search for: tuechler

...ounced.] 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> wrote : >Dear Developers! > >Using special characters I found a strange behaviour in R 2.0.1 and equally >in >R : Copyright 2004, The R Foundation for Statistical Computing >Version 2.0.1 (2004-11-15), ISBN 3-900051-07-0 > >Operating System: Win...

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

Dear All, to produce output of several columns of a data frame, I tried to use lapply and also l_ply. In both cases, I would like to print a header line containing also the name of the respective column in the data frame. For example, I would like the following lapply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x)))) to produce: [1] "a" [1] "b" and

...04-04-12")) content <- c('Neujahr', 'Hl 3 K.', 'Ostern') df1 <- data.frame(datum, content) print(df1) datum content 1 2004-01-01 Neujahr 2 2004-01-06 Hl 3 K. 3 2004-04-12 Ostern Can I get this "table" without 1, 2, 3 ? Thanks in advance Heinz Tuechler

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, to my surprise as.factor does not accept a levels argument. Maybe I did not read the documentation well enough. See the example below. I wanted to use ch1 as factor in the newdata argument of survfit, so I assumed that I could write as.factor(ch1, levels=ch1), since the order should be kept. But as.factor(ch1, levels=ch1) results in the error: Error in as.factor(ch1, levels = ch1)

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

Dear All, where can I find information about how to write an assigment form of a function? For curiosity I tried to write a different form of the levels()-function, since the original method for factor deletes all other attributes of a factor. Of course, the simple method would be to use instead of levels(x) <- newlevels, attr(x, 'levels') <- newlevels. I tried the following: ##

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, 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! 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)

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

...ft, df$right)) df left right 1 1 6 2 2 7 3 3 8 4 4 9 5 5 10 df.combined comb 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Heinz Tuechler Sent: Monday, April 3, 2023 4:39 PM To: r-help at r-project.org Subject: Re: [R] Simple Stacking of Two Columns Jeff Newmiller wrote/hat geschrieben on/am 03.04.2023 18:26: > unname(unlist(NamesWide)) Why not: NamesWide <- data.frame(Name1=c("Tom","Dick"),Name2=c("L...

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, 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

...rnative that is apriori highly unlikely takes a lot of shifting. John Maindonald email: john.maindonald at anu.edu.au<mailto:john.maindonald at anu.edu.au> On 3/04/2016, at 22:00, r-help-request at r-project.org<mailto:r-help-request at r-project.org> wrote: From: Heinz Tuechler <tuechler at gmx.at<mailto:tuechler at gmx.at>> Subject: Re: [R] p values from GLM Date: 3 April 2016 11:00:50 NZST To: Bert Gunter <bgunter.4567 at gmail.com<mailto:bgunter.4567 at gmail.com>>, Duncan Murdoch <murdoch.duncan at gmail.com<mailto:murdoch.duncan at gmail...

> 09.01.2023 18:05:58 akshay kulkarni <akshay_e4 at hotmail.com>: > > We are living in the 21st century world, and the R-core team might,I suppose, have a definite reason ... > Maybe compatibility reasons with S and R-versions from the 20st century? But maybe, you would have expected some reason even then. best regards, Heinz

Dear All, Does someone have, or know of survexp ratetables for european contries, especially Austria and Germany? I know only about slopop in the package relsurv. Thanks in advance Heinz T??chler