我正在尝试创建一个转换String中字符的方法,特别是将所有'0'转换为''。这是我正在使用的代码:
def removeZeros(s: String) = {
val charArray = s.toCharArray
charArray.map( c => if(c == '0') ' ')
new String(charArray)
}
有更简单的方法吗?此语法无效:
def removeZeros(s: String) =
new String(s.toCharArray.map( c => if(c == '0') ' '))
答案 0 :(得分:37)
您可以直接映射字符串:
def removeZero(s: String) = s.map(c => if(c == '0') ' ' else c)
或者您可以使用replace
:
s.replace('0', ' ')
答案 1 :(得分:24)
很简单:
scala> "FooN00b".filterNot(_ == '0')
res0: String = FooNb
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
用一些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
答案 2 :(得分:1)
/**
how to replace a empty chachter in string
**/
val name="Jeeta"
val afterReplace=name.replace(""+'a',"")
答案 3 :(得分:1)
执行此操作的首选方法是使用s.replace('0', ' ')
出于训练目的,您也可以使用尾递归来实现。
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"
更普遍的是:
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"
答案 4 :(得分:0)
根据Ignacio Alorre的说法,如果您要将其他字符替换为字符串:
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")