similar to: segmentation fault with 1D array (PR#500)

I get a segmentation fault as follows: > tmp <- array(list(), c(3, 4)) > tmp[[1, 2]] NULL # as expected > tmp[[1, 2]]$fred <- 1:10 > tmp # as expected [,1] [,2] [,3] [,4] [1,] "NULL" "List,1" "NULL" "NULL" [2,] "NULL" "NULL"

Hi Everyone, The following rather extreme claim on memory causes a segmentation fault on my installation: fred <- matrix(1:1200, 20, 60) littleOP <- function(x, y) { z <- as.vector(x) %*% t(as.vector(y)) dim(z) <- c(dim(x), dim(y)) z } littleOP(fred, fred) # this is OK littleOP(littleOP(fred, fred), fred) # whoops! Segmentation fault What's a bit strange

I was surprised to find that the variance of a scalar, using var(), is NA. Surely this should be zero? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc os = solaris2.7 system = sparc, solaris2.7 status = Patched major = 0 minor = 99.0 year = 2000 month = February day = 9 language = R Search Path: .GlobalEnv,

Dear R users, Suppose i have an A square matrix rxr. I want to obtain a block matrix B (pxr,pxr) where the p diagonal blocks are A and the others values are 0. I would like to do something like : diag(A,ncol=pr, nrow=pr) How can i do it ?? Thanks in advance, Olivier. -- -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- Olivier MARTIN PhD student phone: (33) 04

The following commands z <- matrix(seq(0, 1, len=1000)) image(z, 1, z, zlim=c(0, 1), col=gray(1:10/10)) appear to violate the image functionality as described on the help page. As I understand it, by specifying a zlim of c(0, 1) and a colour vector of length 10, the colours should correspond to the intervals (0, 0.1), (0.1, 0.2), ..., (0.9, 1.0), this being 10 equally-spaced intervals on (0,

I find this a bit puzzling ... > solve(matrix(c(5, 2, 3, 1), 2, 2), c(Inf, 3)) [1] NaN Inf > solve(matrix(c(5, 2, 3, 1), 2, 2)) %*% c(Inf, 3) [,1] [1,] -Inf [2,] Inf I would expect the answer to be c(-Inf, Inf), so why has the -Inf been replaced by NaN in solve? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc

Might it be possible to tighten the definition of "is.qr". I noticed that after I mistakenly typed example(lm) # make lm object named lm.D9 qr.Q(lm.D9) which exhausted the heap memory and produced two warning messages. As an object of class "lm" has a "qr" component, "is.qr" failed to detect that "lm.D9" was not a "qr" object. The

In the optim() documentation, the control parameter "maxit" says "There is no other stopping criteria." That should be "are", or "criterion" ... While I'm at it -- poking at the code a little bit, it looks as if the initial simplex is set from the initial point by displacing each parameter value by max(0.1,0.1*pmax(fabs(Bvec))), where Bvec is the

Hi all, Running R 0.90.1 on a RH 6.1 system. Installation of the integrate_2.1-2 package went smoothly. My code contains a loop in which integrate() is called several times in each pass. I get a segmentation fault after what seems to be a random number of calls to integrate(). Debug output shows: Program received signal SIGSEGV, Segmentation fault. promiseArgs (el=0x40276414,

Another one from the student computer practicals: it is possible to create a dataframe that cannot be printed. Thus > fred <- data.frame(happy=c(TRUE, FALSE, TRUE), sad=7:9) > tmp <- fred[c(FALSE, FALSE, TRUE, TRUE)] # no error message! > tmp Error in as.matrix.data.frame(x) : dim<- length of dims do not match the length of object > print.default(tmp) $"NA" NULL

Dear all, I think I must be going crazy.... If I do > mean(rpois(1000000, 14)) I get: [1] 13.50420 and again: > mean(rpois(1000000, 14)) [1] 13.49896 > mean(rpois(1000000, 14)) [1] 13.50161 > mean(rpois(1000000, 15)) [1] 14.49250 > mean(rpois(1000000, 15)) [1] 14.49897 > mean(rpois(1000000, 14.5)) [1] 13.99689 > mean(rpois(1000000, 14.5)) [1] 13.9963 This was on a platform

I'm just back from a computer practical where we ran into > "f" <- function(x) 2(x^2) # for 2 * x^2 This definition was taken as legitimate, although using the function generated the error message > f(3) Error in f(3) : attempt to apply non-function Am I right in thinking that in prevous versions of R this definition would have been declared as a syntax error? Surely a

The following gives an immediate segmentation fault, v 0.90.1 "fred" <- function(x) { print(x); Recall(x-1)} fred(0) # segmentation fault at recursion 831 Obviously this recursion is non-terminating, but shouldn't it be caught and reported at some limit? In 0.64.1 we had Error: protect(): stack overflow at recursion 831. Cheers, Jonathan. Jonathan Rougier

This situation arose in a computer practical this morning. A student overwrote "c" with a function. This messes up lots of things, but it is difficult to remove the function "c" from the global environment because it is used as a function in both "rm" and "remove". [The solution was to reassign "c" as a number, and then remove it.] Would it be

The following seems to be odd behaviour: fred <- 1:10 mode(fred) # "numeric" exists("fred") # TRUE exists("fred", mode="numeric") # FALSE Unless I have misunderstood, the final evaluation should be TRUE, as an object called "fred" exists and its mode is "numeric". In the same category, I have just noticed that get("fred")

I would like to suggest a slight modification to diag(). In the case where x is a matrix with both row names and column names the same, it would be reasonable if the resulting vector also had those names. I often use diag() on variance matrices, where this modification is helpful. The modification requires replacing if (is.matrix(x) && nargs() == 1) return(c(x)[1 +

If so, I''ve had trouble finding it. Can anyone help?

I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, "workers", sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks

Dear all, I have the following output generated by linear regression. Since there is only one regression intercept and one slope for one set of data, what is the meaning of std. error for intercept and that of slope? Thanks in advance. Sincerely, Minghua > data(thuesen) > attach(thuesen) > lm(short.velocity~blood.glucose) Call: lm(formula = short.velocity ~ blood.glucose)

Dear List, I'd like to run a discriminate analysis on a data set, but have no idea how to go about this in R. I have attempted to locate info in the manuals, but may not be consulting the right sections or documents. Can anyone point me to appropriate documentation if such exists. Many thanks, David S. David White sdavidwhite at bigfoot.com Columbus, Ohio