Hi Florent,That is great, I was working on solving the recurrence equation and this was part of that equation. Now I know how to put everything together, thanks for all the help e everybody!
Cheers, On 27/11/2011 2:05 p.m., Florent D. wrote:
You can make things a lot faster by using the recurrence equation y[n] = alpha * (y[n-1]+x[n]) loopRec<- function(x, alpha){ n<- length(x) y<- numeric(n) if (n == 0) return(y) y[1]<- alpha * x[1] for(i in seq_len(n)[-1]){ y[i]<- alpha * (y[i-1] + x[i]) } return(y) } On Sun, Nov 27, 2011 at 5:17 AM, Michael Kao<mkao006rm...@gmail.com> wrote:Dear Enrico, Brilliant! Thank you for the improvements, not sure what I was thinking using rev. I will test it out to see whether it is fast enough for our implementation, but your help has been SIGNIFICANT!!! Thanks heaps! Michael On 27/11/2011 10:43 a.m., Enrico Schumann wrote:Hi Michael here are two variations of your loop, and both seem faster than the original loop (on my computer). require("rbenchmark") ## your function loopRec<- function(x, alpha){ n<- length(x) y<- double(n) for(i in 1:n){ y[i]<- sum(cumprod(rep(alpha, i)) * rev(x[1:i])) } y } loopRec(c(1, 2, 3), 0.5) loopRec2<- function(x, alpha){ n<- length(x) y<- numeric(n) for(i in seq_len(n)){ y[i]<- sum(cumprod(rep.int(alpha, i)) * x[i:1]) } y } loopRec2(c(1, 2, 3), 0.5) loopRec3<- function(x, alpha){ n<- length(x) y<- numeric(n) aa<- cumprod(rep.int(alpha, n)) for(i in seq_len(n)){ y[i]<- sum(aa[seq_len(i)] * x[i:1]) } y } loopRec3(c(1, 2, 3), 0.5) ## Check whether value is correct all.equal(loopRec(1:1000, 0.5), loopRec2(1:1000, 0.5)) all.equal(loopRec(1:1000, 0.5), loopRec3(1:1000, 0.5)) ## benchmark the functions. benchmark(loopRec(1:1000, 0.5), loopRec2(1:1000, 0.5), loopRec3(1:1000, 0.5), replications = 50, order = "relative") ... I get test replications elapsed relative user.self sys.self 2 loopRec2(1:1000, 0.5) 50 0.77 1.000000 0.76 0.00 3 loopRec3(1:1000, 0.5) 50 0.86 1.116883 0.85 0.00 1 loopRec(1:1000, 0.5) 50 1.84 2.389610 1.79 0.01 Regards, Enrico Am 27.11.2011 01:20, schrieb Michael Kao:Dear R-help, I have been trying really hard to generate the following vector given the data (x) and parameter (alpha) efficiently. Let y be the output list, the aim is to produce the the following vector(y) with at least half the time used by the loop example below. y[1] = alpha * x[1] y[2] = alpha^2 * x[1] + alpha * x[2] y[3] = alpha^3 * x[1] + alpha^2 * x[2] + alpha * x[3] ..... below are the methods I have tried and failed miserably, some are just totally ridiculous so feel free to have a laugh but would appreciate if someone can give me a hint. Otherwise I guess I'll have to give RCpp a try..... ## Bench mark the recursion functions loopRec<- function(x, alpha){ n<- length(x) y<- double(n) for(i in 1:n){ y[i]<- sum(cumprod(rep(alpha, i)) * rev(x[1:i])) } y } loopRec(c(1, 2, 3), 0.5) ## This is a crazy solution, but worth giving it a try. charRec<- function(x, alpha){ n<- length(x) exp.mat<- matrix(rep(x, each = n), nc = n, byrow = TRUE) up.mat<- matrix(eval(parse(text = paste("c(", paste(paste(paste("rep(0, ", 0:(n - 1), ")", sep = ""), paste("cumprod(rep(", alpha, ",", n:1, "))", sep = "") , sep = ","), collapse = ","), ")", sep = ""))), nc = n, byrow = TRUE) colSums(up.mat * exp.mat) } vecRec(c(1, 2, 3), 0.5) ## Sweep is slow, shouldn't use it. matRec<- function(x, alpha){ n<- length(x) exp.mat<- matrix(rep(x, each = n), nc = n, byrow = TRUE) up.mat<- sweep(matrix(cumprod(rep(alpha, n)), nc = n, nr = n, byrow = TRUE), 1, c(1, cumprod(rep(1/alpha, n - 1))), FUN = "*") up.mat[lower.tri(up.mat)]<- 0 colSums(up.mat * exp.mat) } matRec(c(1, 2, 3), 0.5) matRec2<- function(x, alpha){ n<- length(x) exp.mat<- matrix(rep(x, each = n), nc = n, byrow = TRUE) up.mat1<- matrix(cumprod(rep(alpha, n)), nc = n, nr = n, byrow = TRUE) up.mat2<- matrix(c(1, cumprod(rep(1/alpha, n - 1))), nc = n, nr = n) up.mat<- up.mat1 * up.mat2 up.mat[lower.tri(up.mat)]<- 0 colSums(up.mat * exp.mat) } matRec2(c(1, 2, 3), 0.5) ## Check whether value is correct all.equal(loopRec(1:1000, 0.5), vecRec(1:1000, 0.5)) all.equal(loopRec(1:1000, 0.5), matRec(1:1000, 0.5)) all.equal(loopRec(1:1000, 0.5), matRec2(1:1000, 0.5)) ## benchmark the functions. benchmark(loopRec(1:1000, 0.5), vecRec(1:1000, 0.5), matRec(1:1000, 0.5), matRec2(1:1000, 0.5), replications = 50, order = "relative") Thank you very much for your help. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
