similar to: R-beta: sum of squares and NAs

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 tried to ?ms, ?nls and apparently these aren't implemented on R yet. However I seem to remember postings on this list having to do with fitting nonlinear models (no I don't mean GLM type fits, I have a REAL nonlinear model: y=ax^b + c). So please tell me if it is possible to fit nonlinear models in R (by least squares or ML). Thanks! Bill Simpson

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

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

Here is a programming question. The code I am using is quite slow and I was wondering if there is a way to get rid of the for loop. I am dealing with "interaction" in 2x2 table, and am using Edwards's G_I (Likelihood, p. 194). I label the cells in the table as follows stim response "y" "n" total -------------------------------- y hit miss nsignal

I know how to first set postscript output, then execute plot or image command to create a postscript file of a plot or image. I was wondering if it is possible to first create the plot or image on screen so it can be viewed, THEN save it to file? Without doing plot() or image() again? (I ask because I have a complicated image computation that takes about 1-2 hrs to complete. I would like to