使用1:n中的唯一值创建n×n矩阵

Question

我想在R中生成随机n×n矩阵,离散值范围从1到n.棘手的部分是我希望每个值在行和列中都是唯一的.

例如,如果n=3矩阵看起来像:

1 2 3 
2 3 1 
3 1 2 

或者看起来像这样:

2 3 1 
1 2 3 
3 1 2 

任何人都知道如何生成这种矩阵？

Answer 1

你想要什么被称为拉丁方.这里有一个函数(来自Cookbook for R ;另见这里和一些其他在线搜索结果)允许生成它们:

latinsquare <- function(len, reps=1, seed=NA, returnstrings=FALSE) {

    # Save the old random seed and use the new one, if present
    if (!is.na(seed)) {
        if (exists(".Random.seed"))  { saved.seed <- .Random.seed }
        else                         { saved.seed <- NA }
        set.seed(seed)
    }

    # This matrix will contain all the individual squares
    allsq <- matrix(nrow=reps*len, ncol=len)

    # Store a string id of each square if requested
    if (returnstrings) {  squareid <- vector(mode = "character", length = reps) }

    # Get a random element from a vector (the built-in sample function annoyingly
    #   has different behavior if there's only one element in x)
    sample1 <- function(x) {
        if (length(x)==1) { return(x) }
        else              { return(sample(x,1)) }
    }

    # Generate each of n individual squares
    for (n in 1:reps) {

        # Generate an empty square
        sq <- matrix(nrow=len, ncol=len) 

        # If we fill the square sequentially from top left, some latin squares
        # are more probable than others.  So we have to do it random order,
        # all over the square.
        # The rough procedure is:
        # - randomly select a cell that is currently NA (call it the target cell)
        # - find all the NA cells sharing the same row or column as the target
        # - fill the target cell
        # - fill the other cells sharing the row/col
        # - If it ever is impossible to fill a cell because all the numbers
        #    are already used, then quit and start over with a new square.
        # In short, it picks a random empty cell, fills it, then fills in the 
        # other empty cells in the "cross" in random order. If we went totally randomly
        # (without the cross), the failure rate is much higher.
        while (any(is.na(sq))) {

            # Pick a random cell which is currently NA
            k <- sample1(which(is.na(sq)))

            i <- (k-1) %% len +1       # Get the row num
            j <- floor((k-1) / len) +1 # Get the col num

            # Find the other NA cells in the "cross" centered at i,j
            sqrow <- sq[i,]
            sqcol <- sq[,j]

            # A matrix of coordinates of all the NA cells in the cross
            openCell <-rbind( cbind(which(is.na(sqcol)), j),
                              cbind(i, which(is.na(sqrow))))
            # Randomize fill order
            openCell <- openCell[sample(nrow(openCell)),]

            # Put center cell at top of list, so that it gets filled first
            openCell <- rbind(c(i,j), openCell)
            # There will now be three entries for the center cell, so remove duplicated entries
            # Need to make sure it's a matrix -- otherwise, if there's just 
            # one row, it turns into a vector, which causes problems
            openCell <- matrix(openCell[!duplicated(openCell),], ncol=2)

            # Fill in the center of the cross, then the other open spaces in the cross
            for (c in 1:nrow(openCell)) {
                # The current cell to fill
                ci <- openCell[c,1]
                cj <- openCell[c,2]
                # Get the numbers that are unused in the "cross" centered on i,j
                freeNum <- which(!(1:len %in% c(sq[ci,], sq[,cj])))

                # Fill in this location on the square
                if (length(freeNum)>0) { sq[ci,cj] <- sample1(freeNum) }
                else  {
                    # Failed attempt - no available numbers
                    # Re-generate empty square
                    sq <- matrix(nrow=len, ncol=len)

                    # Break out of loop
                    break;
                }
            }
        }

        # Store the individual square into the matrix containing all squares
        allsqrows <- ((n-1)*len) + 1:len
        allsq[allsqrows,] <- sq

        # Store a string representation of the square if requested. Each unique
        # square has a unique string.
        if (returnstrings) { squareid[n] <- paste(sq, collapse="") }

    }

    # Restore the old random seed, if present
    if (!is.na(seed) && !is.na(saved.seed)) { .Random.seed <- saved.seed }

    if (returnstrings) { return(squareid) }
    else               { return(allsq) }
}

Answer 2

mats是此类矩阵的列表。它用于r2dtable生成N随机 nxn 矩阵，其元素选自 0, 1, ..., n-1，其边距分别由 给出margin。然后，它过滤掉所有列都为 0:(n-1) 的列，并将每个矩阵加 1 以给出结果。返回的矩阵数量可能会有所不同，并且N随着 n 变大，您必须生成大量矩阵才能获得少量矩阵。当我尝试下面的 n <- 3 时，mats列出了 100 个矩阵中的 24 个矩阵，但当 n <- 4 时，它只找到了 100 个矩阵中的 1 个。

set.seed(123)
N <- 100 # no of tries
n <- 3 # rows of matrix (= # cols)

check <- function(x) all(apply(x, 2, sort) == seq_len(nrow(x))-1)
margin <- sum(seq_len(n))-n
margins <- rep(margin, n)
L <- r2dtable(N, r = margins, c = margins)
mats <- lapply(Filter(check, L), "+", 1)