Question

我有一个MutableList，我想从中删除一个元素，但我找不到合适的方法。有一种方法可以从ListBuffer中删除元素：

val x = ListBuffer(1, 2, 3, 4, 5, 6, 7, 8, 9)
x -= 5

我无法在MutableList上找到等效的方法。

Answer 1

MutableList缺少-=和--=，因为它没有延伸Shrinkable特征。可以找到各种动机here。

MutableList确实有diff，filter和其他方法可以帮助您，以防您重新分配变量（或实例化新变量）可能是选项和性能问题并不是最重要的：

var mylist = MutableList(1, 2, 3)
mylist = mylist diff Seq(1)
val myNewList = mylist.filter(_ != 2)
val indexFiltered = mylist.zipWithIndex.collect { case (el, ind) if ind != 1 => el }

您经常可以使用ListBuffer代替MutableList，这将解锁所需的-=和--=方法：

val mylist = ListBuffer(1, 2, 3)
mylist -= 1 //mylist is now ListBuffer(2, 3)
mylist --= Seq(2, 3) //mylist is now empty

Answer 2

这不是答案，只是为了警告你有关问题（至少在2.11.x中）：

//street magic
scala> val a = mutable.MutableList(1,2,3)
a: scala.collection.mutable.MutableList[Int] = MutableList(1, 2, 3)
scala> a += 4
res7: a.type = MutableList(1, 2, 3, 4)
scala> a
res8: scala.collection.mutable.MutableList[Int] = MutableList(1, 2, 3, 4)
scala> a ++= List(8,9,10)
res9: a.type = MutableList(1, 2, 3, 4, 8, 9, 10)
scala> val b = a.tail
b: scala.collection.mutable.MutableList[Int] = MutableList(2, 3, 4, 8, 9, 10)
scala> b.length
res10: Int = 6
scala> a.length
res11: Int = 7
scala> a ++= List(8,9,10)
res12: a.type = MutableList(1, 2, 3, 4, 8, 9, 10, 8, 9, 10)
scala> b += 7
res13: b.type = MutableList(2, 3, 4, 8, 9, 10, 7)
scala> a
res14: scala.collection.mutable.MutableList[Int] = MutableList(1, 2, 3, 4, 8, 9, 10, 7) 
scala> b
res15: scala.collection.mutable.MutableList[Int] = MutableList(2, 3, 4, 8, 9, 10, 7)
scala> a ++= List(8,9,10)
res16: a.type = MutableList(1, 2, 3, 4, 8, 9, 10, 7)

这个例子来自一些要点 - 我已经用#devid_blein #street_magic标签在facebook上发布了它，但在互联网上找不到原始链接。

从Scala中的MutableList中删除元素

2 个答案: