search for: seq_len

...derstanding how negative indexing works but >> >> 1) This is right. >> >> (1:10)[-1] >> #[1] 2 3 4 5 6 7 8 9 10 >> >> 2) Are these right? They are at least surprising to me. >> >> (1:10)[-0] >> #integer(0) >> >> (1:10)[-seq_len(0)] >> #integer(0) >> >> >> It was the last example that made me ask, seq_len(0) whould avoid an >> if/else or something similar. > > I think it's ok, because there is no negative zero integer, so -0 is 0. Ok, this makes sense, I should have thought about t...

Thanks. This is exactly the doubt I had. Rui Barradas ?s 05:26 de 05/08/2018, Kenny Bell escreveu: > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)] > #> numeric(0) > Created on 2018-08-05 by the reprex package (v0.2.0.9000). &g...

El dom., 5 ago. 2018 a las 6:27, Kenny Bell (<kmbell56 at gmail.com>) escribi?: > > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)] > #> numeric(0) > Created on 2018-08-05 by the reprex package (v0.2.0.9000). I...

Pat Burns makes a good point. -Peter -------- Original Message -------- Subject: Re: [R] Using seq_len() vs 1:n Date: Fri, 12 Feb 2010 09:01:20 +0000 From: Patrick Burns <pburns at pburns.seanet.com> To: Peter Ehlers <ehlers at ucalgary.ca> References: <4B746AEF.10900 at ucalgary.ca> If you want your code to be compatible with S+, then 'seq_len' isn't going to work. O...

In a post on R-devel, Prof Ripley add the following comment | > BTW, 1:dim(names)[1] is dangerous: it could be 1:0. That was the | > motivation for seq_len. I use the dim(names)[1] and dim(x)[2] along with length(x) with varying levels of frustration depending on the object which I am trying to get the dimensions. I found the reference to seq_len interesting since it is a function that I have never seen (probably just missed it reading the docs)...

R-people, Duncan Murdoch's response in https://stat.ethz.ch/pipermail/r-help/2010-February/227869.html reminded me of something I had been meaning to ask. A while ago I started using for(i in seq_len(v)) {....} in preference to for(i in 1:n) {....} Duncan's post shows that if n can be zero, there is an advantage to using seq_len. Is there ever a *dis*advantage? Peter Ehlers University of Calgary

...quot;,sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4 ?res2<-t(sapply(split(as.data.frame(A),((seq_len(nrow(A))-1)%/%2)+1),colProds)) ?identical(res1,res2) #[1] TRUE #or ?t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),function(x) apply(x,2,prod))) #or library(plyr) ?as.matrix(ddply(as.data.frame(A),.(as.numeric(gl(nrow(A),2,6))),colProds)[,-1]) #???? V1 V2 V3 #[1,]? 4 10 18 #[2,] 63...

...give the same y-axis scales within each row (see the fourth row). Further, it "separates" the panels. Cheers, Marius ## minimal example: library(lattice) ## build example data set dim <- c(100, 6, 2, 3) # n, groups, methods, attributes dimnames <- list(n=paste("n=", seq_len(100), sep=""), groups=paste("group=", seq_len(6), sep=""), methods=paste("method=", seq_len(2), sep=""), attr=paste("attribute=", seq_len(3), sep="")) set.seed(1) data <- rexp(prod(dim)) arr <...

Hello, Maybe I am not understanding how negative indexing works but 1) This is right. (1:10)[-1] #[1] 2 3 4 5 6 7 8 9 10 2) Are these right? They are at least surprising to me. (1:10)[-0] #integer(0) (1:10)[-seq_len(0)] #integer(0) It was the last example that made me ask, seq_len(0) whould avoid an if/else or something similar. Thanks in advance, Rui Barradas

This should more clearly illustrate the issue: c(1, 2, 3, 4)[-seq_len(4)] #> numeric(0) c(1, 2, 3, 4)[-seq_len(3)] #> [1] 4 c(1, 2, 3, 4)[-seq_len(2)] #> [1] 3 4 c(1, 2, 3, 4)[-seq_len(1)] #> [1] 2 3 4 c(1, 2, 3, 4)[-seq_len(0)] #> numeric(0) Created on 2018-08-05 by the reprex package (v0.2.0.9000). On Sun, Aug 5, 2018 at 3:58 AM Rui Barradas <rui...

...is Circle 8..1.13 of the R Inferno. On 05/08/2018 06:57, Rui Barradas wrote: > Thanks. > This is exactly the doubt I had. > > Rui Barradas > > ?s 05:26 de 05/08/2018, Kenny Bell escreveu: >> This should more clearly illustrate the issue: >> >> c(1, 2, 3, 4)[-seq_len(4)] >> #> numeric(0) >> c(1, 2, 3, 4)[-seq_len(3)] >> #> [1] 4 >> c(1, 2, 3, 4)[-seq_len(2)] >> #> [1] 3 4 >> c(1, 2, 3, 4)[-seq_len(1)] >> #> [1] 2 3 4 >> c(1, 2, 3, 4)[-seq_len(0)] >> #> numeric(0) >> Created on 2018-08-05...

...matrixStats. /Henrik On Sun, Aug 5, 2018 at 3:42 AM I?aki ?car <i.ucar86 at gmail.com> wrote: > > El dom., 5 ago. 2018 a las 6:27, Kenny Bell (<kmbell56 at gmail.com>) escribi?: > > > > This should more clearly illustrate the issue: > > > > c(1, 2, 3, 4)[-seq_len(4)] > > #> numeric(0) > > c(1, 2, 3, 4)[-seq_len(3)] > > #> [1] 4 > > c(1, 2, 3, 4)[-seq_len(2)] > > #> [1] 3 4 > > c(1, 2, 3, 4)[-seq_len(1)] > > #> [1] 2 3 4 > > c(1, 2, 3, 4)[-seq_len(0)] > > #> numeric(0) > > Created on...

I have a dataset named DM with p1, p2, ...., p9 (9 columns, numerical values) I would like to calculate to multify each pair of columns (p1p2, p1p3,... p1p9, p2p3, p2p4.... p8p9) and assign them in p1p2, p1p3,... p1p9, p2p3, p2p4.... p8p9 In SAS, l=0; p_int_sum=0; do i=1 to 8; do j=(i+1) to 9; l=l+1; p{i}p{j}=p{i}*p{j}; end; end; I would like to know how to assign them in R I tried for

...s) && !is.na(col.names) && + col.names==TRUE if(is.matrix(x)) { ## fix up dimnames as as.data.frame would p <- ncol(x) d <- dimnames(x) if(is.null(d)) d <- list(NULL, NULL) - if(is.null(d[[1]])) d[[1]] <- seq_len(nrow(x)) - if(is.null(d[[2]]) && p > 0) d[[2]] <- paste("V", 1:p, sep="") + if (is.null(d[[1]]) && makeRownames) d[[1]] <- seq_len(nrow(x)) + if(is.null(d[[2]]) && p > 0 && makeColnames) + d[[2]] <-...

in a long program, I ran into ?*** caught segfault *** address 0xdc3f9b48, cause 'memory not mapped' Traceback: ?1: rep.int(seq_len(nx), rep.int(rep.fac, nx)) ?2: rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) ?3: expand.grid(seq_len(nx), seq_len(ny)) ?4: merge.data.frame(d, ss) ?5: merge(d, ss) ?6: valid.range(opt) ?7: eval.with.vis(expr, envir, enclos) ?8: eval.with.vis(ei, envir) ?9: source("fut-into-opts.R&q...

...been around in R for a long time. It is used in packages eRm, extRemes, hydrosanity, klaR, seas. Who knows what people have in private code, so I don't see any compelling case to change it. If people want a different version, it would only take a minute to write (see below). We could make seq_len take a vector argument, but as you point out in a followup that makes it slower in the common case. It also changes its meaning if a length > 1 vector is supplied, and would speed matter in the long-vector case? What does sequence0 <- function (nvec) { s <- integer(0) for (...

...I not expect to be able to print 1.999999 seconds? This seems to be caused by the following code fragment in format.POSIXlt: np <- getOption("digits.secs") if (is.null(np)) np <- 0L else np <- min(6L, np) if (np >= 1L) for (i in seq_len(np) - 1L) if (all(abs(secs - round(secs, i)) < 1e-06)) { np <- i break } Specifically for (i in seq_len(np) - 1L) if (all(abs(secs - round(secs, i)) < 1e-06)) Which in the case of 1.999999 seconds will...

In the Value section of the 'seq' help page it says 'seq_along' and 'seq_length' always return an integer vector. I believe it should be 'seq_along' and 'seq_len' always return an integer vector. as there are no seq_length function? Best, Niels -- Niels Richard Hansen Web: www.math.ku.dk/~richard Associate Professor...

Hi all, A couple of ideas for improving seq_along: * It would be really useful to have a second argument dim: seq_along(mtcars, 1) seq_along(mtcars, 2) # equivalent to seq_len(dim(mtcars)[1]) seq_len(dim(mtcars)[2]) I often find myself wanting to iterate over the rows or column of a data frame, and there isn't a particularly nice idiom if you want to avoid problems with zeros - you have to use seq_len(nrow(df)) etc * To me, it would seem be very natural to ha...

foo <- rnorm(30*34*12) dim(foo) <- c(30, 34, 12) I want to make a data.frame out of this three-dimensional array. Each dimension will be a variabel (column) in the data.frame. I know how this can be done in a very slow way using for loops, like this: x <- rep(seq(from = 1, to = 30), 34) y <- as.vector(sapply(1:34, function(x) {rep(x, 30)})) month <- as.vector(sapply(1:12,