从文本中获取矩阵的功能方法

Question

我正在尝试解决一些Google Code Jam问题,其中输入矩阵通常以这种形式给出:

2 3 #matrix dimensions
1 2 3 4 5 6 7 8 9 # all 3 elements in the first row
2 3 4 5 6 7 8 9 0 # each element is composed of three integers

其中矩阵的每个元素由三个整数组成.所以这个例子应该转换为

#!scala
Array(
     Array(A(1,2,3),A(4,5,6),A(7,8,9),
     Array(A(2,3,4),A(5,6,7),A(8,9,0),
)

一种必要的解决方案就是形式

#!python
input = """2 3
1 2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 0
"""
lines = input.split('\n')
class Aclass:
    def __init__(self,a,b,c):
        pass

print lines[0]
m,n = (int(x) for x in lines[0].split())
array = []
row = []
A = []
for line in lines[1:]:
    for elt in line.split():
        A.append(elt)
        if len(A)== 3:
            row.append(Aclass(A[0],A[1],A[2]))
            A = []
    array.append(row)
    row = []

from pprint import pprint
pprint(array)

我想到的功能解决方案是

#!scala
def splitList[A](l:List[A],i:Int):List[List[A]] = {
  if (l.isEmpty) return List[List[A]]()
  val (head,tail) = l.splitAt(i)
  return head :: splitList(tail,i)
}

def readMatrix(src:Iterator[String]):Array[Array[TrafficLight]] = {
  val Array(x,y) = src.next.split(" +").map(_.trim.toInt)
  val mat = src.take(x).toList.map(_.split(" ").
          map(_.trim.toInt)).
          map(a => splitList(a.toList,3).
                map(b => TrafficLight(b(0),b(1),b(2))
                 ).toArray
                ).toArray
    return mat
  }

但我真的觉得这是错误的方式,因为:

1. 我正在使用List每一行的功能结构,然后将其转换为数组.整个代码似乎不那么有效
2. 我发现它比python解决方案更不优雅,更不易读.由于它们都使用相同的语义,因此哪些map函数对哪些函数进行操作更难.

什么是正确的功能方式呢？

Answer 1

val x = """2 3
1 2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 0
"""

val a = x split "\n" map (_.trim.split(" "))
val rows = a(0)(0).toInt
val columns = a(0)(1).toInt

val matrix = (a drop 1) map (_ grouped columns toList) toList

并打印结果:

matrix.map(_.map(_.mkString("(",",",")")).mkString("(",",",")")).mkString("\n")

res1: String =
((1,2,3),(4,5,6),(7,8,9))
((2,3,4),(5,6,7),(8,9,0))

假设:

assert(rows == matrix.length)
assert(matrix.forall(_.forall(_.size == columns))) 

生产阵列tabulate更合适:

val a = x split "\n" map (_.trim.split(" "))
val rows = a(0)(0).toInt
val columns = a(0)(1).toInt
val matrix = Array.tabulate(rows, a(1).size / columns, columns)(
  (i,j,k) => a(i +  1)(j * columns + k))