similar to: parallel mle/optim and instability

[ Not sure why, but the first time I sent this it never seemed to go through; apologies if you're seeing this twice ... ] I have some fully functional code that I'm guessing can be done better/quicker with some savvy R vector tricks; any help to make this run a bit faster would be greatly appreciated; I'm particularly stuck on how to calculate using "row-wise" vectors

Dear R-help list, I have grouped data, looking like this: cases <- c(23,12,56,81) total <- c(123,234,248,390) x1 <- c(0,0,1,1) x2 <- c(0,1,0,1) Data <- as.data.frame(cbind(cases,total,x1,x2)) Data I would like to run a logistic regression with group weights on these, where cases and (total-cases) are equal to the group weights (w). My final data would look

I verified with my vendor that indeed we are using an ATI ES1000 Video Card, not an Nvidia as previously thought. I am getting the following error -any suggestions? [38] -1 0 0x0000b400 - 0x0000b403 (0x4) IX[B] [39] -1 0 0x0000b480 - 0x0000b487 (0x8) IX[B] [40] -1 0 0x0000b800 - 0x0000b803 (0x4) IX[B] [41] -1 0 0x0000b880 - 0x0000b887

Full_Name: Naoki Takebayashi Version: 0.65.0+R-release.diff (Oct 6, 1999) OS: Linux/Alpha Submission from: (NULL) (129.79.224.171) This will fix the "problem 2 (crash in fft)" in Bug ID #277 On Linux/Alpha, make check failed because R could not handle the following example in base-Ex.R ##___ Examples ___: # The Old Faithful geyser data data(faithful) : : ## Missing values: x <-

What is going on here? In the lines ending in #### the inputs and outputs are identical yet one gives a warning and the other does not. a1 <- `rownames<-`(anscombe[1:3, ], NULL) a2 <- anscombe[1:3, ] ix <- 5:8 # input arguments to #### are identical in both cases identical(stack(a1[ix]), stack(a2[ix])) ## [1] TRUE identical(a1[-ix], a2[-ix]) ## [1] TRUE res1 <-

They differ in whether the row names are "automatic": > .row_names_info(a1) [1] -3 > .row_names_info(a2) [1] 3 Best, -Deepayan On Tue, 14 Nov 2023 at 08:23, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > What is going on here? In the lines ending in #### the inputs and outputs > are identical yet one gives a warning and the other does not. > >

In that case identical should be FALSE but it is TRUE identical(a1, a2) ## [1] TRUE On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > > They differ in whether the row names are "automatic": > > > .row_names_info(a1) > [1] -3 > > .row_names_info(a2) > [1] 3 > > Best, > -Deepayan > > On Tue, 14 Nov

Also why should that difference result in different behavior? On Tue, Nov 14, 2023 at 9:38?AM Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > In that case identical should be FALSE but it is TRUE > > identical(a1, a2) > ## [1] TRUE > > > On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar > <deepayan.sarkar at gmail.com> wrote: > > > >

On Tue, 14 Nov 2023 at 09:41, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > Also why should that difference result in different behavior? That's justifiable, I think; consider: > d1 = data.frame(a = 1:4) > d2 = d3 = data.frame(b = 1:2) > row.names(d3) = c("a", "b") > data.frame(d1, d2) a b 1 1 1 2 2 2 3 3 1 4 4 2 > data.frame(d1,

Hey all Im having a problem... Ok, this is the situation... I have a small product image, on which ive floated a div over it, and defined it as draggable. Now, this works, and i can drag it around no problem. great. but then, what i want to do is, based on the location of that dragable element from the top and the left, to move (using the Effect.MoveBy method) a larger image - so effective

Does sort.int(c(2,NA,4), index.return=TRUE, na.last=NA, method="radix")$ix give the intended result, because I get: > sort.int(c(2,NA,4), index.return=TRUE, na.last=NA, method="radix") $x [1] 2 4 $ix [1] 1 3 With method="shell" and method="quick" in R devel, I get: > sort.int(c(2,NA,4), index.return=TRUE, na.last=NA, method="shell") $x

I have got tdbsam as backend in smb.conf passdb backend = tdbsam When user change password from windows XP file passdb.tdb schould change date because was updated, but I have still the same date IX 18 10:30. [root@serwer private]# ls -al razem 76 drwx------ 2 root root 4096 IX 11 20:25 . drwxr-xr-x 7 root root 4096 XI 2 15:14 .. -rw------- 1 root root 36864 IX 25 07:57 passdb.tdb -rw-------

AdvisoRs: Is the following a bug, feature, hinky error message, or dumb Bert? > mtest <- matrix(1:12,nr=4) > dftest <- data.frame(mtest) > ix <- cbind(1:2,2:3) > mtest[ix] <- NA > mtest [,1] [,2] [,3] [1,] 1 NA 9 [2,] 2 6 NA [3,] 3 7 11 [4,] 4 8 12 ## But ... > dftest[ix] <- NA Error in `[<-.data.frame`(`*tmp*`, ix, value

Earl Chew wrote: > I'm a little reluctant to introduce another compiled program when there are > so many other options that will work well enough out of the box. > > Here are two ideas: > > 1. Use bc(1) to compute the raw samples > 2. Use perl(1) to compute the raw samples > > To generate raw unsigned samples using bc(1) for example: > > samplerate = 1000;

I have data.frame's with IDs and multiple columns. B/c some of IDs showed up more than once, I need sum up colum values to creat a new dataframe with unique ids. I hope there are some cheaper ways of doing it... Because the dataframe is huge, it takes almost an hour to do the task. Thanks so much in advance! Young # ------------------------- examples are here and sum.dup.r is at the

Hi Folks, Given a series of n independent Bernoulli trials with outcomes Yi (i=1...n) and Prob[Yi = 1] = Pi, I want P = Prob[sum(Yi) = r] (r = 0,1,...,n) I can certainly find a way to do it: Let p be the vector c(P1,P2,...,Pn). The cases r=0 and r=n are trivial (and also are exceptions for the following routine). For a given value of r in (1:(n-1)), library(combinat) Set <- (1:n)

Hi all: I've been looking at the boot package to "bootstrap" sample my data in a particular way. I haven't figured out how to set this up using the boot() command and thus have resorted to trying to write my own script (although I'd prefer if I could get boot() to work for this problem!) The dataset is set up in the following way: ix(factor) value 1 5.73 1 6.99 1

Hi All, I'm a little stumped by the following problem. I've got a dataset with the following structure: idxy ix iy country (other variables) 1 1 1 c1 x1 2 1 2 c1 x2 3 1 3 c1 x3 . . . . . 3739 55 67 c7 x3739 3740 55 68 c7 x3740 where ix and

Hi, I may have a use-case for IR liveness analysis, although it's in the context of debuginfo. Using the sample code from this bug report [0], which is a fairly trivial loop: int foo(int count) { int result = 0; for (unsigned long long ix = start; ix != count; ++ix) result += external(ix); return result; } On x86_64 the 32-bit "count" comparison

New function below is a bit faster due to more efficent memory handling. for-loop FTW! directionless_circular_permutations2 <- function( n ) { n1 <- n - 1L v <- seq.int( n1 ) ix <- combinations( n1, 2L ) jx <- permutations( n-3L, n-3L ) jxrows <- nrow( jx ) jxoffsets <- seq.int( jxrows ) result <- matrix( n, nrow = factorial( n1 )/2L, ncol = n ) k