similar to: Fwd: Re: [EXTERNAL] Re: backquotes and term.labels

>>>>> Ben Bolker <bbolker at gmail.com> >>>>> on Thu, 8 Mar 2018 09:42:40 -0500 writes: > Meant to respond to this but forgot. > I didn't write a new terms() function -- I added an attribute to the > terms() (a vector of the names > of the constructed model matrix), thus preserving the information at > the point when

Thanks to Bill Dunlap for the clarification. On follow-up it turns out that this will be an issue for many if not most of the routines in the survival package: a lot of them look at the terms structure and make use of the dimnames of attr(terms, 'factors'), which also keeps the unneeded backquotes. Others use the term.labels attribute. To dodge this I will need to create a

A user reported a problem with the survdiff function and the use of variables that contain a space.? Here is a simple example.? The same issue occurs in survfit for the same reason. lung2 <- lung names(lung2)[1] <- "in st"?? # old name is inst survdiff(Surv(time, status) ~ `in st`, data=lung2) Error in `[.data.frame`(m, ll) : undefined columns selected In the body of the code

Meant to respond to this but forgot. I didn't write a new terms() function -- I added an attribute to the terms() (a vector of the names of the constructed model matrix), thus preserving the information at the point when it was available. I do agree that it would be preferable to have an upstream fix ... On Thu, Mar 8, 2018 at 9:39 AM, Therneau, Terry M., Ph.D. via R-devel <r-devel

I believe this has to do terms() making "term.labels" (hence the dimnames of "factors") with deparse(), so that the backquotes are included for non-syntactic names. The backquotes are not in the column names of the input data.frame (nor model frame) so you get a mismatch when subscripting the data.frame or model.frame with elements of terms()$term.labels. I think you can

That's more or less right. We wrote a terms.merMod method, which accesses the terms component of the @frame slot, which we have modified upstream ... Do you mean term.labels rather than term.names? BTW ?terms.object says (under the "term.labels" element): Non-syntactic names will be quoted by backticks: this makes it easier to re-construct the formula from the term

This is more of a general question without data. After doing 'survdiff', from the 'survival' package, on strata including four groups (so 4 curves on a Kaplan Meier curve) you get a chi squared p-value whether to reject the null hypothesis or not. Is there a method to followup with pairwise testing on the respective groups? I have searched the library but have come up with

Dear Prof. Therneau I was impressed to discover that the 'survConcordance' now handles Surv() objects in counting format (example below to clarify what I mean). This is not documented in the help page for the function. I am very curious to see how a c-index is estimated in this case, using just the linear predictors. It was my impression that with left truncation the ordering of

The coxme package has a ranef() method, as does lme4. I'm having trouble getting them to play together, as shown below. (The particular model in the example isn't defensible, but uses a standard data set.) The problem is that most of the time only one of lme4 or coxme will be loaded, so each needs to define the basic ranef function as well as a method for it. But when loaded together

Full_Name: Gabor Grothendieck Version: R version 1.9.1, 2004-08-03 OS: Windows XP Submission from: (NULL) (207.35.143.52) The following gives an error: > `a(b)` <- 1:4 > `c(d)` <- (1:4)^2 > lm(`a(b)` ~ `c(d)`) Error in model.matrix.default(mt, mf, contrasts) : model frame and formula mismatch in model.matrix() To fix it replace this line in model.matrix.default:

I'm testing out a new version of coxme and R CMD check fails with "could not find function ranef" (or random.effects or fixef, or anything from nlme). The NAMESPACE file has the line below importFrom(nlme, ranef, random.effects, fixef, fixed.effects, VarCorr) and nlme is declared in the DESCRIPTION file as an import. I feel that I must be staring at some obvious (but

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed

1. I often like to put bits of the output into the manual pages. (We can have a discussion of the value of this elsewhere -- I think it is sometimes a good thing.) In R I need to surround these with \dontrun{} for the sake of the tester, which is fine. But the printed output contains ## Not run and ## End (not run) comments, which defeats the purpose of the lines by breaking them

Hi, I have a data frame: > class(B27.vec) [1] "data.frame" > head(B27.vec) AGE Gend B27 AgeOn DD uveitis psoriasis IBD CD UC InI BASDAI BASFI Smok UV 1 57 1 1 19 38 2 1 1 1 1 1 5.40 8.08 NA 1 2 35 1 1 33 2 2 1 1 1 1 1 1.69 2.28 NA 1 3 49 2 1 40 9 1 1 1 1 1 1 8.30 9.40 NA

The number of recent questions from umn.edu makes me wonder if there's homework involved.... Simpler for your example is to use get and subset. dat <- structure(..... as found below var.to.test <- names(dat)[4:6] #variables of interest nvar <- length(var.to.test) chisq <- double(nvar) for (i in 1:nvar) { tfit <- survdiff(Surv(time, completion==2) ~

I read through Harrington and Fleming (1982) but it is beyond my statistical comprehension. I have survival data for insects that have a very finite expiration date. I'm trying to test for differences in survival distributions between different groups. I understand that the medical field is most often dealing with censored data and that survival analysis, at least in the package survival,

Hi Does anyone know if there is a function like survdiff which can also handle left-truncated and right-censored data? When I use it on left-truncated and right-censored data I get an error message saying Right censored data only. Many thanks Rajen [[alternative HTML version deleted]]

I've moved nlme from Depends to Imports in my coxme package. However, a few of the examples for lmekin use one of the data sets from nlme. This is on purpose, to show how the results are the same and how they differ. If I use data(nlme::ergoStool) the data is not found, data(nlme:::ergoStool) does no better. If I add importFrom(nlme, "ergoStool") the error message is that

Hi, I would like to compare the hazard functions of two samples using the Cox proportional hazards model. For sample 1 I have individual time-to- event data. For sample 2 I don't have individual data, but grouped data that allows to obtain a hazard function. I am wondering if there is an R function that allows to obtain a hazard ratio of the two hazard funtions (under the

Hello, I was hoping someone could answer a question for me that may either be statistical or script related. I don't come from a statistics background, so I am not positive if I am using the correct nomenclature or even the correct procedure. Is it possible to model "random coefficients" in a mixed effects cox-regression using coxme from the Kinship package? For example, using