R：分割矩阵成块

Question

的任意数量的说我有值

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

的矩阵我想此矩阵近似相等大小的范​​围内计算近似正方形的街区一些统计信息。这两种类型的输出会做：

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" 

其他近似也行，因为我总是可以旋转矩阵。然后，我可以使用这些邻里矩阵计算使用tapply()统计，像这样：

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

我要的是一个功能，可以让我任意维度的矩阵与块状街区像N1或任意数量的N2。我很难找出这样一个函数如何处理所需块数不是偶数的情况。 N1和N2有4个街区，但说我想5对一些输出是这样的：

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" 

有谁知道现有的功能，可以做这样的分裂，或者对如何使一个任何想法？谢谢！

[编辑] 我的最终功能，考虑到文森特的建议是：近youree中心

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

你可以尝试用row和col功能发挥： 他们问题缩小到一维的。 以下定义最多2 * 2的大小块。

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

Answer 2

你可以选择你bdeep（行规范）和bwide（CO-SPEC）的参数以您喜欢的任何方式创建矩阵尺寸，并使用这个简单的函数来构建矩阵。只要bwide和bdeep相等，并且nrow == ncol，你应该得到方形子矩阵。

mkblk <- function(bwide, bdeep, nrow, ncol){ 
    bstr1 <- c(rep("A", bdeep), rep("B", nrow-bdeep)) 
    bstr2 <- c(rep("C", bdeep), rep("D", nrow-bdeep)) 
    matrix(c(rep(bstr1, bwide), rep(bstr2, ncol-bwide)), ncol=ncol, nrow=nrow)} 
mkblk(2,2,5,5) 

    [,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" 

#Test of your strategy 
tapply(A, mkblk(2,2,5,5), mean) 
     A   B   C   D 
0.6201744 0.5057402 0.4574495 0.5594227