在Scala中打印数组

Question

我对最基本的Scala操作有问题,这让我发疯.

val a = Array(1,2,3)

println(a)   and result is [I@1e76345

println(a.toString()) and result is [I@1e76345

println(a.toString) and result is [I@1e76345

任何人都可以告诉我如何打印数组而不编写我自己的功能,因为那是愚蠢的.谢谢!

Answer 1

mkString将集合(包括Array)逐个元素转换为字符串表示.

println(a.mkString(" "))

可能就是你想要的.

Answer 2

你可以做正常的事情(参见Rex或Jiri的回答),或者你可以:

scala> Array("bob","sue")
res0: Array[String] = Array(bob, sue)

嘿,不公平!REPL打印出来真的很棒.

scala> res0.toString
res1: String = [Ljava.lang.String;@63c58252

没有快乐,直到:

scala> runtime.ScalaRunTime.stringOf(res0)
res2: String = Array(bob, sue)

scala> runtime.ScalaRunTime.replStringOf(res0, res0.length)
res3: String = 
"Array(bob, sue)
"

scala> runtime.ScalaRunTime.replStringOf(res0, 1)
res4: String = 
"Array(bob)
"

我想知道REPL中是否有宽度设置.更新:没有.它固定在

val maxStringElements = 1000  // no need to mkString billions of elements

但我不会尝试数十亿:

scala> Array.tabulate(100)(identity)
res5: Array[Int] = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99)

scala> import runtime.ScalaRunTime.replStringOf
import runtime.ScalaRunTime.replStringOf

scala> replStringOf(res5, 10)
res6: String = 
"Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
"

scala> res5.take(10).mkString(", ")
res7: String = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

等等,让我们说:

scala> res5.take(10).mkString("Array(", ", ", ")")
res8: String = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

这可能是显而易见的:

scala> var vs = List("1")
vs: List[String] = List(1)

scala> vs = null
vs: List[String] = null

scala> vs.mkString
java.lang.NullPointerException

所以与其:

scala> import runtime.ScalaRunTime.stringOf
import runtime.ScalaRunTime.stringOf

scala> stringOf(vs)
res16: String = null

此外,数组不需要很深,以便从其stringPrefix中受益:

scala> println(res0.deep.toString)
Array(bob, sue)

无论您喜欢哪种方法,都可以将其包装起来:

implicit class MkLines(val t: TraversableOnce[_]) extends AnyVal { 
  def mkLines: String = t.mkString("", EOL, EOL)
  def mkLines(header: String, indented: Boolean = false, embraced: Boolean = false): String = { 
    val space = "\u0020"
    val sep = if (indented) EOL + space * 2 else EOL
    val (lbrace, rbrace) = if (embraced) (space + "{", EOL + "}") else ("", "")
    t.mkString(header + lbrace + sep, sep, rbrace + EOL)
  } 
} 

但是数组需要特殊的转换,因为你没有得到ArrayOps:

implicit class MkArrayLines(val a: Array[_]) extends AnyVal {
  def asTO: TraversableOnce[_] = a
  def mkLines: String = asTO.mkLines
  def mkLines(header: String = "Array", indented: Boolean = false, embraced: Boolean = false): String =
    asTO.mkLines(header, indented, embraced)
}

scala> Console println Array("bob","sue","zeke").mkLines(indented = true)
Array
  bob
  sue
  zeke

Answer 3

这有两种方法.

一个是使用foreach:

val a = Array(1,2,3)
a.foreach(println)

另一种是使用mkString:

val a = Array(1,2,3)
println(a.mkString(""))