similar to: taking variables from data.frame

Hi all, I'd like to fit model where the terms both are in the data.frame, mydata say, and are vectors *not in the data.frame*. >obj<-glm(y~x, data=mydata) #works >Z<-pmax(mydata$x-20,0) >(length(Z)==length(obj$y)) >[1] TRUE >update(obj,.~.+Z) #works However for some subset it doesn't works: >obj<-glm(y~x, data=mydata, subset=f==1) #works

Dear all, I would like to write a function like: myfun<-function(x,fn) {xx<-exp(x); x*fn(xx)} where fn is a symbolic description of any function with its argument to be specified. Therefore myfun(5,"2+0.3*y^2") should return 5*(2+0.3*exp(5)^2), myfun(5,"log(y)") should return 5*log(exp(5)) and so on. I tried with "expression" and others, but without success.

Hi all, I am experiencing the following quite strange (at least in my knowledge) problem in a simple function like the following: fn<-function(y,v=2){ n<-length(y) y<-y[(v+1):(n-v)] plot(y,type="l",lty=3,xlab="Time",ylab=deparse(substitute(y))) } fn(rnorm(50)) #look at the plot!!! The plot appears with numbers on the left side! If I delete the

Hi All, I have tried every fix on my try or tryCatch that I have found on the internet, but so far have not been able to get my R code to continue with the "for loop" after the lmer model results in an error. Here is two attemps of my code, the input is a 3D array file, but really any function would do.... metatrialstry<-function(mydata){ a<-matrix(data=NA, nrow=dim(mydata)[3],

Dear all, I'm writing a package where the main function fn(), say, returns a glm-like object, with three additional components. > obj<-glm(....) > out<-fn(obj,....) out is substantially an updated version of obj, with the same response and some (two or three) additional variables in the linear predictor. My problem is that out is much bigger than obj, and the three additional

Dear members, here is my trouble: My data consists of counts of trapped insects in different attractive traps. I usually use GLMs with a poisson error distribution to find out the differences between my traitments (and to look at other factor effects). But for some dataset where one traitment contains only zeros, GLM with poisson family fail to find any difference between this particular traitment

Dear all, The help file for the generic function vcov states "Classes with methods for this function include: 'lm', 'glm', 'nls', 'lme', 'gls', 'coxph' and 'survreg' (the last two in package 'survival')." Since, I am not able to use vcov.coxph(), I am wondering whether I am missing something (as I suspect..) regards, vito

dear all, Given a matrix A, say, I would like to apply a bivariate function to each combination of its colums. That is if myfun<-function(x,y)cor(x,y) #computes simple correlation of two vectors x and y then the results should be something similar to cor(A). I tried with mapply, outer,...but without success Can anybody help me? many thanks in advance, vito

> 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)) >

Dear all, Are there any packages can estimate the partial linear model. Or any one can give me any suggestions. Many thanks in advance. Jin

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

Dear all, I'm looking for someone that help me to write an R function to simulate survival data under complex situations, namely time-varying hazard ratio, marginal distribution of survival times and covariates. The algorithm is described in the reference below and it should be not very difficult to implement it. However I tried but without success....;-( Below there the code that I used; it

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),

Hi all, I am using the package ?Segmented? to estimate logistic regression models with unknown breakpoints (see Muggeo 2003 Statistics in Medicine 22:3055-3071). In the documentation it suggests that it might be possible to include several variables with breakpoints in the same model: ?Z = a vector or a matrix meaning the (continuous) explanatory variable(s) having segmented relationships with

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, dear R users I am a newbie in R and I wantto use the method of meximum likelihood to fit a Weibull distribution to my survival data. I use "optim" as follows: optim(c(1, 0.25),weibull.like,mydata=mydata,method="L-BFGS-B",hessian = TRUE) My question is: how do I setup the constraints so that the two parametrs of Weibull to be pisotive? Or should I use other function

Hi, I'm new to using R and I'm having problems in trying to do some quantile regressions with the package quantreg. When I try to use rq [specifically, I type in: rq(lnwagehr~lnmin1 + lncpi + ue03 + ue35 + ue57 + ue79, tau=.05, data=mydata, weights=pworwgt, method="fn", na.omit) ] , the computer churns for about 15 minutes and then spits out the following error message:

Dear all, It appears that MASS::polr() and Design::lrm() return the same point estimates but different st.errs when fitting proportional odds models, grade<-c(4,4,2,4,3,2,3,1,3,3,2,2,3,3,2,4,2,4,5,2,1,4,1,2,5,3,4,2,2,1) score<-c(525,533,545,582,581,576,572,609,559,543,576,525,574,582,574,471,595, 557,557,584,599,517,649,584,463,591,488,563,553,549) library(MASS) library(Design)

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