similar to: Standard deviation from MANOVA??

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

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 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.

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

> 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, I would like to conduct a MANOVA. I know that there 's the manova() funciton and the summary.manova() function to get the appropriate summary of test statistics. I just don't manage to specify my model in the manova() call. How to specify a model with multiple responses and one explanatory factor? If I type:

Hello, I had a couple questions about manova modeling in R. I have calculated a manova model, and generated a summary.manova output using both the Wilks test and Pillai test. The output is essentially the same, except that the Wilks lambda = 1 - Pillai. Is this normal? (The output from both is appended below.) My other question is about the use of MANOVA. If I have one variable which has a

Hi all, I am experimenting the function "manova" in R. I tried it on a few data sets, but I did not understand the result: I used "summary(manova_result)" and "summary(manova_result, test='Wilks')" and they gave a bunch of numbers... But I need the Sum-of-Squares of BETWEEN and WITHIN matrices... How do I read off from the R's manova results? Any

I'm trying to learn to use manova(), and don't understand why none of the following work: > data(iris) > fit <- manova(~ Species, data=iris) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : incompatible dimensions > fit <- manova(iris[,1:4] ~ Species, data=iris) Error in model.frame(formula, rownames, variables, varnames, extras,

Can R do a repeated measures MANOVA and tell what dimensionality the statistical variance occupies? I have been using MATLAB and SPSS to do my statistics. MATLAB can do ANOVAs and MANOVAs. When it performs a MANOVA, it returns a parameter d that estimates the dimensionality in which the means lie. It also returns a vector of p-values, where each p_n tests the null hypothesis that the mean

Hi all, I have received the following error from summary.manova: Error in summary.manova(manova.test, test = "Pillai") : residuals have rank 36 < 64 The data is simulated data for 64 variables. The design is a 2*2 factorial with 10 replicates per treatment. Looking at the code for summary.manova, the error involves a problem with qr(). Does anyone have a suggestion as to how to

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

Dear colleagues, I had a question wrt the car package. How do I evaluate whether a simpler multivariate regression model is adequate? For instance, I do the following: ami <- read.table(file = "http://www.public.iastate.edu/~maitra/stat501/datasets/amitriptyline.dat", col.names=c("TCAD", "drug", "gender", "antidepressant","PR",

Hi there, Does anyone know how to extract elements from the table returned by Manova()? Using the univariate equivalent, Anova(), it's easy: a.an<-Anova(lm(y~x1*x2)) a.an$F This will return a vector of the F-values in order of the terms of the model. However, a similar application using Manova(): m.an<-Manova(lm(Y~x1~x2)) m.an$F Returns NULL. So does any attempt at calling the

Hi, Does anyone know what the error message mean? > Boot2.Pillai <- function(x, ind) { + x <- as.matrix(x[,2:ncol(x)]) + boot.x <- as.factor(x[ind, 1]) + boot.man <- manova(x ~ boot.x) + summary(manova(boot.man))[[4]][[3]] + } > > man.res <- manova(as.matrix(pl.nosite) ~ + as.factor(plankton.new[,1]))$residuals > boot2.plank <-

Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =

Dear All, I am trying to perform MANOVA. I have table with 504 columns(species) and 36 rows) with two grouping (season and location) Zx <- Z[c(4:504)] Zxm <- as.matrix(Z) m<- manova(Zxm~Season*location, data=Z) when I do summary.aov, I get respond for each species but summary.manova summary.manova(m) :" residuals have rank" 24<501. What can it be the reason for this error

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

Dear all, I need to do a Manova but I have an unbalanced design. I have morphological measurements similar to the iris dataset, but I don't have the same number of measurements for all species. Does anyone know a procedure to do Manova with this kind of input in R? Thank you very much, Naiara. -------------------------------------------- Naiara S. Pinto Ecology, Evolution and Behavior 1

Hello all, I have a question on manova in R: I'm using the function "manova()" from the stats package. Is there anything like a stepwise (backward or forward) manova in R (like there is for regression and anova). When I enter: step(Model1, data=Mydata) R returns the message: Error in drop1.mlm(fit, scope$drop, scale = scale, trace = trace, k = k, : no 'drop1'