similar to: predict() still not yet "safe".. (PR#1840)

Hello, I have two dataframes DF1 and DF2 that should be identical but are not (DF1 has some rows that aren't in DF2, and vice versa). I would like to produce a new dataframe DF3 containing rows in DF1 that aren't in DF2 (and similarly DF4 would contain rows in DF2 that aren't in DF1). I have a solution for this problem (see self contained example below) but it's awkward and

?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 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

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 <-

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",

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

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:

I am fitting a regression model with a bs term and then making predictions based on the model. According to some info on the internet at http://www.stat.auckland.ac.nz/~yee/smartpred/DummiesGuide.txt there are some problems with using predict.lm when you have a model with terms such as bs,ns,or poly. However when I used one of the examples they said would illustrate the problems I get virtually

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

Hello everyone. I am doing a logistic gam (package mgcv) on a pretty large dataframe (130.000 cases with 100 variables). Because of that, the gam is fitted on a random subset of 10000. Now when I want to predict the values for the rest of the data, I get the following error: > gam.basis_alleakti.1.pr=predict(gam.basis_alleakti.1, +

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

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

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

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

Hi, I'm working with gamm models of this sort: 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 if I use plot(gm1$gam), I obtain 3 different surface plots, one for each level of my factor but I would like to create more complex contour plots for those 3

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

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 <-