search for: seq_along

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) ?

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 'seq.int', 'seq_along' and 'seq...

...el. 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 gives an error: ave(dat$ID, dat$ID, FUN = seq_along) The only difference between the two calls is that the first one uses seq_along and the second uses function(x) seq_along(x) in its place. Does anyone know why the second gives an error? Is this...

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 partic...

...are not called when there is no explicit class attribute. E.g., % R-3.6.0 --vanilla --quiet > print.function <- function(x, ...) { cat("Function with argument list "); cat(sep="\n ", head(deparse(args(x)), -1)); invisible(x) } > f <- function(x, ...) { sum( x * seq_along(x) ) } > f function(x, ...) { sum( x * seq_along(x) ) } > print(f) Function with argument list function (x, ...) Previous to R-3.6.0 autoprinting did call such methods % R-3.5.3 --vanilla --quiet > print.function <- function(x, ...) { cat("Function with argument list "); cat(...

...exed on class Date in the latest versions, versus 2. It seems to be related to changes to/from daylight savings time, happens those weekends. Is it not intended that class Date be used like this, or is this new behaviour incorrect? Giles Example: > a<-as.Date(15423:15426) > x<-xts(seq_along(a),a) > print(x) [,1] 2012-03-24 1 2012-03-25 2 2012-03-25 3 2012-03-26 4 > print(index(x)) [1] "2012-03-24" "2012-03-25" "2012-03-25" "2012-03-26" > print(as.numeric(index(x))) [1] 15423 15424 15424 15425 #for reference, zoo...

Dear R Helpers, I am missing something very elementary here, and I don't seem to get it from the help pages of the ave, seq and seq_along functions, so I wonder if you could offer a quick help. To use an example from an earlier post on this list, I have a dataframe of this kind: dat = data.frame(name = rep(c("Mary", "Sam", "John"), c(3,2,4))) dat$freq = ave(seq_along(dat$name), dat$name, FUN = seq_alo...

...''data.frame'' and then transform it as output? (fun1 and fun1 is just for illustration). ? Thanks a lot, OV ? code: input <- data.frame(x1 = rnorm(20), x2 = rnorm(20), x3 = rnorm(20)) fun1 <- function(x) { ??? ID <- NULL; minimum <- NULL; maximum <- NULL ??? for(i in seq_along(names(x)))?? { ??????? ID[i]?????? <- names(x)[i] ????????? minimum[i]? <- min(x[, names(x)[i]]) ??????????? maximum[i]? <- max(x[, names(x)[i]]) ??????????????????????????????????? } ??? output <- structure(list(ID, minimum, maximum), row.names = seq_along(names(x)), .Names = c("I...

Dear R developers, Reviewing my code, I have realized that about 80% of the time in the lapply I need to access the names of the objects inside the loop. In such cases I iterate over indexes or names: lapply(names(x), ... [i]), lapply(seq_along(x), ... x[[i]] ... names(x)[i] ), or for(i in seq_along(x)) ... which is rather inconvenient. How about an argument to lapply which would specify the [ or [[ subseting to use in the splitting of the vector? Or may be a different set of functions lapply1, sapply1? I believe this pattern is rathe...

...ear useRs,while trying to plot the yearly curves of 1000 stations and overlapping each set of curves with mean curve and then saving it automatically in a pdf file, i tried the following commands >Path = "C:\\R\\003.pdf">pdf(file=Path) for (i in seq(1:1000) >a<-lapply(seq_along(tcp), function(x) tcp[[x]][,-1]) >b<-lapply(seq_along(a), function(a) matrix(rowMeans(tcp[[a]]),ncol=1)) >lapply(seq_along(tcp), function(i) (matplot(tcp[[i]][,-1], type="l",col="grey") ) >lines(b, lwd=2, type="l")) } >dev.off()although i was suc...

...., > > > % R-3.6.0 --vanilla --quiet > >> print.function <- function(x, ...) { cat("Function with argument > list "); > > cat(sep="\n ", head(deparse(args(x)), -1)); invisible(x) } > >> f <- function(x, ...) { sum( x * seq_along(x) ) } > >> f > > function(x, ...) { sum( x * seq_along(x) ) } > >> print(f) > > Function with argument list function (x, ...) > > > Previous to R-3.6.0 autoprinting did call such methods > > % R-3.5.3 --vanilla --quiet > &...

.... the name of each plot is contained in "names" file. when i run this loop, i get an error. "Error in plot.new() : Unable to open file 'C:/R/SAVEHERE/myplot_Tak.jpg' for writing". could you please correct the mistake in the loop? >names<-(names(sp)) >for(a in seq_along(names)){ >mypath <- file.path("C:","R","SAVEHERE",paste("myplot_", names[a], ".jpg",sep = "")) >jpeg(file=mypath) >for (i in seq(1)){ >b<-lapply(res,function(x) {if(is.data.frame(x[,-1])) rowMeans(x[,-1]) else mean(x[,-1]...

Hi, by replacing 'll' with 'wh' in the source code for base::which() one gets ~20% speed up for *named logical vectors*. CURRENT CODE: which <- function(x, arr.ind = FALSE) { if(!is.logical(x)) stop("argument to 'which' is not logical") wh <- seq_along(x)[ll <- x & !is.na(x)] m <- length(wh) dl <- dim(x) if (is.null(dl) || !arr.ind) { names(wh) <- names(x)[ll] } ... wh; } SUGGESTED CODE: (Remove 'll' and use 'wh') which2 <- function(x, arr.ind = FALSE) { if(!is.logical(x))...

Hi, My english isn't brilliant and my problem is very difficult to describe but I try ;) My first question is: May I write loop "for" like this or similar - for (i in sth : sth[length(sth)], k in sth_else : length(sth_else) ) - I'd like to have two independent conditions in the same loop "for". My secound question depend on program below. I'd like to write every

...t;- pmatch(i, rows, duplicates.ok = TRUE) } xj <- .subset2(.subset(xx, j), 1L) return(if (length(dim(xj)) != 2L) xj[i] else xj[i, , drop = FALSE]) } if (any(is.na(cols))) stop("undefined columns selected") nxx <- structure(seq_along(xx), names = names(xx)) sxx <- match(nxx[j], seq_along(xx)) } else sxx <- seq_along(x) rows <- NULL if (is.character(i)) { rows <- attr(xx, "row.names") i <- pmatch(i, rows, duplicates.ok = TRUE) } for (j in seq_along(x)) { xj <- xx[[sxx[j]]]...

...wing is a piece of R code that implements the procedure. > ### Systematic resampling > set.seed(2) > x <- LETTERS[1 : 6] # labels > w <- rexp(6) > w <- w / sum(w) # probabilities > W <- c(0, cumsum(w)) # cdf > u <- (1 : 6 + runif(1)) / 6 > indNew <- sapply(seq_along(u), + function(i) (sum(W[i] <= u & u < W[i + 1]))) > (y <- rep(x, indNew)) [1] "A" "B" "D" "D" "F" > ## graphically... > plot(stepfun(seq_along(u), W), xaxt = "n") > axis(1, at = seq_along(u), x) &...

...which are arranged into a matrix layout. Below is my current version, followed by a few questions. e = expression(alpha,"testing very large width", hat(beta), integral(f(x)*dx, a, b)) rowMax.units <- function(u, nrow){ # rowMax with a fake matrix of units matrix.indices <- matrix(seq_along(u), nrow=nrow) do.call(unit.c, lapply(seq(1, nrow), function(ii) { max(u[matrix.indices[ii, ]]) })) } colMax.units <- function(u, ncol){ # colMax with a fake matrix of units matrix.indices <- matrix(seq_along(u), ncol=ncol) do.call(unit.c, lapply(seq(1, ncol), function(ii) { max(u[ma...

...lit.default(1:5, 1:2) : data length is not a multiple of split variable > x = structure(1:5, class="A") > split(x, 1:2) $`1` [1] 1 $`2` [1] 2 Also, this is inconsistent with split<-, which does have recycling > split(x, 1:2) <- 1:2 Warning message: In split.default(seq_along(x), f, drop = drop, ...) : data length is not a multiple of split variable > x [1] 1 2 1 2 1 attr(,"class") [1] "A" A solution is to change a call to seq_along(f) toward the end of split.default to seq_along(x). @@ -32,7 +32,7 @@ lf <- levels(f) y <-...

...he function into a single one. I tried to test this by > f <- function(x) 1+x > g <- function(x) f(x) > x <- rnorm(1e6) > system.time(sapply(x,f)) [1] 11.315 0.157 11.735 0.000 0.000 > system.time(sapply(x,g)) [1] 8.850 0.140 9.283 0.000 0.000 > system.time(for (i in seq_along(x)) f(x[i])) [1] 2.466 0.036 2.884 0.000 0.000 > system.time(for (i in seq_along(x)) g(x[i])) [1] 3.548 0.045 4.165 0.000 0.000 but I find that hard to interpret -- the overhead looks significant in the first case, but something strange (at least to my limited knowledge) is happening with sappl...

...each file in both directories is a vector. I am not sure how to tell R to correlate the first column in dir1 to the correspond column from dir2. we will finally get only one spatial correlation map. I tried to this: # calculate the correlation so we will get a correlation map for (.files in seq_along(dir1)){ results[[length(results) + 1L]]<- cor(file1 ,file2) } I got error:Error in cor(file1, file2) : allocMatrix: too many elements specified` -- View this message in context: http://r.789695.n4.nabble.com/How-to-calculate-the-spatial-correlation-of-several-f...