similar to: List of lm objects

My previous question put in a simpler way: How would I pass a value of a variable to a function such as lm(Effect1~Trt*Dose, data = x, contrasts = list(Trt = contr.sum, Dose = contr.sum))? Here, 'Effect' is a column name in my data matrix, and I want "Effect1" to be replaced by "Effect2" and so on (my other column names in the data frame) for successive anova

Hi R-Helpers: I?m trying to use the function GLMM from lme4 package, (R-1.8.1, Windows 98),and I get the following error: > pd5 = GLMM(nplant~sitio+ + fert+ + remo+ + sitio:fert+ + remo:sitio+ + remo:fert+ + remo:fert:sitio + data=datos, + family=binomial, +

Hola todos: acabo de actualizar R a R 2.10.1, y cuando actualice Rcmdr (Versión 1.5-4) me cargo la versión en inglés, no me acuerdo que hice antes para tener Rcmdr en español (trabajo con windows XP). alguien podría recordarme lo que hay que hacer? Gracias Gabriela. ______________________________ Lic. María Gabriela Cendoya Magíster en

Estimados me ha surgido una duda con un simple stem and leaf el mismo es el siguiente: > stem.leaf(Datos$ej1.21) 1 | 2: represents 0.12 leaf unit: 0.01 n: 40 LO: 0.72 0.85 3 10 | 9 11 | 4 12 | 4 5 13 | 7 7 14 | 710 (*) 9 15 | 18 16 16 | 3447899 (6) 17 | 045599 (***) 18 18 | 2568 14 19 | 23710 (**) 10 20 | 389 7 21

Hi: Does R have a function as gsorth is SAS, that perform a the Gram-Schmidt orthonormal factorization of the m ?n matrix A, where m is greater than or equal to n? That is, the GSORTH subroutine in SAS computes the column-orthonormal m ?n matrix P and the upper triangular n ?n matrix T such that A = P*T. or any other version of Gram-Schmidt orthonormal factorization? I search the help, but I

Hi: I'm using function lmer from package lme4, and I get this message: " There were 12 warnings (use warnings() to see them)" So I checked them: Warnings 1 to 11 said: 1: optim returned message ERROR: ABNORMAL_TERMINATION_IN_LNSRCH in: "LMEoptimize<-"(`*tmp*`, value = structure(list(maxIter = 50, ... and Warning 12 said: 12: IRLS iterations for glmm did

Dear R-devel: The following code seems to allow one function to permanently modify a calling function. I did not expect this would be allowed (short of more creative gymnastics) and wonder if it is really intended. (I can see other ways to accomplish the intended task of this code [e.g. via match.call instead of substitute below] that do not trigger the problem, but I don't think that is

Hello there! Somebody tried the meetme|b which initiates the conf-background AGI. Actually I want to originate another call from a conference.my AGI originates the call and connects it to the conference, but the calleeee is nowhere My extension exten => 21,1,meetme(21|pb) and my AGI **************************************************************************** #!/usr/bin/perl -w

Hello everybody!!! I really need your help guys, I am using the AGI mode in meetme application, and I want that AGI should wait for an input from the client/user i.e. a digit and then proceed, but I have used that AGI function agi->wait_for_digit(), but no use....my agi just passes, or ignores this function, where AGI should stop here and wait for the input.... .....my extension in my

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

Hi all, assume the following problem: a function call takes a function object and a data variable and calls this function with this data on a remote host. It uses serialization to pass both the function and the data via a socket connection to a remote host. The problem is that depending on the way we call the same construct, the function may be serialized to include the data, which was not

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

Hi: I was trying to download the package MCMCglmm and it give me this message: Aviso: dependency ?tensorA? is not available probando la URL 'http://www.vps.fmvz.usp.br/CRAN/bin/windows64/contrib/2.11/MCMCglmm_2.06.zip' Content type 'application/zip' length 8988896 bytes (8.6 Mb) URL abierta downloaded 8.6 Mb package 'MCMCglmm' successfully unpacked and MD5 sums checked my

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

Buen día a todos el problema que tengo es el siguiente: tengo una matriz 0 0 0 1 5 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 6 0 0 0 0 0 0 0 0 0 1 4 0 1 0 7 2 9 0 1 necesito hacer el primer conteo de números hasta que encuentre el primer cero. para la fila 1: tiene 1 y 5 son dos elementos fila 2: solo tiene 1 elemento fila 3: tiene 1 fila 4: 1 y 6 dos elementos fila 5: no tiene elementos distintos de

I'm drawing a fitted normal distribution over a histogram. The use case is trivial (fitting normal distributions on densities) but I want to extend it to other fitting scenarios. What has stumped me so far is how to take the list that is returned by nls() and use it for curve(). I realize that I can easily do all of this with a few intermediate steps for any specific case. But I had expected

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

I haven't followed your example closely, but can't you use the predict() method for this? To draw a curve, the function that will be used in curve() sets up a newdata dataframe and passes it to predict(fit, newdata= ...) to get predictions at those locations. Duncan Murdoch On 17/10/2020 5:27 a.m., Boris Steipe wrote: > I'm drawing a fitted normal distribution over a

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

Hi, I have problem getting the standard deviation from the manova output. I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1 + x2 + x3, data=mydata) . I tried to get the predicted values and their standard deviation by using: predict(myfit, type="response", se.fit=TRUE) But the problem is that I don't get the standard deviation values, I only