similar to: Windows Front end-crash error

Dear List: I am running into a memory issue that I haven't noticed before. I am running a simulation with all of the code used below. I have increased my memory to 712mb and have a total of 1 gb on my machine. What appears to be happening is I run a simulation where I create 1,000 datasets with a sample size of 100. I then run each dataset through a gls and obtain some estimates. This works

Dear List: A few weeks ago I posted some questions regarding data simulation and received some very helpful comments, thank you. I have modified my code accordingly and have made some progress. However, I now am facing a new challenge along similar lines. I am attempting to simulate 250 datasets and then run the data through a linear model. I use rm() and gc() as I move along to clean up the

Hi, I've a data frame like this: > as.data.frame(cbind(rnorm(1:12),rnorm(1:12))) V1 V2 1 -1.30849402 -0.52094136 2 0.96157302 0.76217871 3 -0.44223351 -1.72630871 4 -0.10432438 -1.04732942 5 -1.38748914 0.95877311 6 -0.63965975 0.65494811 7 -0.24058318 0.19496830 8 -0.11172988 1.01680655 9 0.08065333 0.22168589 10 0.25196536 0.84619914 11

Dear R users: I am not entirely convinced that clogit gives me the correct result when I use pspline() and maybe you could help correct me here. When I add a constant to my covariate I expect only the intercept to change, but not the coefficients. This is true (in clogit) when I assume a linear in the logit model, but the same does not happen when I use pspline(). If I did something similar

Hi,I have a result from polr which I fit a univariate variable (of ordinal data) with probit function. What I would like to do is to overlay the plot of my fitted values with the different intercept for each level in my ordinal data. I can do something like:lines(rep(intercept1, 1000), seq(from=0,to=max(fit),by=max(fit)/1000))where my intercept1 is, for example, the intercept that breaks between

You also need to reply-all so the mailing list stays in the loop. -- Sent from my phone. Please excuse my brevity. On December 19, 2017 4:00:29 PM PST, Timothy Axberg <axbergtimothy at gmail.com> wrote: >Sorry about that. Here is the code typed directly on the email. > >qe = (Qmax * Kl * ce) / (1 + Kl * ce) > >##The data >ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)

Hello, R users, I applied segmented regression method contributed by Muggeo and got different slope estimates depending on the initial break points. The results are listed below and I'd like to know what is a reasonable approach handling this kinds of problem. I think applying various initial break points is certainly not a efficient approach. Is there any other methods to deal with segmented

I appreciate to Roger Koenker, Achim Zeileis and Vito Muggeo for their informative answers. Listed below is unedited replies I got followed by the question I posted. Kyong 1. Roger Koenker: You might try rqss() in the quantreg package. It gives piecewise linear fits for a nonparametric form of median regression using total variation of the derivative of the fitted function as a penalty

Should I repost the question with reply-all? On Tue, Dec 19, 2017 at 6:13 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > You also need to reply-all so the mailing list stays in the loop. > -- > Sent from my phone. Please excuse my brevity. > > On December 19, 2017 4:00:29 PM PST, Timothy Axberg < > axbergtimothy at gmail.com> wrote: > >Sorry about

Hello R users, I've used the following help two compare two regression line slopes. Wanted to test if they differ significantly: Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope

Hi, for testing if the slope of experimental data differs from a given slope I'm using the function "test_regression_against_slope" (see below). I am now confronted with the problem that I have data which requires a modelII regression (also called reduced major axes regression (RMA) or geometric mean regression). For this I use the function "modelII" (see below). What

This isn't a question about R, but I'm hoping someone will be willing to help. I've been looking at calibration plots in multiple regression (plotting observed response Y on the vertical axis versus predicted response [Y hat] on the horizontal axis). According to Frank Harrell's "Regression Modeling Strategies" book (pp. 61-63), when making such a plot on new data

Hi, sry if i hijack this, but maybe it's helpful for other gluster users... > pure NVME-based volume will be waste of money. Gluster excells when you have more servers and clients to consume that data. > I would choose LVM cache (NVMEs) + HW RAID10 of SAS 15K disks to cope with the load. At least if you decide to go with more disks for the raids, use several (not the built-in ones)

Hello, I have some trouble using step() and stepAIC() functions. I'm predicting recruitment against several factors for different plant species using a negative binomial glm. Sometimes, summary(step(model)) or summary(stepAIC(model) does not select the best model (lowest AIC) but just stops before. For some species, step() works and stepAIC don't and in others, it's the opposite.

Actually, pure NVME-based volume will be waste of money. Gluster excells when you have more servers and clients to consume that data. I would choose? LVM cache (NVMEs) + HW RAID10 of SAS 15K disks to cope with the load. At least if you decide to go with more disks for the raids, use several? (not the built-in ones) controllers. @Martin, in order to get a more reliable setup, you will have to

Hello, I would like to write a loop to 1) run 100 linear regressions, and 2) compile the slopes of all regression into one vector. Sample input data are: y1<-rnorm(100, mean=0.01, sd=0.001) y2<-rnorm(100, mean=0.1, sd=0.01) x<-(c(10,400)) #I have gotten this far with the loop for (i in 1:100) { #create the linear model for each data set model1<-lm(c(y1[i],y2[i])~x)

Dear, i am a student and I need help in comparing between different slopes and finding whther there is a significant difference between them? Thanks a lot [[alternative HTML version deleted]]

Hi all, I am trying to implement the following matlab code with Ryacas : syms U x x0 C d1=diff(U/(1+exp(-(x-x0)/C)),x); pretty(d1) d2=diff(U/(1+exp(-(x-x0)/C)),x,2); pretty(d2) solx2 = solve(d2 == 0, x, 'Real', true) pretty(solx2) slope2=subs(d1,solx2) I have tried the following : library(Ryacas) x <- Sym("x");U <- Sym("U");x0 <-

Have you studied the "Introduction to Ryacas" vignette that come with the package? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Sep 19, 2017 at 2:37 AM, Vivek Sutradhara <viveksutra at gmail.com> wrote:

Thanks for the response. Yes, I did study the vignette but did not understand it fully. Anyway, I have tried once again now. I am happy to say that I have got what I wanted. library(Ryacas) x <- Sym("x");U <- Sym("U");x0 <- Sym("x0");C <- Sym("C") my_func <- function(x,U,x0,C) { return (U/(1+exp(-(x-x0)/C)))} FirstDeriv <-