可以直接Map字符串：

def removeZero(s: String) = s.map(c => if(c == '0') ' ' else c)

或者，您可以使用replace：

s.replace('0', ' ')

很简单：

scala> "FooN00b".filterNot(_ == '0')
res0: String = FooNb

要将某些字符替换为其他字符：

scala> "FooN00b" map { case '0' => 'o'  case 'N' => 'D'  case c => c }
res1: String = FooDoob

用任意数量的字符替换一个字符：

scala> "FooN00b" flatMap { case '0' => "oOo"  case 'N' => ""  case c => s"$c" }
res2: String = FoooOooOob

根据伊格纳西奥Alorre的说法，如果你想把其他字符替换成字符串：

def replaceChars (str: String) = str.map(c => if (c == '0') ' ' else if (c =='T') '_' else if (c=='-') '_' else if(c=='Z') '.' else c)

val charReplaced = replaceChars("N000TZ")

执行此操作的首选方法是使用s.replace('0', ' ')
出于训练的目的，您也可以使用尾递归来实现这一点。

def replace(s: String): String = {
  def go(chars: List[Char], acc: List[Char]): List[Char] = chars match {
    case Nil => acc.reverse
    case '0' :: xs => go(xs, ' ' :: acc)
    case x :: xs => go(xs, x :: acc)
  }

  go(s.toCharArray.toList, Nil).mkString
}

replace("01230123") // " 123 123"

并且更一般地：

def replace(s: String, find: Char, replacement: Char): String = {
  def go(chars: List[Char], acc: List[Char]): List[Char] = chars match {
    case Nil => acc.reverse
    case `find` :: xs => go(xs, replacement :: acc)
    case x :: xs => go(xs, x :: acc)
  }

  go(s.toCharArray.toList, Nil).mkString
}

replace("01230123", '0', ' ') // " 123 123"

/**
 * how to replace a empty character in string
 */
val name = "Jeeta"
val afterReplace = name.replace("" + 'a', "")