similar to: Function Relevel DOE NOT FOUND

Dear colleagues, How can I do to "relevel" the intercept? I need that the treatment "Db" be the intercept, and have p-values for the comparisons with the others treatments. I used the function "relevel" but it did not work out to have what I want. Thanks for your help, Xavier T1 <- read.table(file.choose(), h=T) > head(T1) treatment replicate Time

Dear list, I am running R code which produces a data.frame with variable rows. For plotting purposes I need to relevel the column called "names" but since the dimension of the data.frame can vary I do not know how to dynamically revel the data.frame. Here is an example data.frame called df to illustrate what I try to achieve: > df <-

Dear list, I'm having second thoughts after solving a very trivial problem: I want to extend the relevel() function to reorder an arbitrary number of levels of a factor in one go. I could not find a trivial way of using the code obtained by getS3method("relevel","factor"). Instead, I thought of solving the problem in a recursive manner (possibly after reading

Something that bit me: The function relevel takes a factor, and a reference level to be promoted to the first place. If ?ref? is a character this level is promoted, if it?s a numeric the ?ref?-th level is promoted. Which turns out to be very confusing if you have factor with numeric values (e.g. when reading in a csv with some dirty numeric columns and stringsAsFactors TRUE) For example:

Can anyone help me with the necessary code to relevel a numeric*factor interaction term in a linear model? I would like to report the estimate, std. error and t-value for the reference factor. First, I estimated a linear model with dummy variables and was able to retrieve model estimates for the reference factor using relevel. for example: > summary(update(mod.mod, . ~ . - dummy + +

To the R developers, The discussion below reminded me that I think it might be a good idea to take the Relevel function from the Lexis package and replace relevel in stats with it. This is really nothing special for epidemiology. It is fully compatible with the existing relevel (it actually contains the relevel code almost verbatim as a subset), but it has the extra functionality of combining

I m in front of a problem of simultaneity bias between two equations and would like to apply the two stage level least square....Is there a special command in R ? I didn't found something about that in the R help. Thanks for your help !!!! ************************************************************** Mathieu Vuilleumier - collaborateur scientifique Institut de recherches ??conomique et

Hi! My boss asked me to be able to share some directories on a FreeBSD Samba server with users already created (and used) on a NT 4.0 PDC server. So I followed http://www.sugoi.org/bits/index.php?bit_id=10 I just replaced the Windows 2k part by adding the NetBIOS name of the machine to the NT PDC. When I did the smbpasswd trick it worked like charm: testsmb# smbpasswd -j CH-DOMAIN -r PDC

Greetings, I am revisiting code from several different files I have saved from the past and all used to run flawlessly; now when I run any of the svyglm related functions, I am coming up with an error: Error in model.frame.default(formula = F3ATTAINB ~ F1PARED, data = data, : the ... list does not contain 4 elements The following is a minimal reproducible example: library(RCurl)

Hi All, I'm perplexed by the way the unclass function displays a factor whose labels have been swapped with the relevel function. I realize it won't affect any results and that the relevel did nothing useful in this particular case. I'm just doing it to learn ways to manipulate factors. The display of unclass leaves me feeling that the relevel had failed. I've checked three books

hello, I'd like know how to sort a data frame in R for example how I should do to sort by Catholic with swiss data frame like below thanks Fertility Agriculture Examination Education Catholic Infant.Mortality Courtelary 80.2 17.0 15 12 9.96 22.2 Delemont 83.1 45.1 6 9 84.84 22.2

I need to automate changing the reference value for factors in analysis, and it is my understanding that the following two sets of code should produce identical results, but they do not: 1)dataset$method 2)eval(parse(text=paste("dataset",IVcat[k],sep="$"))) (in this case I have IVcat[k]=method, which is why they are equal) I even tested them in R, which says that these

hi, i am not hitting an error when i copy and paste your code into a fresh console. maybe compare your sessionInfo() to mine? > sessionInfo() R version 3.4.1 (2017-06-30) Platform: x86_64-w64-mingw32/x64 (64-bit) Running under: Windows Server 2008 R2 x64 (build 7601) Service Pack 1 Matrix products: default locale: [1] LC_COLLATE=English_United States.1252

I am using the following model: lm <- lm(mydata[[variableName]] ~ Age + Gender + Group, data=mydata) There are 5 groups in "Group": nonc (the control), c1,c2,c3 and c4. How do I contrast nonc vs the others? and How do I contrast c1 vs other c's (ie c2,c3,c4 as a subgroup)? I have looked at the contrasts option in lm and model.matrix and am really none the wiser. Though it

I have a function (see below) that does some bootstrapping (I am happy to expand offline why I could use existing functions.) I put my results into and empty matrix and add a row of results with each iteration. My problem is i am a new user to R and I don't understand data frames, matrices, elements, and vectors well. What I would like is to have a data frame I can manipulate outside of the

Hello! I would like to know if there is a simple way to reorder levels of a given factor.Let's say that the vector testf<-factor(c("red","red","red","blue","blue","white")) levels(testf) : blue red white should have reordered levels such as levels(testf) : red blue white (this is for presentation purposes) I guess

I need to use contr.sum and observe that some levels are not statistically different from the overall mean of zero. What is the proper way of forcing the zero estimate? It seems the column corresponding to that level should become a column of zeros. Is there a way to achieve that without me constructing the design matrix? Thank you. Stephen Bond [[alternative HTML version deleted]]

G'morning What does the error message "Error in x %*% coef(object) : non- conformable arguments" indicate when calculating the response values for newdata with a model from bigglm (in package biglm), and how can I debug it? I am attempting to do Monte Carlo simulations, which may explain the loop in the code that follows. After the code I have included the output, which shows that

Hello everyone, I have a variable with several categories and I want to convert this into dummy variables and do logistic regression on it. I used model.matrix to create dummy variables but it always picked the smallest one as the reference. For example, model.matrix(~.,data=as.data.frame(letters[1:5])) will code 'a' as '0 0 0 0'. But I want to code another category as

Hello. First post, though I have read scores. Mostly i can solve my problems reading the archives, but I lack the vocabulary to ask this one right. Any social faux pais, I apologize. I am sure this is a very basic question and I am embarrassed to ask, but I spent several hours scouring the archives and trying various suggestions, but I am too new to get any of them to work. I am trying to