Scala 中二维向量的 indexOf [英] indexOf for 2D Vector in Scala

问题描述

假设我想找到一个元素在 2D 向量中第一次出现的索引.

Say I want to find the indices of the first occurrence of an element in a 2D Vector.

val table = Vector.tabulate(10, 10)((x,y) => 10*x + y)
val row = table indexWhere (_.indexOf(42) != -1) // row = 4
val col = 
  if(row == -1) -1
  else table(row) indexOf 42 // col = 2

这似乎有点低效，因为 indexOf 在包含该元素的行上被调用了两次.有没有更好的方法来做到这一点，而不必求助于命令式代码?

This seems a tad inefficient, as indexOf gets called twice on the row containing the element. Is there a better way to do this without having to resort to imperative code?

推荐答案

这不仅效率低下，而且还很危险.如果我们稍微修改一下呢?

It's not only inefficient, it's also dangerous. What if we modified it slightly?

val row = table indexWhere (_.indexOf(101) != -1) 
val col = table(row) indexOf 42 //Uh-oh, IndexOutOfBounds!

这看起来真的很适合 for 表达式:

This really seems suited to a for-expression:

val z =
  for {
    i <- 0 until table.length
    j <- 0 until table(i).length
    if (table(i)(j) == 42)
  } yield (i, j)

z.headOption.getOrElse(-1, -1)

这可能过于强制了，但在幕后，这一切都被转换为 flatMap 和 filter.你可以用它来写它，但这更易读.

This might be too imperative, but it all gets transformed to flatMap and filter under the hood. You could write it with that, but this is far more readable.

如果您想要快速失败的行为，递归解决方案将符合要求:

If you want fail-fast behaviour, a recursive solution will fit the bill:

def findElement(table: Vector[Vector[Int]], elem: Int): (Int, Int) = {
  @tailrec
  def feRec(row: Int, col: Int): (Int, Int) = {
    if (row == table.length) (-1, -1)
    else if (col == table(row).length) feRec(row + 1, 0)
    else if (table(row)(col) == elem) (row, col)
    else feRec(row, col + 1)
  }
  feRec(0, 0)
}

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