similar to: seq_along and rep_along

Using the IRanges package from Bioconductor and somewhat recent R-2.9.1. ov = IRanges(1:3, 4:6) length(ov) # 3 seq(along = ov) # 1 2 3 as wanted seq_along(ov) # 1! I had expected that the last line would yield 1:3. My guess is that somehow seq_along don't utilize that ov is an S4 class with a length method. The last line of the *Details* section of ?seq has a typeo. Currently it is

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

I'm trying to understand the behavior of seq_along in the following example: x <- 1:5; sum(x) y <- 6:10; sum(y) data <- c(x,y) S <- sum( data[seq_along(x)] ) S T <- sum( data[seq_along(y)] ) T Why is T != sum(y) ?

I am moving this from r-help to r-devel. Based on offline communications with Jim, suppose dat is defined as follows: set.seed(123) dat <- data.frame(ID= c(rep(1,2),rep(2,3), rep(3,3), rep(4,4), rep(5,5)), var1 =rnorm(17, 35,2), var2=runif(17,0,1)) # Then this ave call works as expected: ave(dat$ID, dat$ID, FUN = function(x) seq_along(x)) # but this apparently identical calculation

This can be vectorized. Try ix <- seq_along(vec2) S_diff2 <- sapply(seq_len(N1-(N2-1)*ratio_sampling), \(j) sum((vec1[(ix-1)*ratio_sampling+j] - vec2[ix])**2)) On Sun, Jun 16, 2024 at 11:27?AM Laurent Rhelp <laurentRHelp at free.fr> wrote: > > Dear RHelp-list, > > I try to use the package comprehenr to replace a for loop by a list > comprehension. > > I

Hello, Another way would be n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D5 <- data.frame(distance=integer(n),difference=integer(n)) D5[] <- do.call(rbind, lapply(seq_len(nrow(D1)), function(i) t(sapply(seq_len(nrow(D2)), function(j){ c(distance=sqrt(sum((D1[i,1:2]-D2[j,1:2])^2)),difference=(D1[i,3]-D2[j,3])^2) } )))) identical(D3, D5) In my first answer I forgot to say that

Dear all, I have a list of n matrices which all have the same dimension (r x s). What would be a fast/elegant way to calculate the element wise average? So result[1, 1] <- mean(c(raw[[1]][1, 1] , raw[[2]][1, 1], raw[[...]][1, 1], raw[[n]][1, 1])) Here is my attempt. #create a dummy dataset n <- 3 r <- 5 s <- 6 raw <- lapply(seq_len(n), function(i){ matrix(rnorm(r * s), ncol =

Hello, colnames seems to be not optimized well for data.frame. It escapes processing for data.frame in if (is.data.frame(x) && do.NULL) return(names(x)) but only when do.NULL true. This makes huge difference when do.NULL false. Minimal edit to `colnames`: if (is.data.frame(x)) { nm <- names(x) if (do.NULL || !is.null(nm)) return(nm) else

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

A private reply by Martin made me realize that I was wrong about stopifnot(exprs=TRUE) . It actually works fine. I apologize. What I tried and was failed was stopifnot(exprs=T) . Error in exprs[[1]] : object of type 'symbol' is not subsettable The shortcut assert <- function(exprs) stopifnot(exprs = exprs) mentioned in "Warning" section of the documentation similarly fails

Dear Jesus, The difference is marginal when each code chunk does the same things. Your for loop does not yields the same output as the lapply. Here is the cleaned version of your code. n<-10000 set.seed(123) x<-rnorm(n) y<-x+rnorm(n) rand.data<-data.frame(x,y) k<-100 samples <- split(sample(n), rep(seq_len(k),length=n)) library(microbenchmark) microbenchmark( "for"

Hello, Sorry, but I think my first answer is wrong. You probably want something along the lines of sp <- split(priceStore_Grps, priceStore_Grps$StorePC) res <- lapply(seq_along(sp), function(i){ sp[[i]]$StoreID <- paste("Store", i, sep = "_") sp[[i]] }) res <- do.call(rbind, res) row.names(res) <- NULL Hope this helps, Rui Barradas On 5/26/2018

Hi, In the example you showed: m1<- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or ?m1[!upper.tri(m1)] <-? rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1<- c(3,4,5) ?m2<- matrix(0,length(vec1),max(vec1)) ?indx <- cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]<- 1 ?m2 #???? [,1] [,2] [,3] [,4] [,5]

Hallo all, I`ve read a lot of things in this forum about an Excel export via R. It is no problem to export my data frames via write.table or write.xls (xls or csv), but some things are not very convenient for me: I always have to adjust the column with to see all the numbers or the text and there is no frame between the cells. And I missing the possibility to make some headers bold or coloured.

Hi, str() on raster objects fails for certain dimensions. For example: > str(as.raster(0, nrow=1, ncol=100)) 'raster' chr [1, 1:100] "#000000" "#000000" "#000000" "#000000" ... > str(as.raster(0, nrow=1, ncol=101)) Error in `[.raster`(object, seq_len(max.len)) : subscript out of bounds This seems to do with how str() and

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Here is a case where the current scheme fails: > with(datasets::mtcars, xyplot(mpg~wt|gear)$call) xyplot(substitute(expr), data, enclos = parent.frame()) Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, May 11, 2017 at 1:09 AM, Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > On Wed, May 10, 2017 at 2:36 AM, William Dunlap via R-devel > <r-devel at

Some formula methods for S3 generic functions use the idiom returnValue$call <- sys.call(sys.parent()) to show how to recreate the returned object or to use as a label on a plot. It is often followed by returnValue$call[[1]] <- quote(myName) E.g., I see it in packages "latticeExtra" and "leaps", and I suspect it used in "lattice" as well. This idiom

I am trying to plot a graph but the points on the graph should be different symbols and colors. It should represent what is in the legend. I tried using the points command but this does not work. Is there another command in R that would allow me to use different symbols and colors for the points? Thank you kindly. data(mtcars) plot(mtcars$wt,mtcars$mpg,xlab= "Weight(lbs/1000)",

I am looking for ways to add points to three different plots in parallel. I generate three scatter plots and name them as s3d1, s3d2 and s3d3 s3d1<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20) s3d2<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20)