similar to: Ever see a stata import problem like this?

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

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

Hi R-people: When I use the command to fit a model with an intercept, only: glm ( formula=haspdata ~ 1, data=dat, family=binomial, weights= dat$hy.wgt.s, subset=(dat$haspdat0!=3) ) I get the message: Warning message: non-integer #successes in a binomial glm! in: eval(expr, envir, enclos) Does anyone know what this means?? The data for this command is listed below. Thanks, Phil Smith CDC

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

Hi everyone, I'm sorry if this turns out to be more a statistical question than one specifically about R - but would greatly appreciate your advice anyway. I've been using a logistic regression model to look at the relationship between a binary outcome (say, the odds of picking n white balls from a bag containing m balls in total) and a variety of other binary parameters:

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

Hi. I'm trying to create a 3d plot for a teaching example of finding a least-squares estimate of the parameters to fit a line to some data. I was hoping to get a nice plot with a clear, single minima where the derivative of the surface is zero. No matter how much I tinker, I can't seem to get a simple straightforward plot. Am I doing something wrong? Thanks in anticipation, Ross-c x

Hello, We use drc to fit dose-response curves, recently we discovered that there are quite different standard error values returned for the same dataset depending on the drc-version / R-version that was used (not clear which factor is important) On R 2.9.0 using drc_1.6-3 we get an IC50 of 1.27447 and a standard error on the IC50 of 0.43540 Whereas on R 2.7.0 using drc_1.4-2 the IC50 is