similar to: Problems with eval() in connection with match.call()

After many hours of debugging code, I came to the conclusion that I have a fundamental misunderstanding regarding eval, and hope that someone here can explain to me, why the following code acts as it does: foo <- function(expr) { eval(substitute(expr), envir=list(a=5), enclos=parent.frame()) } bar <- function(er) { foo(er) } > foo(a) [1] 5 > bar(a) Error in eval(expr,

Dear list, Given a linear mixed model (from lme4) I want to 1) first change the input dataset and then 2) change the model formula. I want this to happen in a function call; Please see below. Options 1) and 2) below work whereas 3) fails with the message > foo() Error in is.data.frame(data) : object 'beets2' not found Question: What is it one must to in case 3) to have R look

Dear all, I want to update an lm (or glm) object by changing the response variable and I want to do so inside a function. Doing the update outside of a function is straight forward: x <- 1:5 y <- c(1,2,3,3,6) mm <- lm(y~x) y2 <- c(1,3,3,4,6) mm2<- update(mm, y2 ~ .) But I want to make the update inside a function (where the new response variable is declared inside the

I am trying to build a package binary, and get the message below. Can anyone point me to a solution to that problem. Thanks in advance S?ren .... installing data files installing man source files installing indices Error in load(zfile, envir = envir) : input has been corrupted, with LF replaced by CR Execution halted make[2]: *** [indices] Error 1 make[1]: *** [all] Error 2

Dear all, I was trying to see how the function 'confint' is defined. Doing > confint function (object, parm, level = 0.95, ...) UseMethod("confint") <environment: namespace:stats> does not really enlighten me. How can I get to see the implementation (I guess it should be possible according to the general philosophy of the R project)? Thanks in advance S??ren

my understanding of `eval' behaviour is obviously incomplete. my question: usually `eval(expr)' and `eval(expr, envir=parent.frame())' should be identical. why does the last `eval' (yielding `r3') in this code _not_ evaluate in the local environment of function `f' but rather in the global environment (if `f' is called from there)?

I don't have time now to investigate myself, and I'm not feeling like deciding myself if the following is a bug: myplot <- function(dat, cex = 1.2, ...) { if(!is.data.frame(dat <- as.data.frame(dat))) stop("`dat' must be a data.frame") if(any(is.na(match(c("x","y"), names(dat))))) stop("`dat' must have a `x' and a

I'm trying to evaluate a pre-built expression using eval(), e.g. dataset <- data.frame(y = runif(30, 50,100), x = gl(5, 6)) # one like this mf <- expression(model.frame(y~x)) eval(mf, dataset, parent.frame()) # rather than this eval(expression(model.frame(y~x)), dataset, parent.frame()) In the example above there is no problem, the problem comes when I try to do a similar thing

Dear developeRs, I maintain a package 'pls', which has a main fit function mvr(), and functions plsr() and pcr() which are meant to take the same arguments as mvr() and do exactly the same, but have different default values for the 'method' argument. The three functions are all exported from the name space. In the 'pre namespace' era, I took inspiration from lm() and

Dear all, One of my students (whom I am trying to convince to use R) wants to get a fairly large SAS dataset into R (about 150mB). An obvious and simple thing she tried was to write the dataset as a .csv-file and then read that into R, but that takes forever (or something close to that..). The dataset is so large, that exporting it as an Excel file from SAS is not feasible (more than 65000 lines).

Thanks. I wonder if there is a general way of extracting var(u) and var(e), which would be needed to simulate u and e. Clearly, one can get the estimated parameters, but is there a clever way of 'setting up' the matrices?? Best S??ren On 6/19/05, S??ren H??jsgaard <Soren.Hojsgaard at agrsci.dk> wrote: > I would like to draw a sample from a linear mixed model y=Xb+Zu+e which has

>From time to time people request symbolic computations beyond what D() and deriv() etc can provide. A brief look at the internet shows that there are many more or less developed computer algebra packages freely available. Therefore, I wondered if it would be an idea to try to 'integrate' one of these packages in R, which I guess can be done in more or less elegant ways... I do not know

# The result I am after is the result after a substitution in an expression, such as substitute(expression(a+b+c), list(a=1)) expression(1 + b + c) # However, the way I want to do it is for a an expression "stored as a variable" as (expr <- expression(a+b+c)) expression(a + b + c) # a) The following does not work (expr2 <- substitute(expr, list(a=1))) expr # b) - whereas this

Hi R-listers, I am trying to make a trellis boxplot with the HSuccess (y-axis) in each Rayos (beach sections) (x-axis), for each Aeventexhumed (A, B, C) - nesting event. I am not able to do so and keep receiving: Error in eval(expr, envir, enclos) : object 'Rayos' not found Please advise, Jean require(plyr) resp <- read.csv("ABC Arribada R File Dec 12 Jean

I am absolutely new to R and I am aware of only a few basic command lines. I was running a robust regression in R, using the following command line library (MASS) rfdmodel1 <- rlm (TotalEmployment_2004 ~ MISSISSIPPI + LOUISIANA + TotalEmployment_2000 + PCWhitePop_2004 + UnemploymentRate_2004 + PCUrbanPop2000 + PCPeopleWithACollegeDegree_2000 + PCPopulation.of.or.over.65.years.of.age_2004)

> datafilename="E:/my documents/r/sex/bysex1.csv" > data.sex=read.table(datafilename,header=T) > data.sex y.sex.age.region.c.n 1 1980,F,A,N,-18.15,13.61 2 1980,F,A,N,-18.61,13.04 3 1980,F,A,N,-18.81,12.32 4 1990,F,A,N,-21.12,11.7 5 1990,F,A,N,-20.77,11.58 6 1990,F,A,N,-21.6,13.34 7 1990,F,A,N,-21.78,12.6 > model.anova<-aov(c~age*sex,data=data.sex)

Dear friends, I am testing glm as at page 514/515 of THE R BOOK by M.Crawley, that is on proportion data. I use glm(y~x1+,family=binomial) y is a proportion in (0,1), and x is a real number. I get the error: In eval(expr, envir, enclos) : non-integer #successes in a binomial glm! But that is exactly what was suggested in the book, where there is no mention of a similar warning. Where am I

Dear all, One of my students have installed R1.9.0 on windows, and gets the fatal error 'invalid HOMEDRIVE' Can anyone help her/me out on that one? Thanks in advance Søren Højsgaard [[alternative HTML version deleted]]

Dear all; I would like to draw a graph with vertices and edges, and I guess tcl/tk would be appropriate. I don't know tcl/tk but have googled for a 10-page (or so) introduction to 'getting started with tcl/tk in R' but without any luck. - Does anyone know of the existence of such a document or any other web-based material on the subject? - Does anyone have an (informative) piece of

Dear List, Having used R on Windows for years, I have recently installed Ubuntu as a virtual machine. I admit that I am quite new to the linux world. Installing R (current version = 2.15.1) went smoothly, but I would like to have both the current version of R and R-devel installed. I've searched the net for info on doing this but so far without luck. Can anyone help me getting this up and