Venables & Ripley (2nd ed, p 165) say ppoints() should return (i-1/2)/n for n>=11; (i-3/8)/(n+1/4) for n<=10. The version below should work as described: ppoints<-function (x) { n <- length(x) if (n <= 10) (1:n - 0.375)/(n + 0.25) else (1:n - 0.5)/n

Hello R-helpers, I would like to fit a non-linear function to data (Discrete X axis, over-dispersed Poisson values on the Y axis). I found the functions gnlr in the gnlm package from Jim Lindsey: this can handle nonlinear regression equations for the parameters of Poisson and negative binomial distributions, among others. I also found the function nls2 in the software package

Hello colleagues, I am attempting to determine the nonlinear least-squares estimates of the nonlinear model parameters using nls. I have come across a common problem that R users have reported when I attempt to fit a particular 3-parameter nonlinear function to my dataset: Error in nls(r ~ tlm(a, N.fix, k, theta), data = tlm.data, start = list(a = a.st, : step factor 0.000488281