similar to: working with eval and environments

Hi, I've got a problem building a contrast matrix for the Dunnet contrast in package multcopm. The following works fine: > summary(simtest(adiff ~ trial)) Simultaneous tests: Dunnett contrasts Data: adiff by trial Contrast matrix: trial1 trial2 trial3 trial4 trial5 trial2-trial1 -1 1 0 0 0 trial3-trial1 -1 0 1 0 0

Hi again, yesterday I mailed this query however I could not see this on the mail list. Therefore, I am reposting it again. I was using package.skeleton() function to create the skeleton of my package in windows. Here is my attempt: rm(list = ls()) setwd("F:/R_PackageBuild") package.skeleton("trial1", namespace = TRUE, code_files = "F:/R_PackageBuild/trial.r") In

i have a function that returns a list containing a variety of variable types i am trying to run the function multiple times and return the output into a variable with a semi-consistent naming pattern i.e., for ten trials i want to return the list into variables trial1,trial2,...trial10 is there a generic way to get this to happen i have a similar process that does the same thing to an external

Hi, I am wondering if anyone can suggest how to test the equality of 2 proportions. The caveat here is that the 2 proportions were calculated from the same number of samples using 2 different tests. So essentially we are comparing 2 accuracy rates from same, say 100, samples. I think this is like a paired test, but don't know if really we need to consider the "paired" nature of the

Dear R-ers, I hope there is a really simple solution to my problem. I've written a function that I saved in an .r file. I source this file in my code. For a while it worked fine. But then when I run the line: source("F mylineplot.r") I started getting a warning: In readLines(file) : incomplete final line found on 'F mylineplot.r' I have no idea why - I tried to check and

Hi R-Helpers: I?m trying to fit the same linear model to a bunch of variables in a data frame, so I was trying to adapt the codes John Fox, Spencer Graves and Peter Dalgaard proposed and discused yesterday on this e-mail list: for (y in df[, 3:5]) { mod = lm(y ~ Trt*Dose, data = x, contrasts = list(Trt = contr.sum, Dose = contr.sum)) Anova(mod, type = "III") } ## by John Fox or for

> 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)) >

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

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

Hi all, Can anyone who is familar with CART tell me what I missed in my tree code? library (MASS) myfit <- tree (y ~ x1 + x2 + x3 + x4 ) # tree.screens () # useless plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0) # tile.tree (myfit, fgl$type) # useless # close.screen (all= TRUE) # useless My current tree plot resulted from above code shows as:

Dear List, I think I'm going crazy here...can anyone explain why do I get the same predictions in train and test data sets below when the second has a missing input? y <- rnorm(1000) x1 <- rnorm(1000) x2 <- rnorm(1000) train <- data.frame(y,x1,x2) test <- data.frame(x1) myfit <- glm(y ~ x1 + x2, data=train) summary(myfit) all(predict(myfit, test) == predict(myfit, train))

Hi again, I was using package.skeleton() function to create the skeleton of my package in windows. Here is my attempt: rm(list = ls()) setwd("F:/R_PackageBuild") package.skeleton("trial1", namespace = TRUE, code_files = "F:/R_PackageBuild/trial.r") In the trial.r file, there are 2 objects, one is a function and another is data. Here they are: fn1 <-

Hi, I am trying to estimate the effects of covariates on the hazard function, rather than on the survival. I know this is actually the same thing. For example, using the survival package, and doing: > myfit <- survreg( Surv(time, event) ~ mymodel ) all I have to do to get the quantities of my interest is > -myfit$coefficients/myfit$scale The standard erros are easily worked out, as

Hello R users, A beginners question which I could not find the answer to in earler posts. My thought process: Here "z" is a 119 x 15 data matrix Step 1: start at column one, bind every column with column 1 Step2: use the new matrix, "test", in the fitCopula package Step3: store each result in myfit, bind each result to "answer" Step4: return "answer"

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)) return(prd)} and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05), lower = c(1000,0),