search for: leasts

I am running a Xen installation on a Centos-4.4 based test box, my desktop unit, and I am having intermittent problems installing certain pieces of software. Some time back I posted to this list regarding my inability to install Adobe Acrobat on this unit and was advised to essentially repackage the software and try again. This is beyond my present means of knowledge or time to acquire same and

I have some output from a survey shown below and would like to insert total rows for each category to then be able to use it as a data set for further analysis. Is it possible to do this in R. In this example I would like to insert a row between rows 5 and 6, 10 and 11, 15 and 16, 20 and 21 and put in the sum of the rows 1-5, 6-10, 11-15, 16-20 and to delete row 21 which has NA in it. I would

Hello R-helpers, I would like to calculate least square means after having built a GLM with quasipoisson errors. In my model the dependent variable is continuous, I have one continuous independent variable and one categorical independent variable (that is the variable for which I would like to calculate the least square means). I've looked around for the command to calculate the least

Rick, Mark asked "why" you want the least amount of compression. I think we still do not understand your goals. If you can give us a hint at what you're trying to accomplish in your search for the least amount of compression, that might help us find the right answer. Brian Willoughby Sound Consulting On Jul 12, 2007, at 13:26, Rick wrote: > I try to make it easier for

I would really appreciate it if someone can give suggestions on how to do spectra fitting in R using ordinary least square fitting and non-negativity constraints. The lm() function works well for ordinary least square fitting, but how to specify non-negativity constraints? It wouldn't make sense if the fitting coefficients coming out as negative in absorption spectra deconvolution. Thanks.

Friends, Is there a Partial Least Squares package implemented in R? Thanks, Lana [[alternative HTML version deleted]]

This is a simple problem, but a mystery to me. I'm trying to grab $Family "Scelionidae" from one dataframe and put it into another dataframe occupied with NA in $Family. The result is a "158" ends up there instead of Scelionidae. Simply put fam$Family[1] <- least$Family[1] If I have made a mistake here, can somebody point it out. I've included the simple

I'm wondering if english is not Rick's best language ("string setting" is just not quite the correct phrase). So, given that, I also wonder if by "least compression" he means "compressed into least number of bytes"??? If that is the case, then he actually wants "best compression" (most compression) or "flac --best" and his question makes

Dear all, Does any one know if any R package or function can do Ordinary Least Products regression? Many thanks! Bill

I am considering buying a small, and therefore easily portable, computer as an alternative to the laptop I already have. Obviously it would not have battery, a screen, nor a keyboard etc. but more or less be an easily portable computing unit to move between offices where a keyboard and monitor(s) could then be connected. I want to run CentOS 7, later CentOS 8. The smaller, the better, however,

Hi, I know ways to do this but they all seem awkward and I somehow believe that there is a convenient shortcut. If I have a data.frame with many columns, how can I request all rows for which at least one column satisfy an expression? For instance, all rows where at least one column is negative. Many thanks, Werner

Dear all, I'm trying to select from a dataframe all rows which correspond to a factor (the id variable) for which there exists at least one positive value of a certain variable. As an example: x <- data.frame(matrix(c(rep(11,4),rep(12,3),rep(13,3),rep(0,3),1,rep(0,4),rep(1,2)),ncol=2)) > x X1 X2 1 11 0 2 11 0 3 11 0 4 11 1 5 12 0 6 12 0 7 12 0 8 13 0 9 13 1 10

My synapses are rather fried after a long few days of debugging other problems, so perhaps I'm being lazy in sending this to the general list, but I can't think straight about it. Forgive me if there is an overly obvious solution to this. I have a list of phone numbers that are SIP extensions. I'd like to dial them via SIP if ${EXTEN} is equal to one of those numbers. If

What fun. I don't often stoke up the boiler and actually compile open-source code, but here I am at the beginning of just such a joyous and wondrous adventure. I'm running the latest level of AIX (V5.3...V4.3 has been obsolete for at least, what, 5 years) and using the IBM compiler (xlc is similarly obsolete, from back when the compiler came with the O/S...oh, for the halcyon days of

Greetings, I would like to fit a multiple linear regression model in which the residuals are expected to follow a multivariate normal distribution, using weighted least squares. I know that the data in question have biases that would result in correlated residuals, and I have a means for quantifying those biases as a covariance matrix. I cannot, unfortunately, correct the data for these biases.

I need to find out the difference between the way R calculates weighted regression and standard regression. I want to plot a 95% confidence interval around an estimte i got from least squares regression. I cant find he documentation for this ive looked in ?stats ?lm ?predict.lm ?weights ?residuals.lm Can anyone shed light? thanks Chris. -- View this message in context:

Dear all, When doing nonlinear regression, we normally use nls if e are iid normal. i learned that if the form of the variance of e is not completely known, we can use the IRWLS (Iteratively Reweighted Least Squares ) algorithm: for example, var e*i =*g0+g1*x*1 1. Start with *w**i = *1 2. Use least squares to estimate b. 3. Use the residuals to estimate g, perhaps by regressing e^2 on

Hello, I have the following model: lm.7 <- lm(Y ~ F + C1 + C2 , data = EM4) F is a 4-level factor, the rest are covariates centered at their mean (Y is a two-column matrix). I have tried to find functions to give the model-adjusted means (adjusted at the covariates'means) and their standard deviations for each. (That is, what I believe is called in SAS "least square or LS-means,

Hi, am very new to R and I've written an optim function, but can't get it to work least.squares.fitter<-function(start.params,gr,low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F) { result<-optim(par=start.params,method=c('Nelder-Mead'),fn=least.squares.fit,lower=low.constraints,upper=high.constraints,data=data,scale=scale,ploton=ploton)

Your data frame fam contains factors. Turn it into character strings using fam$Family = as.character(fam$Family) and try again. It may be helpful if you read up on R's factors, see ?factor. HTH, Peter On Tue, Nov 21, 2017 at 2:14 PM, Glen Forister <gforister at gmail.com> wrote: > This is a simple problem, but a mystery to me. > I'm trying to grab $Family