similar to: Trouble with glm() .... non-integer #successes in a binomial glm

Hi: I am using step.glm() as follows: form1 <- as.formula(haspdata ~ 1) lg.mod1 <- glm ( formula=form1, data=st.mtx, family=binomial , na.action=na.omit ) upper <- as.formula( haspdata ~ ( c5+childnm+educ1+incpov1+marital+msa+racekid+racemom+sex+shotcard )^2) lower <- as.formula(haspdata~1) lst <- list( upper=upper , lower=lower ) form1a.step <- step(lg.mod1, scope=lst

Hi ? This is giving me a headache. I?m trying to do a relatively simple optimization ? actually trying to approximate the output from the Excel Solver function but at roughly 1000x the speed. ? The optimization parameters look like this. The only trouble is that I want to add a constraint that sum(wgt.vect)=1, and I can?t figure out how to do that in optim. Mo.vect <-

Here is a runable program. When I plot Day and Wgt, it graphs all the data points. All I need is daily.sub1 plotted. I also need each "Tanks" to have its own col or pch. When I run it with the line with pch, it gives me nothing. rm(list=ls()) Trial<-rep(c(1,2),each=12) Tanks=rep(c("a3","a4","c4","h4"),each=3,2) Day=rep(c(1:12),2)

Hi Michael, A few comments 1. To add the constraint sum(wgt.vect=1) you would use the method of Lagrange multipliers. What this means is that in addition to the w_i (the components of the weight variables) you would add an additional variable, call it lambda. Then you would modify your optim.fun() function to add the term lambda * (sum(wgt.vect - 1) 2. Are you sure that you have defined

You can't -- at least as I read the docs for ?optim (but I'm pretty ignorant about this, so maybe there's a way to tweak it so you can). See here: https://cran.r-project.org/web/views/Optimization.html for other R optimization capabilities. Also, given your credentials, the r-sig-finance list might be a better place for you to post your query. Cheers, Bert Bert Gunter

Greetings Everybody: I generated a 1.2MB dta file based on the general social survey with Stata8 for linux. The file can be re-opened with Stata, but when I bring it into R, it says all the values are missing for most of the variables. This dataset is called "morgen.dta" and I dropped a copy online in case you are interested http://www.ku.edu/~pauljohn/R/morgen.dta looks like this

Thanks Bert. But everyone on that forum wants to use finance tools rather than general optimization stuff! And I am not optimizing a traditional Markowitz mean-variance problem. Plus, smarter people here. :-) > On May 3, 2018, at 3:01 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > You can't -- at least as I read the docs for ?optim (but I'm pretty > ignorant

On Thu, May 3, 2018 at 2:03 PM, Michael Ashton <m.ashton at enduringinvestments.com> wrote: > Thanks Bert. But everyone on that forum wants to use finance tools rather than general optimization stuff! And I am not optimizing a traditional Markowitz mean-variance problem. Plus, smarter people here. :-) > I'm very confused by these statements. Most of the "finance tools"

Is the code below showing a bug in Crosstable()? My expectation was that the values produced by xtabs were rounded instead of truncated: library(gmodels) abc <- c("a", "a", "b", "b", "c", "c") def <- c("d", "e", "f", "f", "d", "e") wgt <- c(0.8, 0.6, 0.4, 0.5, 1.4, 1.3)

OK, I need help plotting. I have column headings of Day, Wgt, Foodin, Rep, Grp and Tanks. Rep=c(1,2,3) and Tanks=c(a1,a2,a3,a4,a5,a6, c1,c2,c3,c4,c5,c6, h1,h2,h3,h4,h5,h6). I created a subset where I only would like Rep=2, and Tanks=c(a4,c4,h4) and would like to graph (points) of Wgt and Day. I would think that I only need 3 colors, but when I run with only 3, only 2 lines show up. When I add

I've noticed that R has a number of very useful functions for obtaining descriptive statistics on groups of variables, including summary {stats}, describe {Hmisc}, and describe {psych}, but none that I have found is able to provided weighted descriptives of subsets of a data set (ex. descriptives for both males and females for age, where accurate results require use of sampling

Hi, You can use the projectLinear argument in BB::spg to optimize with linear equality/inequality constraints. Here is how you implement the constraint that all parameters sum to 1. require(BB) spg(par=p0, fn=myFn, project="projectLinear", projectArgs=list(A=matrix(1, 1, length(p0)), b=1, meq=1)) Hope this is helpful, Ravi [[alternative HTML version deleted]]

Hello, I'm using the R package called mvpart, which is about the multivariate regression trees. The function I wrote is: mrt1<- mvpart(coefmat~sChip+sScreen+sMem,data=mixdata, xv="pick", plot.add=TRUE,uniform=TRUE,which=4,all=TRUE,xadj=2,yadj=2,rsq=TRUE,big.pts=TRUE,wgt.ave.pca=TRUE,legend=TRUE,bars=F, pca=TRUE) where "coefmat" is a matrix(of dimension N*K) to store

Hi, Thanks for answering my question on the new smbmount not working. I tried all variety of combinations on the commands you suggested, but they all simply do this: % smbmount -c //server/share 'mount /mnt/samba-share -n CLIENT -U user' Usage: smbmount service <password> [-p port] [-d debuglevel] [-l log] Version 2.0.3 -p port connect to the specified port

hi all, I am fitting a logistic regression model on binary data. I care about the fitted probabilities, so I am not worried about infinite (or non-existent) MLEs. I use: > glm(Y~., data=X, weights=wgt, family=binomial(link=logit), maxit=250) I understand the three ways to fit model, and in my case Y is a factor, one column > Y <- c(rep("A",679), rep("B",38))

hi all, I am fitting a logistic regression model on binary data. I care about the fitted probabilities, so I am not worried about infinite (or non-existent) MLEs. I use: > glm(Y~., data=X, weights=wgt, family=binomial(link=logit), maxit=250) I understand the three ways to fit model, and in my case Y is a factor, one column > Y <- c(rep("A",679), rep("B",38))

Hello All, I noticed when I generated some boxplots, the data is presented in alphabetical order along the x-axis (the data in this case was the four quandrants of a sample area (NE,NW, SE, SW) that was my first column of data). Is there a way to have R plot the data in a different order? I imagine you could use a dummy variable, but didn't know if there might be a simple argument that

Dear R users, I'm using the "reldist" add-on package to calculate relative distribution in R as part of my research project. The subject is a general mental health score ranging from 0 to 12 (integer values only) with 0 indicating no mental health problem and positive values meaning some or sever mental health problem. When I run the programme to compare the scores of population

Greetings all, I''m forwarding a copy of an email I sent reporting attempted break-ins on my main server, earth.terran.org. I am forwarding this because I think it is relevant that folks watch for this kind of activity in their logs to catch people who "try doorknobs" in the middle of the night. After sending this email, I sent a talk request to the user, who was still logged

Hello, Anyone have information on whether RedHat-5.0+ is affected by the recent (today's) CERT advisory regarding QPOP? thanks, -bp -- B. James Phillippe <bryan@terran.org> Linux Software Engineer, WGT Inc. http://earth.terran.org/~bryan