Arrays 如何在无回路的Scala中重塑二维线性化阵列

我想知道如何在Scala中无循环地重塑二维线性化阵列，即使用函数编程

假设一个6x12=72个元素的二维数组。我需要的是调整这个数组的大小，使它有8x16=128个元素，假设新元素初始化为零

以下是我使用循环的方式：

val a = (1 to 72).toArray
a: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72)

scala> val h = 6
h: Int = 6

scala> val w = 12
w: Int = 12

scala> val H = 8
H: Int = 8

scala> val W = 16
W: Int = 16

scala> val b = Array.ofDim[Int](H * W)
b: Array[Int] = Array(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

for (i <- 0 until h)
  for (j <- 0 until w)
    b(i * W + j) = a(i * w + j)

b: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 0, 0, 0, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 0, 0, 0, 0, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 0, 0, 0, 0, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 0, 0, 0, 0, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 0, 0, 0, 0, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

vala=（1到72）。toArray
a:数组[Int]=数组（1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,42,43,44,45,47,48,49,50,51,52,54,56,58,59,60,61,62,63,64,65,66,68,71,72）
scala>val h=6
h:Int=6
scala>val w=12
w:Int=12
scala>val H=8
H:Int=8
scala>val W=16
W:Int=16
scala>valb=Array.ofDim[Int]（H*W）
b:数组[Int]=数组(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
为了
结果（以及从代码中复制的字符串，用于比较）：

a.grouped(w).map(_.padTo(W, 0)).padTo(H, Array.fill(W)(0)).flatten.toArray

result        : Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 0, 0, 0, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 0, 0, 0, 0, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 0, 0, 0, 0, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 0, 0, 0, 0, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 0, 0, 0, 0, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
b: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 0, 0, 0, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 0, 0, 0, 0, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 0, 0, 0, 0, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 0, 0, 0, 0, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 0, 0, 0, 0, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)