similar to: how to write assignment form of function

Dear All! For descriptive purposes I would like to add attributes to objects. These attributes should be kept, even if by indexing only part of the object is used. I noted that some attributes like levels and class of a factor exist also after indexing, while others, like comment or label vanish. Is there a way to make an arbitrary attribute to be kept after indexing? This would be especially

Dear All, after searching on CRAN I got the impression that there is no standard way in R to label values of a numerical variable. Since this would be useful for me I intend to create such an attribute, at the moment for my personal use. Still I would like to choose a name which does not conflict with names of commonly used attributes. Would value.labels or vallabs create conflicts? The

Hello! A problem with special characters seemed to me to be a bug. I sent a mail to R-windows at r-project.org concerning the problem (see below). How can I find out, if this is considered as a bug or an error of myself? Which part of FAQs or documentation did I miss to find the answer? thanks in advance Heinz T??chler -------------------- copy of abovementioned mail ---------- to: R-windows

Hello, it seems that the main results of survival analysis with package survival are shown only as side effects of the print method. If I compute e.g. a Kaplan-Meier estimate by > km.survdur<-survfit(s.survdur) then I can simply print the results by > km.survdur Call: survfit(formula = s.survdur) n events median 0.95LCL 0.95UCL 100.0 58.0 46.8 41.0 79.3 Is

Dear All, Assume I have a data.frame that contains also factors and I would like to get another data.frame containing the factors as numeric vectors, to apply functions like sapply(..., median) on them. I read the warning concerning as.numeric or unclass, but in my case this makes sense, because the factor levels are properly ordered. I can do it, if I write for each single column

Dear All, a Surv object I put in a data frame behaves somehow unexpected (see example). If I do a Cox regression on the original Surv object it works. If I put it in a data.frame and do the regression on the data frame it does not work. Seemingly it has to do with the class attribute, because if I change the class attribute to let "Surv" appeare first, again it works. Is this known?

[I tried to send this message privately, but the return address bounced.] I think this has been fixed in R-patched, but I doubt if the fix has been tested in Win98. Could you please download a copy from <http://cran.r-project.org/bin/windows/base/rpatched.html> and confirm that it has been fixed? Duncan Murdoch On Sat, 27 Nov 2004 23:31:23 +0100, Heinz Tuechler <tuechler at gmx.at>

Hello, still I have difficulties with variable names in functions. I know the famous example form help for deparse/substitute but I will give a simpler one to explain my problem. I know from Reid Huntsinger (Tue, 8 Feb 2005 12:39:32 -0500) that: "Semantically, R is pass-by-value, so you don't really have the names, just the values. In implementation, though, R *does* pass names, in part

Dear All, it seems to me that the function read.spss of package foreign changed its behaviour regarding factors. I noted that in version 0.8-8 variables with value labels in SPSS were transformed in factors with the labels in alphabetic order. In version 0.8-10 they seem to be ordered preserving the order corresponding to their numerical codes in SPSS. However I could not find a description of

Dear All, it seems to me that the function read.spss of package foreign changed its behaviour regarding factors. I noted that in version 0.8-8 variables with value labels in SPSS were transformed in factors with the labels in alphabetic order. In version 0.8-10 they seem to be ordered preserving the order corresponding to their numerical codes in SPSS. However I could not find a description of

Hello, applying a function to a list of variables I face the following problem: Let's say I want to compute tables for several variables. I could write a command for every single table, like bravo<-c(1,1,2,3,5,5,5,);charly<-c(7,7,4,4,2,1) table(bravo); table(charly) > table(bravo); table(charly) bravo 1 2 3 5 2 1 1 3 charly 1 2 4 7 1 1 2 2 The results are two tables with the

Dear All! If I try to compare the attributes of two objects, I find a surprising behaviour of attr.all.equal(). With identical attributes I receive the answert NULL. If the attributes differ, the answer is as expecxted and differences are shown. all.equal(attributes(), attributes()) instead returns TRUE, if attributes are equal. See example: v <- 1:5 attr(v, 'testattribute')

Dear All! Reading character strings containing an "umlaut" from a csv-file I find a (to me) surprising behaviour in R 2.8.0, that I did not notice in R 2.7.2. A comparison by "==" results in FALSE, while grep does find the aggreement. See the example below. The crucial line is x=="div 1-2 Ver?nderungen", with the result [1] FALSE in R 2.8.0 but [1] TRUE in R

Dear All, is there a stratified version of the Wilcoxon test (also known as van Elteren test) available in R? I could find it in the survdiff function of the survival package for censored data. I think, it should be possible to use this function creating a dummy censoring indicator and setting it to not censored, but may be there is a better way to perform the test. Thanks, Heinz T??chler

Are you sure that you want to read in the output_file in text <- readLines(output_file, warn = FALSE)? best regards, Heinz Thierry Onkelinx wrote/hat geschrieben on/am 06.09.2017 11:41: > Dear all, > > I'm trying to write a post_processor() for a custom rmarkdown format. The > goal of the post_processor() is to modify the latex file before it is > compiled. For some

Dear All! If I pass an object to an assignment function I cannot get it's name by deparse(substitute(argument)), but I get *tmp* and I found no way to get the original name, in the example below it should be "va1". Is there a way? Thanks, Heinz ## example 'fu1<-' <- function(var, value) { print(c(name.of.var=deparse(substitute(var))))} fu1(va1) <- 3 name.of.var

Dear all, Last December there was a thread regarding the famous FAQ 7.21 "How can I turn a string into a variable?" and asking what people want to do with these strings. My, certainly trivial application would be as follows: Assume I have a data.frame containing besides others also the columns f1, f2, ..., fn and I want to create a command like: apply(sapply(list(f1,f2,f3),is.na),2,sum)

Hi List, A friend of mine recently asked the same question as Heinz T?chler. Since I've already written the code I'd like to share with the list. # x is an object returned by "survfit"; # "smed" returns a matrix of 5 columns of # n, events, median, 0.95LCL, 0.95UCL. # The matrix returned has rownames as the # group labels (eg., treatment arms) if any. smed <-

Dear All, trying to write a data.frame, containing Surv objects to a csv-file I get "Error in dimnames(X) <- list(dn[[1L]], unlist(collabs, use.names = FALSE)) : length of 'dimnames' [2] not equal to array extent". See example below. May be, I overlooked something, but I expected that also data.frames containing Surv objects may be written to csv files. Is there a

Dear all, I'm trying to write a post_processor() for a custom rmarkdown format. The goal of the post_processor() is to modify the latex file before it is compiled. For some reason the post_processor() is not run. The post_processor() does work when I run it manually on the tex file. Any suggestions on what I'm doing wrong? Below is the relevant snippet of the code. The full code is