scala - 如何通过右侧的分隔符拆分字符串？

Question

如何从右侧通过分隔符拆分字符串？

例如

scala> "hello there how are you?".rightSplit(" ", 1)
res0: Array[java.lang.String] = Array(hello there how are, you?)

Python 有一个.rsplit()方法，这是我在 Scala 中所追求的：

In [1]: "hello there how are you?".rsplit(" ", 1)
Out[1]: ['hello there how are', 'you?']

Answer 1

我认为最简单的解决方案是搜索索引位置，然后根据它进行拆分。例如：

scala> val msg = "hello there how are you?"
msg: String = hello there how are you?

scala> msg splitAt (msg lastIndexOf ' ')
res1: (String, String) = (hello there how are," you?")

并且由于有人评论lastIndexOf返回-1，因此解决方案非常好：

scala> val msg = "AstringWithoutSpaces"
msg: String = AstringWithoutSpaces

scala> msg splitAt (msg lastIndexOf ' ')
res0: (String, String) = ("",AstringWithoutSpaces)

Answer 2

您可以使用普通的旧正则表达式：

scala> val LastSpace = " (?=[^ ]+$)"
LastSpace: String = " (?=[^ ]+$)"

scala> "hello there how are you?".split(LastSpace)
res0: Array[String] = Array(hello there how are, you?)

(?=[^ ]+$)表示我们将提前 ( ) 查找一组至少有 1 个字符长度?=的非空格 ( ) 字符。[^ ]最后，这个空格后面跟着这样的序列必须在字符串的末尾：$.

如果只有一个令牌，此解决方案不会中断：

scala> "hello".split(LastSpace)
res1: Array[String] = Array(hello)

Answer 3

scala> val sl = "hello there how are you?".split(" ").reverse.toList
sl: List[String] = List(you?, are, how, there, hello)

scala> val sr = (sl.head :: (sl.tail.reverse.mkString(" ") :: Nil)).reverse
sr: List[String] = List(hello there how are, you?)