similar to: levels() deletes other attributes

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! 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, 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)

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, 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, as shown in the example, unique() deletes names of vector elements. Is this intended? Of course, one can use indexing by !duplicated() instead. Greetings, Heinz ## unique deletes names v1 <- c(a=1, b=2, c=3, e=2, a=4) unique(v1) # names deleted v1[!duplicated(v1)] # names preserved platform i386-pc-mingw32 arch i386

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, 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: ##

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, there seems to be some strange influence of the Design package on data.frame. If I build a data.frame containing a Surv object without loading the package Design, the data frame is usable to coxph. If instead I just load Design and build a data.frame afterwards, the naming of the Surv object is different and it does not work with coxph. (In my real application I loaded Design to use the

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 drop unused factor levels two ways are outlined in R-help. In both cases a label attribute is lost. The same happens, when using car:::recode. Is there a simple way to avoid losing attributes? Thanks, Heinz ## example ff <- factor(substring("statistics", 1:10, 1:10), levels=letters) attributes(ff)$label <- 'test label' attributes(ff)$label gg <- ff[,

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

Dear All, is there a simple way to print a data.frame without its row.names? example: datum <- as.Date(c("2004-01-01", "2004-01-06", "2004-04-12")) 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

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

[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>

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

Hi all, I want to do this kind of function In R enviroment : For example : R <- 4 testString <- "I love $R" then search this testString, when find "$R",replace "$R" to R ,and because the value of R is 4 So the final string I want to get is "I love 4" How can I implement? Thanks advance Michael

dear R experts---I am trying to replicate a perl feature. I want to be able to embed R commands inside a character string, and have the string be printed with the command executed. my perl equivalent is my $a=10; my $teststring = "the expression, $a+1, is ::$a+1::, but add one more for ::$a+2::\n"; $teststring =~ s/::(.*?)::/$1/gee; print $teststring; of course, R does not use