similar to: boot package question: sampling on factor, not row

Hi, I am trying to find m breakpoints of a linear regression model. I used the segmented package. It works fine for small number of predicators and breakpoints.(3 r.v. 3 points). However, my model has 14 variables it even would not work even for just one breakpoints!. The error message is always estimated breakpoints are out of range. Since my problem is time related problem. So I

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, i want to define 3 ideal breaks (bin) for each variable one of those variables is attached in the previous email, i don't want to consider quartile method because quartile is not working ideally for that data set because data distribution is non normal. so i want you to suggest another method so that i can define 3 breaks with the ideal interval for Recency, frequency and monetary to

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

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

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-------

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

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

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;

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, 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

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

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

[ 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