根据其值和另一个数组的值更新一个数组 [英] Update an Array based on its values and the values of another Array

问题描述

我正在尝试根据其他数组的值更改数组的值.特别是这些是我正在使用的数组:

I am trying to change the value of an array based on the value of a different array. In particular these are the arrays I am working with:

val inpoly: Array[Boolean]=Array(false, true, true, false)
val idx1: Array[Boolean]=Array(true, false, false, false)

我想检查数组idx1的位置，并在该位置上将inpoly设置为true，否则，只需保留inpoly已经具有的值即可.

I would like to check the array idx1 and where it is true I would like to set inpoly to true in that specific position, otherwise, just leave the value that inpoly already has.

我的预期结果是拥有这个数组:

My expected result would be to have this array:

final_vector= true, true, true, false

由于idx1的第一个值为true，所以将true分配给inpoly. idx1的所有其他值均为false，因此请保持inpoly不变

since the first value of idx1 is true, assign true to inpoly. All the other values of idx1 are false, so leave inpoly as it is

我尝试了以下代码:

idx1.map({
  case true => true
  case false => inpoly})

但是我得到以下结果:

res308: Array[Any] = Array(true, Array(false, true, true, false), Array(false, true, true, false), Array(false, true, true, false))

任何人都可以帮忙吗?

推荐答案

您似乎想在布尔对上使用或".这是使用||的一种方法，它是Scala的OR函数:

It looks like you'd like to use an "OR" on the pairs of booleans. Here's one approach using || which is Scala's OR function:

val updated = inpoly.zip(idx1).map(pair => pair._1 || pair._2)

这将为您提供

updated: Array[Boolean] = Array(true, true, true, false)

在这里，zip会将两个数组组合成一个成对的数组.然后，您将映射这些对，如果其中一个为true，则返回true；如果两个均为false，则返回false.

Here, zip will combine the two arrays into one array of pairs. Then you will map through those pairs and return a true if either are true and a false if both are false.

您还可以使用模式匹配:

You could also use a pattern match:

val updated = inpoly.zip(idx1).map {
 case (left, right) => left || right
 case _ => None 
}

如果一个数组的长度与另一个数组的长度不同，您将要考虑要做什么.

You'll want to think about what you will do in the case that one Array has a different length than the other.

并且，如果您坚持修改inpoly而不是创建新的数组，请执行以下操作:

And, if you insist on modifying inpoly rather than creating a new array, do the following:

var inpoly: Array[Boolean]=Array(false, true, true, false)
val idx1: Array[Boolean]=Array(true, false, false, false)

inpoly = inpoly.zip(idx1).map(pair => pair._1 || pair._2)

请注意，在这里，我们将inpoly声明为var，而不是val.但是，我建议将inpoly保留为val，并为该数组的更新版本创建一个新的val.在Scala中，建议使用不可变值，尽管这可能需要一些时间来习惯.

Note that here, we declare inpoly to be a var, not a val. However, I would instead suggest keeping inpoly as a val and creating a new val for the updated version of that array. Using immutable values is a recommended practice in Scala, though it can take some getting used to.

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