Dear R-List, I have a set of possibilities I want to sample from:
bases <- list(c('A', 'C'), c('A', 'G'), c('C', 'T')) possibilities <- as.matrix(expand.grid(bases)) >possibilities Var1 Var2 Var3 [1,] "A" "A" "C" [2,] "C" "A" "C" [3,] "A" "G" "C" [4,] "C" "G" "C" [5,] "A" "A" "T" [6,] "C" "A" "T" [7,] "A" "G" "T" [8,] "C" "G" "T" If I want to randomly sample one of these rows. If I do this, I find that it is 25% likely that my choice will have an identical first and last letter (e.g. [1,] "A" "A" "C"). It is also 25% likely that my choice will have an identical first and third letter (e.g. [4,] "C" "G" "C"). It is not likely at all that the second and third letter of my choice could be identical. What I would like to do, is sample one of the rows, but given the constraint that the probability of drawing identical letters 1 and 2 should be 50% or 0.5, and at the same time the probability of drawing identical letters 1 and 3 should be 50%. I am unsure on how to do this, but I know it involves coming up with a modified set of weights for the sample() function. My progress is below, any advice is much appreciated. Best Wishes, Ben Ward, UEA. So I have used the following code to come up with a matrix, which contains weighting according to each criteria: possibilities <- as.matrix(expand.grid(bases)) identities <- apply(possibilities, 1, function(x) c(x[1] == x[2], x[1] == x[3], x[2] == x[3])) prob <- matrix(rep(0, length(identities)), ncol = ncol(identities)) consProb <- apply(identities, 1, function(x){0.5 / length(which(x))}) polProb <- apply(identities, 1, function(x){0.5 / length(which(!x))}) for(i in 1:nrow(identities)){ prob[i, which(identities[i,])] <- consProb[i] prob[i, which(!identities[i,])] <- polProb[i] } rownames(prob) <- c("1==2", "1==3", "2==3") colnames(prob) <- apply(possibilities, 1, function(x)paste(x, collapse = ", ")) This code gives the following matrix: A, A, C C, A, C A, G, C C, G, C A, A, T C, A, T A, G, T C, G, T 1==2 0.25000000 0.08333333 0.08333333 0.08333333 0.25000000 0.08333333 0.08333333 0.08333333 1==3 0.08333333 0.25000000 0.08333333 0.25000000 0.08333333 0.08333333 0.08333333 0.08333333 2==3 0.06250000 0.06250000 0.06250000 0.06250000 0.06250000 0.06250000 0.06250000 0.06250000 Each column is one of the choices from 'possibilities', and each row gives a series of weights based on three different criteria: Row 1, that if it possible from the choices for letter 1 == letter 2, that combined chance be 50%. Row 2, that if it possible from the choices for letter 1 == letter 3, that combined chance be 50%. Row 3, that if it possible from the choices for letter 2 == letter 3, that combined chance be 50%. So: If I used sample(x = 1:now(possibilities), size = 1, prob = prob[1,]) repeatedly, I expect about half the choices to contain identical letters 1 and 2. If I used sample(x = 1:now(possibilities), size = 1, prob = prob[2,]) repeatedly, I expect about half the choices to contain identical letters 1 and 3. If I used sample(x = 1:now(possibilities), size = 1, prob = prob[3,]) repeatedly, I expect about half the choices to contain identical letters 2 and 3. Except that in this case, since it is not possible. Note each row sums to 1. What I would like to do - if it is possible - is combine these three sets of weights into one set, that when used with sample(x = 1:nrow(possibilities, size = 1, prob = MAGICPROB) will give me a list of choices, where ~50% of them contain identical letters 1 and 2, AND ~50% of them contain identical letters 1 and 3, AND ~50% again contain identical letters 2 and 3 (except in this example as it is not possible from the choices). Can multiple probability weightings be combined in such a manner? [[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 http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.