Question

我有一个由12个矩阵组成的数组：

ma = array(sample(0:127,3*4*6,replace=TRUE), c(3,4,12))

让它们命名为A，B，C ...... L

我想创建一个矩阵，其上面的矩阵以4rows * 3列模式排列：

ABC
DEF
GHI
JKL

因此最终矩阵将有12列和12列。

我可以按照以下代码执行此操作：

rbind(cbind(m[,,1],m[,,2],m[,,3]),
      cbind(m[,,4],m[,,5],m[,,6]),
      cbind(m[,,7],m[,,8],m[,,9]),
      cbind(m[,,10],m[,,11],m[,,12]))

但我无法为此编写通用函数：

matbinder(ma,n)

其中ma是输入矩阵数组，n是要放在一行中的初始矩阵数（在本例中为3）。

Answer 1

我将它用作样本矩阵，因为它将按照您喜欢的顺序包含所有字母

ma = array(as.vector(t(outer(letters[1:12],1:12, FUN=paste0))), c(3,4,12))

然后你可以像

那样进行转换

a<-ma
dim(a) <- c(3,12,4)
apply(a,2,c)

产生

      [,1] [,2] [,3] [,4]  [,5] [,6] [,7] [,8]  [,9] [,10] [,11] [,12]
 [1,] "a1" "a4" "a7" "a10" "b1" "b4" "b7" "b10" "c1" "c4"  "c7"  "c10"
 [2,] "a2" "a5" "a8" "a11" "b2" "b5" "b8" "b11" "c2" "c5"  "c8"  "c11"
 [3,] "a3" "a6" "a9" "a12" "b3" "b6" "b9" "b12" "c3" "c6"  "c9"  "c12"
 [4,] "d1" "d4" "d7" "d10" "e1" "e4" "e7" "e10" "f1" "f4"  "f7"  "f10"
 [5,] "d2" "d5" "d8" "d11" "e2" "e5" "e8" "e11" "f2" "f5"  "f8"  "f11"
 [6,] "d3" "d6" "d9" "d12" "e3" "e6" "e9" "e12" "f3" "f6"  "f9"  "f12"
 [7,] "g1" "g4" "g7" "g10" "h1" "h4" "h7" "h10" "i1" "i4"  "i7"  "i10"
 [8,] "g2" "g5" "g8" "g11" "h2" "h5" "h8" "h11" "i2" "i5"  "i8"  "i11"
 [9,] "g3" "g6" "g9" "g12" "h3" "h6" "h9" "h12" "i3" "i6"  "i9"  "i12"
[10,] "j1" "j4" "j7" "j10" "k1" "k4" "k7" "k10" "l1" "l4"  "l7"  "l10"
[11,] "j2" "j5" "j8" "j11" "k2" "k5" "k8" "k11" "l2" "l5"  "l8"  "l11"
[12,] "j3" "j6" "j9" "j12" "k3" "k6" "k9" "k12" "l3" "l6"  "l9"  "l12"

这是有效的，因为通过翻转尺寸，我们基本上构建了一个3 * 12矩阵的4个元素数组，它对应于我们想要的行。然后我们使用apply来折叠额外的维度。

通用matbinder(ma,n)函数看起来像

matbinder <- function(ma,n) {
    d<-dim(ma)
    r<-ceiling(d[3]/n)
    a<-c(ma, rep(NA, (n*r-d[3]) * prod(d[1:2])))
    dim(a)<-c(d[1], n*d[2],r)
    apply(a,2,c)
}

Answer 2

利用我有限的知识，我管理了以下代码：

matbinder2 = function(submat, mats_per_row){
    submat_dims = dim(submat)
    submatr = submat_dims[1]
    submatc = submat_dims[2]
    submatn = submat_dims[3]

    matc = submatc*mats_per_row
    matr = submatr* ceiling(submatn/mats_per_row)
    mat = array(NA, c(matr,matc))

    for(m in 1:submatn){
        r = 1+(floor(submatc*(m-1)/matc))*submatr
        c = 1+((submatc*(m-1))%%matc)
        i=1;j=1;k=0;l=0;
        while(TRUE){
            mat[r+l,c+k] =  submat[i,j,m]
            j=j+1; k=k+1; 
            if(j>submatc){j=1;k=0; i=i+1;l=l+1;if(i>submatr){break;}}
        }
    }
    mat
}

将矩阵并排放置以创建另一个矩阵

2 个答案: