And I missed copying one line: sums <- rowSums(x)
should come between the calculations of x and x0. Duncan Murdoch On 2025-04-26 1:01 p.m., Duncan Murdoch wrote:
On 2025-04-24 3:25 p.m., Bert Gunter wrote:Folks: Unless my wee old brain errs (always a danger), uniform sampling from an n-vector <xi> for which 0 <= ai <= xi <= bi and SUM(xi) = k, a constant, where to ensure that the constraints are consistent (and nontrivial), SUM(ai)< k and SUM(bi) > k, is a simple linear transformation (details left to the reader) of uniform sampling from a standard n-1 dim simplex, for which a web search yielded: https://cs.stackexchange.com/questions/3227/uniform-sampling-from-a-simplexI think for n > 2 it is sometimes a simplex, and sometimes a different shape. For example, with n = 3 and all a_i = 0 and all b_i = 1, it's a simplex if k < 1 or k > 2, but not for 1 < k < 2, where it's a hexagon. You can visualize this with the following code: x <- matrix(runif(30000), ncol = 3) # the full cube x0 <- x[1.45 < sums & sums < 1.55,] # points near k = 1.5 rgl::plot3d(x0)The proposed "solutions" in this thread which rejection sample sequentially from uniforms do *not* produce a uniform distribution on the simplex, as the link and references therein explain.My solution didn't do rejection sampling, but it won't give a uniform density. I didn't realize that was requested.Apologies if I have misunderstood the problem and proposed solutions, although there does seem to be some confusion in the thread about exactly what was sought.Yes, that's true. Duncan MurdochCheers, Bert "An educated person is one who can entertain new ideas, entertain others, and entertain herself." On Thu, Apr 24, 2025 at 10:33 AM Brian Smith <briansmith199...@gmail.com> wrote:Hi Rui, This code is able to generate absolutely correct random sample vector based on the applicable constraints. I have one question though. If I understood the R code correctly then, the first element is drawing random number from Uniform distribution unconditionally, however drawing of sample point for the second element is conditional to the first one. Therefore if we have large vector size instead of current 2, I guess the feasible region for the last few elements will be very small. Will that be any problem? does there any algorithm exist where all (n-1) elements would be drawn unconditionally assuming our vector has n elements? Thanks and regards, On Wed, 23 Apr 2025 at 10:57, Rui Barradas <ruipbarra...@sapo.pt> wrote:Hello, Here are your tests and the random numbers' histograms. one_vec <- function(a, b, s) { repeat{ repeat{ u <- runif(1, a[1], b[1]) if(s - u > 0) break } v <- s - u if(a[2] < v && v < b[2]) break } c(u, v) } gen_mat <- function(m, a, b, s) { replicate(m, one_vec(a, b, s)) } a <- c(0.015, 0.005) b <- c(0.070, 0.045) s <- 0.05528650577311 m <- 10000L set.seed(2025) res <- gen_mat(m, a, b, s) apply(res, 1, min) > a #> [1] TRUE TRUE apply(res, 1, max) < b #> [1] TRUE TRUE # plot histograms of one million 2d vectors system.time( res1mil <- gen_mat(1e6, a, b, s) ) #> user system elapsed #> 3.01 0.06 3.86 old_par <- par(mfrow = c(1, 2)) hist(res1mil[1L,]) hist(res1mil[2L,]) par(old_par) Hope this helps, Rui Barradas Às 23:31 de 22/04/2025, Rui Barradas escreveu:Hello, Inline. Às 17:55 de 22/04/2025, Brian Smith escreveu:i.e. we should have all elements of Reduce("+", res) should be equal to s =0.05528650577311My assertion is that it is not happing here.You are right, that's not what is happening. The output is n vectors of 2 elements each. It's each of these vectors that add up to s. Appparently I misunderstood the problem. Maybe this is what you want? (There is no n argument, the matrix is always 2*m) one_vec <- function(a, b, s) { repeat{ repeat{ u <- runif(1, a[1], b[1]) if(s - u > 0) break } v <- s - u if(a[2] < v && v < b[2]) break } c(u, v) } gen_mat <- function(m, a, b, s) { replicate(m, one_vec(a, b, s)) } set.seed(2025) res <- gen_mat(10000, a, b, s) colSums(res) Hope this helps, Rui BarradasOn Tue, 22 Apr 2025 at 22:20, Brian Smith <briansmith199...@gmail.comwrote:Hi Rui, Thanks for the explanation. But in this case, are we looking at the correct solution at all? My goal is to generate random vector where: 1) the first element is bounded by (a[1], b[1]) and second element is bounded by (a[2], b[2]) 2) sum of the element is s According to the outcome, The first matrix values are bounded by c(a[1], b[1]) & second matrix values are bounded by c(a[2], b[2]) But, regarding the sum, I think we should have sum (element-wise) sum should be equal to s = 0.05528650577311. How could we achieve that then? On Tue, 22 Apr 2025 at 22:03, Rui Barradas <ruipbarra...@sapo.pt>wrote:Às 12:39 de 22/04/2025, Brian Smith escreveu:Hi Rui, Many thanks for your time and insight. However, I am not sure if I could understand the code. Below iswhat Itried based on your code library(Surrogate) a <- c(0.015, 0.005) b <- c(0.070, 0.045) set.seed(2025) res <- mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m), MoreArgs = list(s = 0.05528650577311, n = 2, m = 10000), a, b) res1 = res[[1]] res2 = res[[2]] apply(res1, 1, min) > a ## [1] TRUE TRUE apply(res2, 1, min) > a ## [1] FALSE TRUE I could not understand what basically 2 blocks of res signify?Whichone I should take as final simulation of the vector? If I take the first block then the lower bound condition is fulfilled, but notwiththe second block. However with the both blocks, the total equals sissatisfying. I appreciate your further insight. Thanks and regards, On Mon, 21 Apr 2025 at 20:43, Rui Barradas <ruipbarra...@sapo.pt> wrote:Hello, Inline. Às 16:08 de 21/04/2025, Rui Barradas escreveu:Às 15:27 de 21/04/2025, Brian Smith escreveu:Hi, There is a function called RandVec in the package Surrogate which can generate andom vectors (continuous number) with a fixed sum The help page of this function states that: a The function RandVec generates an n by m matrix x. Each of the m columns contain n random values lying in the interval [a,b]. The argument a specifies the lower limit of the interval. Default 0. b The argument b specifies the upper limit of the interval. Default 1. However in my case, the lower and upper limits are not same. For example, if I need to draw a pair of number x, y, such that x + y = 1, then the lower and upper limits are different. I tried with below code library(Surrogate) RandVec(a=c(0.1, 0.2), b=c(0.2, 0.8), s=1, n=2,m=5)$RandVecOutputThis generates error with message Error in if (b - a == 0) { : the condition has length > 1 Is there any way to generate the numbers with different lowerandupper limits? ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more,seehttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttps://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproduciblecode.Hello, Use ?mapply to cycle through all values of a and b. Note that the output matrices are transposed, the random vectors are the rows.Sorry, this is not true. The columns are the random vectors, as documented. An example setting the RNG seed, for reproducibility. library(Surrogate) a <- c(0.1, 0.2) b <- c(0.2, 0.8) set.seed(2025) res <- mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m), MoreArgs = list(s = 1, n = 2, m = 5), a, b) res #> $RandVecOutput #> [,1] [,2] [,3] [,4] [,5] #> [1,] 0.146079 0.1649319 0.1413759 0.257086 0.1715478 #> [2,] 0.253921 0.2350681 0.2586241 0.142914 0.2284522 #> #> $RandVecOutput #> [,1] [,2] [,3] [,4] [,5] #> [1,] 0.5930918 0.2154583 0.6915523 0.7167089 0.3617918 #> [2,] 0.4069082 0.7845417 0.3084477 0.2832911 0.6382082 lapply(res, colSums) #> $RandVecOutput #> [1] 0.4 0.4 0.4 0.4 0.4 #> #> $RandVecOutput #> [1] 1 1 1 1 1 Hope this helps, Rui Barradaslibrary(Surrogate) a <- c(0.1, 0.2) b <- c(0.2, 0.8) mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m), MoreArgs = list(s = 1, n = 2, m = 5), a, b) #> $RandVecOutput #> [,1] [,2] [,3] [,4] [,5] #> [1,] 0.2004363 0.1552328 0.2391742 0.1744857 0.1949236 #> [2,] 0.1995637 0.2447672 0.1608258 0.2255143 0.2050764 #> #> $RandVecOutput #> [,1] [,2] [,3] [,4] [,5] #> [1,] 0.2157416 0.4691191 0.5067447 0.7749258 0.7728955 #> [2,] 0.7842584 0.5308809 0.4932553 0.2250742 0.2271045 Hope this helps, Rui Barradas-- Este e-mail foi analisado pelo software antivírus AVG para verificar a presença de vírus. www.avg.comHello, The two blocks of res are the two random matrices, one for each combination of (a,b). mapply passes each of the values in itsargumentslist (the ellipses in the help page) and computes the anonymous function with the pairs (a[1], b[1]), (a[2], b[2]). Since a and b are two elements vectors the output res is a twomembersnamed list. Your error is to compare the result of apply(res2, 1,min)to a, when you should compare to a[2]. See the code below. library(Surrogate) a <- c(0.015, 0.005) b <- c(0.070, 0.045) set.seed(2025) res <- mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m), MoreArgs = list(s = 0.05528650577311, n = 2, m = 10000), a, b) res1 = res[[1]] res2 = res[[2]] # first check that the sums are correct # these sums should be s = 0.05528650577311, up to floating-point accuracy lapply(res, \(x) colSums(x[, 1:5]) |> print(digits = 14L)) #> [1] 0.05528650577311 0.05528650577311 0.05528650577311 0.05528650577311 #> [5] 0.05528650577311 #> [1] 0.05528650577311 0.05528650577311 0.05528650577311 0.05528650577311 #> [5] 0.05528650577311 #> $RandVecOutput #> [1] 0.05528651 0.05528651 0.05528651 0.05528651 0.05528651 #> #> $RandVecOutput #> [1] 0.05528651 0.05528651 0.05528651 0.05528651 0.05528651 # now check the min and max apply(res1, 1, min) > a[1L] ## [1] TRUE TRUE #> [1] TRUE TRUE apply(res2, 1, min) > a[2L] ## [1] TRUE TRUE #> [1] TRUE TRUE apply(res1, 1, max) < b[1L] ## [1] TRUE TRUE #> [1] TRUE TRUE apply(res2, 1, max) < b[2L] ## [1] TRUE TRUE #> [1] TRUE TRUE Which one should you take as final simulation of the vector? Both. The first matrix values are bounded by c(a[1], b[1]) with columnsumsequal to s. The second matrix values are bounded by c(a[2], b[2]) with columnsumsalso equal to s. Hoep this helps, Rui Barradas______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide https://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code.-- Este e-mail foi analisado pelo software antivírus AVG para verificar apresença de vírus.www.avg.com______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide https://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.[[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide https://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide https://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
