similar to: names generated in list indexing

Dear R-help, When the result of a matrix subscription degenerates to a scalar the names implied by the dimnames are discarded. > x <- matrix(0, 1, 1, dimnames=list('a', 'x')) ## below I expected result to have names='x', it's not > x[1,] [1] 0 ## below I expected result to have names='a', it's not > x[,1] [1] 0 This is probably a side effect

Dear R-developers, I am trying to figure out a way to call browser() when an error occur, and naturally I want the browser() to be called in the environment of the error. I tried something simple in vain: > f <- function() { x <- 1; stop('ok') } > tryCatch(f(), error=browser()) Called from: tryCatch(f(), error = browser()) ## if browser() was called in the local environment

Dear R users, Please disregard my previous post "converting result of substitute to 'ordidnary' expression". The problem I have has nothing to do with substitute. Consider: > dat <- data.frame(x=1:10, y=1:10) > subsetexp <- expression(5<x) > ## this does work > subset(dat, eval(subsetexp)) x y 6 6 6 7 7 7 8 8 8 9 9 9 10 10 10 > ##

Dear R-devel, The median() function assigns a name, "NA", to its return value if the return value is NA and the input vector has names, otherwise the names attribute is NULL. This looks strange and inconsistent with the behavior of mean(). This inconsistency becomes a problem when median() is used inside user code that relies on consistent naming convention. Thanks, Vadim > foo

Dear R-devel, Is there a reason that lapply(NULL, ...) returns the empty list, rather than NULL? It seems intuitive to expect the latter, and rather counterintuitive that lapply(list(), ... ) returns the same value as lapply(NULL, ...). > lapply(list(), function(x) 1) list() > lapply(NULL, function(x) 1) list() > version _ platform i386-pc-mingw32 arch

Dear R-users, Suppose I have an expression: expr = expression(a>0) and now I want to modify it to expression(a>0 & b>0). The following doesn't work: expr = expression(expr & b>0) What would be a good way of doing this? Thanks, Vadim ________________________________ Note: This email is for the confidential use of the named addressee(s) only and may contain

Dear R users, As substitute() help page points out: Substituting and quoting often causes confusion when the argument is 'expression(...)'. The result is a call to the 'expression' constructor function and needs to be evaluated with 'eval' to give the actual expression object. And indeed I am confused. Consider: > dat <- data.frame(x=1:10,

Dear R users, As substitute() help page points out: Substituting and quoting often causes confusion when the argument is 'expression(...)'. The result is a call to the 'expression' constructor function and needs to be evaluated with 'eval' to give the actual expression object. And indeed I am confused. Consider: > dat <- data.frame(x=1:10,

Dear R-users, I am somewhat puzzled by how R treats data frames with nested data frames. Below are a couple of examples, maybe someone could help explain what the guiding logic here is. ## construct plain data frame > z <- data.frame(x=1) ## add a data frame member > z$y <- data.frame(a=1,b=2) ## puzzle 1: z is apparently different from a straightforward construction of the

Dear R-devel, It seems that lsfit incorrectly reports coefficients when the input matrix 'x' is rank-deficient, see the example below: ## here values of 'b' and 'c' are incorrectly swapped > x <- cbind(a=rnorm(100), b=0, c=rnorm(100)); y <- rnorm(100); lsfit(x, y)$coef Intercept a b c -0.0227787 0.1042860 -0.1729261 0.0000000 Warning

Hi, I ran into a problem of converting a text representation of an expression into parsed expression to be further evaluated inside lm (). > n <- 100 > data <- data.frame(x= rnorm (n), y= rnorm (n)) > data. lm <- lm (y ~ x, data=data) > > ## this works > update(data. lm , subset=x<0) Call: lm (formula = y ~ x, data = data, subset = x < 0)

Hi, I seem to have a problem when passing named parameters to R via Rscript (R2.5.1, bash shell). As soon as I name elements of a list Rscript generates an error. I will appreciate if someone could point to me a correct way of doing this. Thanks, Vadim ## This works bash-3.2$ Rscript.exe -e 'list(1)' [[1]] [1] 1 # and these do not work bash-3.2$ Rscript.exe -e

Hi, I have few high-level questions about the Snow and Rmpi packages . I understand that Snow uses Rmpi as one of possible transport layers, yet my questions about user experience, not technical details: 1. Does Snow install and work well in Windows? 2. Interruptibility. I understand that currently it is impossible to interrupt a running top-level command in Snow ( Ctl-c or the likes), the

Hi, When using evalq to evaluate expressions within a say data.frame context I often wish there was a 'subset' argument, much like in lm() or any ather advanced regression model. I would be grateful for a tip how to do this. Here is an illustration of what I want: n <- 100 data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z)) # this works evalq({ i <- 0<x;

Dear R-users, It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right? I am not sure cat() correctly prints strings which are generated by triple-shQuote(): > shQuote(shQuote("a")) [1] "\"\\\"a\\\"\"" > cat(shQuote(shQuote(shQuote("a"))), '\n')

Dear R-users, It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right? I am not sure cat() correctly prints strings which are generated by triple-shQuote(): > shQuote(shQuote("a")) [1] "\"\\\"a\\\"\"" > cat(shQuote(shQuote(shQuote("a"))), '\n')

Dear R-users, How can I detect a NULL in a recursive list? For a regular list I could use lapply: > lapply(list(x=NULL), is.null) $x [1] TRUE However that doesn't work for structures like list(list(x=NULL)). I tried rapply but it treats NULL as a list and discards them: > rapply(list(a=1, b=list(x=NULL)), is.null) a FALSE Any suggestion? Thank you for your help, Vadim Note:

Dear R Users, Given two vectors, say a = seq(2) and b = seq(3), I want to make an 2*3 array, where (i,j) element is list(a=a[i], b=b[j]). I tried the outer() function but it generates an error message that I don't understand, see below. What do I do wrong? The expan.grid function is not good enough since I need a solution that works when a and b are not atomic, say a=list(list(x=1,

Dear R-devel, The following line crashes R > approx(1, 1, 0, method='const', rule=2, f=0, yleft=NULL, ties='ordered')$y Process R:2 exited abnormally with code 5 at Tue Jul 21 14:18:09 2009 > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 9.1 year

Dear R-users, I am trying to write a wrapper function around save() that will report the file which is being saved to. So I thought that the followintg would do the trick, but it doesn't. I understand that 'y' is somehow not visible inside save.verbose, but don't know how to fix this. save.verbose <- function(..., file) { cat("save.verbose:", file, "\n")