similar to: Help : glm p-values for a factor predictor

It might help if you provided the code you used. It's possible that you didn't use direction="backward" in stepAIC(). Or if you did, it was still running, so whatever else you try will still be slow. The statement "R provides only the pvalues for each level" is wrong: look at the anova() function. Bob On 29 June 2017 at 11:13, Beno?t PELE <benoit.pele at

Hi Team, I using the syntax as: data.df<- data.frame( city= c(rep(c("Delhi", "Bangalore","Chandigarh"),each=5)), population= c(4000:6000,3500:4300,3000:3200) ) But i am getting the error as arguments imply differing number of rows: 15, 3003. Tried searching google but could not understand & find the solution. Thanks, Shivi [[alternative HTML version

Hi Michael, > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Michael > Friendly > Sent: Thursday, June 29, 2017 9:04 AM > To: Beno?t PELE <benoit.pele at acoss.fr>; r-help at r-project.org > Subject: Re: [R] Help : glm p-values for a factor predictor > > On 6/29/17 11:13 AM, Beno?t PELE wrote: > > My question is

4000:6000 gives you 4000, 4001, ..., 6000. I suspect you want population= c(seq(4000, 6000, length=5), seq(3500, 4300, length=5), seq(3000, 3200, length=5)) Bob On 20 September 2017 at 17:07, Shivi Bhatia <shivipmp82 at gmail.com> wrote: > Hi Team, > > I using the syntax as: > > data.df<- data.frame( > city= c(rep(c("Delhi",

The loop never assigns anything to d0, only t. The first line makes t a character string "d0$V1" (or "d0$V2" etc.). The second line assigns either 0 or 1 to t. Looking at this, I don't think you've got into the R psychology (bad news if you want to use R, good news in many other ways). I assume d0 is a list, so could you put the V's into a vector, and then just use

Hi, I am trying to debug the following code: for (i in 1:10){ t <- paste("d0$V",i,sep="") t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0) } and I would like to see what code is actually processing R, how can I do that? More to the point, I am trying to update my variables d0$V1 to d0$V10 according to the presence or absence of some text (contained in the file d1)

Hey I'm not really sure what I should put on here, but I am having trouble with my R code. I am trying to get the p-values, R^2s etc for a number of different groups of variables that are all in one dataset. This is the code: #Stand counter st<-1 #Collections stands<-numeric(67) slopes<-numeric(67) intercepts<-numeric(67) mses<-numeric(67) rsquares<-numeric(67)

Buen día estimados Estoy tratando de hacer un tibble con los resultados de un apply que se muestran en la consola que me da R, no estoy seguro si eso se pueda hacer, pero me gustaría organizar los resultados de esa manera. mi código es: data("mtcars") Mtcars_matriz <- as.matrix(mtcars) apply(Mtcars_matriz, MARGIN =2, FUN = shapiro.test) DF2 <- tibble(Variable = NA, W = NA, Pvalue =

Dear all, I observer a strange behavior of the pvalues of the t-test under the null hypothesis. Specifically, I obtain 2 samples of 3 individuals each from a normal distribution of mean 0 and variance 1. Then, I calculate the pvalue using the t-test (var.equal=TRUE, samples are independent). When I make a histogram of pvalues I see that consistently the bin of the smallest pvalues has a lower

Good day, I'm using a VAR model to forecast sales with some extra variables (google trends data). I have divided my dataset into a trainingset (weekly sales + vars in 2006 and 2007) and a holdout set (2008). It is unclear to me how I should predict the out-of-sample data, because using the predict() function in the vars package seems to estimate my google trends vars as well. However, I want

Hi all, My data genes [,1] [,2] [,3] [,4] [1,] 25 72 23 55 [2,] 34 53 41 33 [3,] 26 43 26 44 [4,] 36 64 64 22 [5,] 47 72 67 34 stu<-t.test(genes[,1:2],genes[,3:4]) > stu$p.value [1] 0.4198002 i get 1 pvalue for the entire col1:col2 Vs col3:col4. I am trying to get 5 p values for the 5 rows i have. I am trying to avoid a for loop coz my

Hi - I am not R expert and I would appreciate your time if you can help me about my xyplot question. I would like to add text (p-value) in a 4 panels xyplot. I thought panel = function{} should work but I am not sure where I did it wrong. The error message from the following code is "Argument subscripts is missing with no default values" xyplot(GLG ~ PD | factor(TRT) , groups =

Hi, #working directory data1 #changed name data to data1.? Added some files in each of sub directories a1, a2, etc. ?indx1<- indx[indx!=""] lapply(indx1,function(x) list.files(x)) #[[1]] #[1] "a1.txt"??????? "mmmmm11kk.txt" #[[2]] #[1] "a2.txt"??????? "mmmmm11kk.txt" #[[3]] #[1] "a3.txt"??????? "mmmmm11kk.txt" #[[4]] #[1]

Thanks for idea but it won't work for me as 'internal domains' can be anything, including gmail.com (and i don't know which of them are really internal/local, this is decided by sending SMTP everytime something is sent, based on MX records). Problem is that Dovecot/Sieve is using wrong SMTP server (one used for receiving e-mails which should NEVER be used for sending [and

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by

Op 28-1-2020 om 19:20 schreef azurit at pobox.sk: > Really no more info? You could do something with the sendmail_path or submission_host settings. Regards, Stephan. > > > > > Cit?t azurit at pobox.sk: > >> Thanks for idea but it won't work for me as 'internal domains' can be >> anything, including gmail.com (and i don't know which of them are

Hi, I became a little bit confused when working with the Wilcoxon test in R. As far as I understood, there are mainly two versions: 1) wilcox.test{stats}, which is the default and an approximation, especially, when ties are involved 2) wilcox_test{coin}, which does calculate the distribution _exactly_ even, with ties. I have the following scenario: #---BeginCode--- # big example size = 60

I made an ANOVA with repeated mesures (aov(Mesure~Distance*Genre*Correct+Error(Sujet/(Distance*Genre*Correct)), data)) and I would like to extract the pvalues. The output is: ----------------------------------------------------------- Error: Sujet Df Sum Sq Mean Sq F value Pr(>F) Residuals 21 97.082 4.623 Error: Sujet:Distance Df Sum Sq Mean Sq F value Pr(>F) Distance

Hello R users, Can someone tell if there is a package in R that can do outlier detection that give outputs simiilar to what I got from SAS below. Many thanks in advance for any help! Outlier Details Approx Chi-

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal