similar to: Getting ordered factor levels from C

Hello, I can get the median for each factor, but I'd like another column to go with each factor. The nm column is a long name for the lvls column. So unique work except for the order can get messed up. Example: x =

I'm having a problem with aggregate.formula when I call it in a function and the function is converted from a string in the funtion I think my problem may also only occur when the left hand side of the formula is cbind(...) Here is example code that generates a dataset and then the error. The first function "agg2" fails > agg2(FALSE) do agg 2 Error in m[[2L]][[2L]] : object

Hi All, I am currently conducting a mixed model. I have 7 repeated measures on a simulated clinical trial. If I understand the model correctly, the outcome is the measure (as a factor) the predictors are clinical group and trial (1-7). The fixed factors are the measure and group. The random factors are the intercept and id and group. I tried using 2 functions to calculate mixed effects.

xtabs does this automatically if your cross classifying variables are factors with levels all the cities (sorted, if you like): > x <- sample(letters[1:5],8, rep=TRUE) > y <- sample(letters[1:5],8,rep=TRUE) > xtabs(~ x + y) y x c d e a 1 0 0 b 0 0 1 c 1 0 0 d 1 1 1 e 1 1 0 > lvls <- sort(union(x,y)) > x <- factor(x, levels = lvls) > y <-

Is there any way that I can see the step by step code for functions in the base package? For instance the dexp function. I am a student working on writing my own function for something that is similar to this dexp function and I would like to see the step by step code. Brittany Laine GTA WVU Statistics Department 331 Hodges

Dear William and Ben, Thank you for your replies and elegant solutions. I am having trouble with the fact that two of the previous locations do not appear in current locations (that is no one moved to OKC and Dallas from other cities), so these two cities are not being included in the output. I have provided a better sample of the data and the ideal output (wide form - a 10x10 bilateral matrix)

Hi Rowland, Thanks for the help and ideally I would like to get rid of the sub domain all together but that is probably not going to happen. So couple comments and please forgive any of my ignorance. For your second question, all users in the subdomain who have access to the SAMBA server do have uidNumber set and it matches the uidNumber set in MASTER. Since this is the case, would the

I really like the ease of use with the boxplot command in R. I would rather have a boxplot that shows the average value and the standard deviation then the median value and the quartiles. Is there a way to do this? Chad Junkermeier, Graduate Student Dept. of Physics West Virginia University PO Box 6315 210 Hodges Hall Morgantown WV 26506-6315 phone: (304) 293-3442 ext. 1430 fax: (304)

The Textile Editor Helper (TEH) is a Javascript-based text formatting toolbar that will be added to all of your text areas that utilize the TEH feature. TEH was developed to provide a more WYSIWYG-ish option for users of our Rails CMS called slate while still letting us use Textile. ====================== = So what are the features? ====================== These are a list of the current features

First of all, I do not understand why conversion to characters are not allowed. That's what Sundar's solution is doing implicitly (but more elegantly). Here's a test of all three. See the function definitions below. > f1 <- factor(sample(letters[1:3], 1e4, replace=TRUE)) > f2 <- factor(sample(letters[3:5], 1e4, replace=TRUE)) > f3 <- factor(sample(letters[5:7],

I have two factors l1, l2, and I'd like to merge them. (Remark: The factors can not be converted to charaters) Function c() does not give me the result I want: > l1 = factor(c('aaaa', 'bbbb')) > l2 = factor(c('ccc', 'dd')) > lMerge = factor(c(l1, l2)) > lMerge [1] 1 2 1 2 Levels: 1 2 > I'd like to merge l1 and l2 and to get lMerge

Hello, I am trying to map a factor variable within a data frame to a new variable whose entries are derived from the content of the original variable and there are fewer factors in the new variable. That is, I'm trying to set up a surjection. After first thinking that this would be a common operation and would have a quite simple interface, I can not seem to find one, nor any similar posts

On Wed, 18 Apr 2018 10:52:12 -0400 Wvu Hpc <wvu.hpc at gmail.com> wrote: > Hi Rowland, > > Thanks for the help and ideally I would like to get rid of the sub > domain all together but that is probably not going to happen. > > So couple comments and please forgive any of my ignorance. No problem. > > For your second question, all users in the subdomain who have

Has anyone tried to create dialog boxes for Windows in R so that one doesn't have to type in so much information but rather enter it in a menu-based format. If not, does anyone plan on doing this in the future if it's possible? Thanks. George (Kelley) George A. Kelley, DA, FACSM Professor & Director, Meta-Analytic Research Group School of Medicine Department of Community Medicine

Hello, I want to do regression or missing value imputation by knn. I searched r-help mailing list. This question was asked in 2005. ksmooth and loess were recommended. But my case is different. I have many predictors (p>20) and I really want try knn with a given k. ksmooth and loess use band width to define neighborhood size. This contrasts to knn's variable band width via fixing a

Hi, I have a character vector as below: a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'

Hi! This is my first time posting. I've read the general rules and guidelines, but please bear with me if I make some fatal error in posting. Anyway, I have a continuous response and 29 predictors made up of continuous variables and nominal and ordinal categorical variables. I'd like to do lasso on these, but I get an error. The way I am using "lars" doesn't allow for the

How to use sub, gsub, etc. to replace "\" in a string to "/"? For example,convert "C:\foo\bar" to "C:/foo/bar". Thanks, Shengqiao Li

I know that I can get a count of histogram bins in base R with plot=FALSE. However, I'd like to do the same thing with lattice. The problem is that I've set up shingles, and I'd like to get the count within each bin within each shingle. plot=FALSE doesn't seem to do it. [[alternative HTML version deleted]]

Hello everyone, I added some code to sftp, which would be useful for me. I add a command "run", which can run any command on the remote shell, in the mean time I want the server send the standard output back. for example, when I : sftp> run ls I want to see the result locally. I thought the stdout of sftp-server has been redirected to the client. But I get "message too