R:将矩阵拆分为任意数量的 block

Question

假设我有一个值矩阵

    set.seed(1)
    A <- matrix(runif(25),ncol=5)

我想计算这个近似相等大小的矩阵内近似正方形邻域的一些统计数据。这两种输出中的任何一种都可以:

    N1 <-  matrix(c(rep(c("A","A","B","B","B"),2),rep(c("C","C","D","D","D"),3)),ncol=5)
    N2 <-  matrix(c(rep(c("A","A","A","B","B"),3),rep(c("C","C","D","D","D"),2)),ncol=5)
    N1
         [,1] [,2] [,3] [,4] [,5]
    [1,] "A"  "A"  "C"  "C"  "C" 
    [2,] "A"  "A"  "C"  "C"  "C" 
    [3,] "B"  "B"  "D"  "D"  "D" 
    [4,] "B"  "B"  "D"  "D"  "D" 
    [5,] "B"  "B"  "D"  "D"  "D" 

    N2
         [,1] [,2] [,3] [,4] [,5]
    [1,] "A"  "A"  "A"  "C"  "C" 
    [2,] "A"  "A"  "A"  "C"  "C" 
    [3,] "A"  "A"  "A"  "D"  "D" 
    [4,] "B"  "B"  "B"  "D"  "D" 
    [5,] "B"  "B"  "B"  "D"  "D" 

其他近似值也可以，因为我总是可以旋转矩阵。然后我可以使用这些邻域矩阵来计算统计数据 tapply() ， 像这样:

    tapply(A,N1,mean)
            A         B         C         D 
    0.6201744 0.5057402 0.4574495 0.5594227

我想要的是一个函数，它可以使我成为具有任意数量的块状邻域的任意维度矩阵，例如 N1或 N2 .我很难弄清楚这样一个函数如何处理所需数量的块甚至不是正方形的情况。 N1和 N2有 4 个街区，但说我想要 5 个这样的输出:

    N3 <-  matrix(c("A","A","B","B","B","A","A","C","C","C","D","D","C","C","C",
            "D","D","E","E","E","D","D","E","E","E"),ncol=5)
         [,1] [,2] [,3] [,4] [,5]
    [1,] "A"  "A"  "D"  "D"  "D" 
    [2,] "A"  "A"  "D"  "D"  "D" 
    [3,] "B"  "C"  "C"  "E"  "E" 
    [4,] "B"  "C"  "C"  "E"  "E" 
    [5,] "B"  "C"  "C"  "E"  "E" 

有谁知道可以进行这种拆分的现有功能，或者对如何制作一个功能有任何想法？谢谢!

[[编辑]]
我的最终功能，考虑到文森特的建议:

    DecideBLocks <- function(A,nhoods){
        nc <- ncol(A)
        nr <- nrow(A)
        nhood_side <- floor(sqrt((nc*nr)/nhoods))
        Neighborhoods <- matrix(paste(ceiling(col(A)/nhood_side), ceiling(row(A)/nhood_side), sep="-"), nc=ncol(A)) 
        nhoods.out <- length(unique(c(Neighborhoods)))
        if (nhoods.out != nhoods){
            cat(nhoods.out,"neighborhoods created.\nThese were on average",nhood_side,"by",nhood_side,"cells\nit's a different number than that stated the function tries to round things to square neighborhoods\n")
        }
        return(Neighborhoods)
    }
    A <- matrix(rnorm(120),12)
    B <- DecideBLocks(A,13)

Answer 1

你可以试试玩row和 col职能:
他们将问题简化为一维问题。
下面定义了大小最多为 2*2 的块。

matrix( 
  paste(
    ceiling(col(A)/2), 
    ceiling(row(A)/2), 
    sep="-"), 
  nc=ncol(A) 
)