similar to: loop and sapply problem, help need

Hi, I have two questions: 1) Is there any way of renaming zpool without export/import ?? 2) If I took hardware snapshot of devices under a zpool ( where the snapshot device will be exact copy including metadata i.e zpool and associated file systems) is there any way to rename zpool name of snapshotted devices ?? without losing data part? Thanks & Regards, sridhar. -- This message posted

Hi R Fans, I stumbled across a strange (I think) bug in R 2.9.1. I have read in a data file with 5934 rows and 9 columns with the commands: daten = data.frame(read.table("C:/fussball.dat",header=TRUE)) Then I needed a subset of the data file: newd = daten[daten[,1]!=daten[,2],] --> two values do not meet the logical specification and are dropped. The strange thing about it:

Hi all, If the question below as been answered before I apologize for the posting. I would like to get the frequencies of occurrence of all values in a given variable in a multivariate dataset. In short for each variable (or field) a summary of values contained with in a value:frequency pair, there can be many such pairs for a given variable. I would like to do the same for several such variables.

 Hi:  Using nls how can I increase the numbers of iterations to go beyond 50.  I just want to be able to predict for the last two weeks of the year.  This is what I have:  weight_random <- runif(50,1,24)  weight <- sort(weight_random);weight weightData <- data.frame(weight,week=1:50)                          weightData plot(weight ~ week, weightData) M_model <- nls(weight ~ alpha +

Dear All   I have the following code set up: Code #1 a <-matrix(seq(0,8, by = sign(8-0)*0.25)) b <-matrix(seq(8,16, by = sign(16-8)*0.25)) c <-runif(1000,50,60) d <-exp(-c*a)+exp(-c*b)   This will give me the obvious error message of lengths not matching. What I am trying to do here is to have 33 rows x 1000 columns d values calculated in total. As an eaxmple for visual, this is what

?s 17:38 de 30/11/2023, Robert Baer escreveu: > I am having trouble using back ticks with the R extractor function > 'predict' and an lm() model.? I'm trying too construct some nice vectors > that can be used for plotting the two types of regression intervals.? I > think it works with normal column heading names but it fails when I have > "special"

Hi, I'm using a loop to evaluate several models by taking adjacent variables from my dataframe. When i try to get predictions for new values, i get an error message about a missing variable in my new dataframe. Below is an example adapted from ?gam in mgcv package library(mgcv) set.seed(0) n<-400 sig<-2 x0 <- runif(n, 0, 1) x1 <- runif(n, 0, 1) x2 <- runif(n, 0, 1) x3 <-

Hey List, I did a multiple regression and my final model looks as follows: model9<-lm(calP ~ nsP + I(st^2) + distPr + I(distPr^2)) Now I tried to predict the values for calP from this model using the following function: xv<-seq(0,89,by=1) yv<-predict(model9,list(distPr=xv,st=xv,nsP=xv)) The predicted values are however strange. Now I do not know weather just the model does not fit

I am having trouble using back ticks with the R extractor function 'predict' and an lm() model.? I'm trying too construct some nice vectors that can be used for plotting the two types of regression intervals.? I think it works with normal column heading names but it fails when I have "special" back-tick names.? Can anyone help with how I would reference these?? Short of

Good day everyone, Can anyone suggest an effective method to solve the following problem: I have 2 dataframes D1 and D2 as follows: D1: dates ws wc pwc 2005-10-19:12:00 10.8 80 81 2005-10-20:12:00 12.3 5 15 2005-10-21:15:00 12.3 3 15 2005-10-22:15:00 11.3 13 95 2005-10-23:12:00 12.3 13 2 2005-10-24:15:00 10.3 2 95 2005-10-25:15:00 10.3 2 2 D2:

Hi Maria If you have problems just start with a small model with predictions and then plot with xyplot the same applies to xyplot Try library(gamm4) spring <- dget(file = "G:/1/example.txt") str(spring) 'data.frame': 11744 obs. of 11 variables: $ WATERBODY_ID : Factor w/ 1994 levels "GB102021072830",..: 1 1 2 2 2 3 3 3 4 4 ... $ SITE_ID

Hi, We have this backup server running for a long time without problem, it run the rsync command from cron and connect to many other servers geting the files from those servers. One of these servers crashed a week before and we restored it, since then the rsync that connect to this server is giving the following erro after it get some files: rsync error: timeout in data send/receive (code

Dear all,? I hope that you can help me on this. I have been struggling to figure this out but I haven't found any solution. I am running a generalised mixed effect model, gamm4, for an ecology project. Below is the code for the model: model<-gamm4(LIFE.OE_spring~s(Q95, by=super.end.group)+Year+Hms_Rsctned+Hms_Poaching+X.broadleaved_woodland? ? ? ? ? ? ?+X.urban.suburban+X.CapWks,

Here's my little discussion example for a quadratic regression: http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R Students press me to know the benefits of poly() over the more obvious regression formulas. I think I understand the theory on why poly() should be more numerically stable, but I'm having trouble writing down an example that proves the benefit of this. I

Executive Summary: This only concerns predict() at new points when the formula contained things like poly() or ns() {in the RHS}. Because I hadn't have time to fix this yet, I submit it. [R version 1.5.1 and at least 1.4.1, probably all versions at all:] For these, R 1.5.x does not differ from 1.4.1 : Inspite of R 1.5.0's "NEWS" entry and the ?Safepredict help page,

Dear all, I'm a fairly new R user having two questions regarding gam: 1. The prediction example on p. 38 in the mgcv manual. In order to get predictions based on the original data set, by leaving out the 'newdata' argument ("newd" in the example), I get an error message "Warning message: the condition has length > 1 and only the first element will be used in: if

I am working on Johansen cointegration test, using urca and var package. in the selection of var, I have got following results. >VARselect(newd, lag.max = 10,type = "none") $selection AIC(n) HQ(n) SC(n) FPE(n) 6 6 6 5 $criteria 1 2 3 4 5 6 7 8 9 AIC(n) -3.818646e+01 -3.864064e+01

Say I fit a logistic model and want to calculate an odds ratio between 2 sets of predictors. It is easy to obtain the difference in the predicted logodds using the predict() function, and thus get a point-estimate OR. But I can't see how to obtain the confidence interval for such an OR. For example: model <- glm(chd ~age.cat + male + lowed, family=binomial(logit)) pred1 <-

thank you for your valuable reply. I have attached my commands, results, and data with this mail..maybe it will be beneficial for you to feedback. On Tue, Nov 21, 2017 at 9:13 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > Your example is incomplete... as the bottom of this and every post says, > we need to be able to proceed from an empty R environment to wherever you

Your example is incomplete... as the bottom of this and every post says, we need to be able to proceed from an empty R environment to wherever you are having the problem (reproducible), in as few steps as possible (minimal). The example needs to include data, preferably in R syntax as the dput function creates... see the howtos referenced below for help with that. [1], [2], [3] You also need to