并排把矩阵侧创建另一个矩阵

我具有由下面的代码给出的12点矩阵的阵列：并排把矩阵侧创建另一个矩阵

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

让它们被命名为A，B，C ... ...你

我想创建一个矩阵，矩阵排列成4行×3列：

ABC 
DEF 
GHI 
JKL

所以最终的矩阵将有12行和12列。

我可以做到这一点下面的代码：

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]))

但我不能写这样的通用功能：

matbinder（MA，N）＃其中MA是输入矩阵阵列和n是要放在一行中的初始矩阵的数量（在这种情况下是3）。

来源

2014-06-18 rnso

@Jilber：感谢编辑：使代码更易读。 – rnso

我将以此作为样品基质，因为它会在订单中所有的字母，你喜欢

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

然后，你可以做改造喜欢

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"

这是可行的，因为通过翻转尺寸，我们基本上构建了一个3 * 12矩阵的4元素数组与我们想要的行相对应的ces。然后我们使用apply来折叠整个额外的维度。

一个通用matbinder(ma,n)功能看起来像

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) 
}

来源

2014-06-18 14:41:53 MrFlick

太好了。这正是我想要的。谢谢。是否可以对其进行修改，使得如果ma中只有11个矩阵，则最终矩阵中的剩余位置将填充“NA”。类似地，该函数可能用于命令matbinder（ma，5），因为它目前会产生一个错误：dims [product 120]与对象[144]的长度不匹配。 – rnso

@rnso我已经更新了我的答案，必要时用NA填充。 – MrFlick

谢谢。这真是令人印象深刻。 – rnso

用我有限的知识，我设法下面的代码：

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 
}

来源

2014-06-21 02:32:22 rnso

并排把矩阵侧创建另一个矩阵

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