similar to: Missing variable in new dataframe for prediction

Dear all, i generated several GLM objects, named myobject1 to myobject25. Now i'd like to update both of them. So i tried: for(ii in 1:25) { assign(paste("myobject.updated", ii, sep=""), update( myobject[ii] ,.~ + VAR2)) } Doesn't work I also tried to get all the names in a vector and update(names.myobject[ii],.~ + VAR2) Stiil doesn't work Any ideas

I''m doing some simulations for which i need to use both S-plus and R. I generate in S+ some random normal distributions to define one dataset by iteration. I need to use the same dataset generated in S-plus in R. I was first thinking to generate in R the same dataset by using the same random number generator with a fixed seed. But It seems that S-plus and R don''t use the same

Hi, suppose i have a gam object gamobject<- gam( Y~ s(X1)+s(X2)+ X3) I would like to extract the predicted partial effect of X3 but selecting it by its name, as it's to be included in a function and i don't always know the exact position of X3. something like predict(gamobject,type="terms")["X3",]. But that doesn't work as there's no name. So, looking

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:  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"

Dear R experts Sorry for this question M1 <- 1:10 lcd1 <- c(11, 22, 33, 44, 11, 22, 33, 33, 22, 11) lcd2 <- c(22, 11, 44, 11, 33, 11, 22, 22, 11, 22) lcd3 <- c(12, 12, 34, 14, 13, 12, 23, 23, 12, 12) #generating variables through sampling pvec <- c("PR1", "PR2", "PR3", "PR4", "PR5", "PR6", "PR7",

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

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

Dear R list, I'm trying to calculate Mahalanobis distances for 'Species' of 'iris' data as obtained below: Squared Distance to Species From Species: Setosa Versicolor Virginica Setosa 0 89.86419 179.38471 Versicolor 89.86419 0 17.20107 Virginica 179.38471 17.20107 0 These distances were obtained with proc 'CANDISC'

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

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

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

Hi list, I'm working with gamm models of this sort, using Simon Wood's mgcv library: gm<- gamm(Z~te(x,y),data=DATA,random=list(Group=~1)) gm1<-gamm(Z~te(x,y,by=Factor)+Factor,data=DATA,random=list(Group=~1)) with a dataset of about 70000 rows and 110 levels for Group in order to test whether tensor product smooths vary across factor levels. I was wondering if comparing those two