R：分割矩陣成塊

的任意數量的說我有值R：分割矩陣成塊

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

的矩陣我想此矩陣近似相等大小的範圍內計算近似正方形的街區一些統計信息。這兩種類型的輸出會做：

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，因爲我總是可以旋轉矩陣。然後，我可以使用這些鄰里矩陣計算使用tapply()統計，像這樣：

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

我要的是一個功能，可以讓我任意維度的矩陣與塊狀街區像N1或任意數量的N2。我很難找出這樣一個函數如何處理所需塊數不是偶數的情況。 N1和N2有4個街區，但說我想5對一些輸出是這樣的：

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"

有誰知道現有的功能，可以做這樣的分裂，或者對如何使一個任何想法？謝謝！

[編輯] 我的最終功能，考慮到文森特的建議是：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)

來源

2012-02-24 tim riffe

你可以嘗試用row和col功能發揮：他們問題縮小到一維的。以下定義最多2 * 2的大小塊。

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

來源

2012-02-24 13:44:17

謝謝文森特，我想我可以通過修補這個東西來得到我需要的東西 – 2012-02-24 16:42:15

你可以選擇你bdeep（行規範）和bwide（CO-SPEC）的參數以您喜歡的任何方式創建矩陣尺寸，並使用這個簡單的函數來構建矩陣。只要bwide和bdeep相等，並且nrow == ncol，你應該得到方形子矩陣。

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

來源

2012-02-24 13:41:34

謝謝迪文，文森特打得更接近一點，但這裏也有不錯的想法。 – 2012-02-24 16:42:06

R：分割矩陣成塊

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