similar to: How to get the name of the first argument in an assignment function?

Dear All, applying some function within a with() function I wanted to use also sapply() and get() to form a data.frame, but did not succede. Below is a simplified example. It is possible to use sapply() within a with() function, it is also possible to use get() within a with() function, but when I try to use get within sapply within with I arrive at "Error in get(x, envir, mode, inherits) :

channel <- odbcConnectExcel("nuova tabella terapia occupazionale mod.xls") > ## list the spreadsheets > sqlTables(channel) TABLE_CAT TABLE_SCHEM TABLE_NAME 1 c:\\TABELLE DEFINITIVE\\nuova tabella terapia occupazionale mod <NA> 'emi tot 2006 OAI_60g TO FU1$' 2 c:\\TABELLE DEFINITIVE\\nuova

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, 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, 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 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! 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, 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

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, 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)

[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>

On Mon, 11 Jun 2012 10:48:18 -0700 Andrew Trick <atrick at apple.com> wrote: > On Jun 11, 2012, at 9:30 AM, Hal Finkel <hfinkel at anl.gov> wrote: > > > I'm considering writing more-detailed itineraries for some PowerPC > > CPUs that use the 'traditional' instruction grouping scheme. In > > essence, this means that multiple instructions will stall

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

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

Hi all, I'm working with a dataset from 10 treatments, each treatment with 30 subjects, each subject measured 5 times. The plot of the dataset suggests that a 3-parameter logistic could be a reasonable function to describe the data. When I try to fit the model using gnls I got the message 'Step halving factor reduced below minimum in NLS step'. I´m using as the initial values of the

Dear All, reposting, because I did not find a solution, maybe someone could check the example below. It's taken from the help page of survdiff. Executing it, gives the error "Error in floor(temp) : Non-numeric argument to mathematical function" best regards, Heinz library(survival) ## Example from help page of survdiff ## Expected survival for heart transplant patients based