scala 将 Seq 转换为 ArrayBuffer
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Convert Seq to ArrayBuffer
提问by classicalist
Is there any concise way to convert a Seqinto ArrayBufferin Scala?
有没有什么简洁的方法可以在 Scala 中将a 转换Seq为ArrayBuffer？
回答by Eastsun
scala> val seq = 1::2::3::Nil
seq: List[Int] = List(1, 2, 3)
scala> seq.toBuffer
res2: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1, 2, 3)
EDITAfter Scala 2.1x, there is a method .to[Coll]defined in TraversableLike, which can be used as follow:
编辑Scala 2.1x 之后，.to[Coll]在TraversableLike 中定义了一个方法，可以如下使用：
scala> import collection.mutable
import collection.mutable
scala> seq.to[mutable.ArrayBuffer]
res1: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3)
scala> seq.to[mutable.Set]
res2: scala.collection.mutable.Set[Int] = Set(1, 2, 3)
回答by Philippe
This will work:
这将起作用：
ArrayBuffer(mySeq : _*)
Some explanations: this uses the apply method in the ArrayBuffer companion object. The signature of that method is
一些解释：这使用了ArrayBuffer 伴随对象中的 apply 方法。该方法的签名是
def apply [A] (elems: A*): ArrayBuffer[A]
meaning that it takes a variable number of arguments. For instance:
这意味着它需要可变数量的参数。例如：
ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8)
is also a valid call. The ascription : _* indicates to the compiler that a Seq should be used in place of that variable number of arguments (see Section 4.6.2 in the Scala Reference).
也是一个有效的调用。说明 : _* 向编译器指示应该使用 Seq 代替可变数量的参数（参见Scala 参考中的第 4.6.2 节）。
