scala - 在元组中拆分字符串

Question

在一个函数中 def a(l: List[(Int, String)]): List[(Int, String)] = ??? 我想把一个String拆分成他们的单词小写。逗号等应该被忽略，所以我想我需要 replaceAll("[^A-Za-z]+", "").toLowerCase() 某处？ Int 值应与句子中的值保持一致。

它应该如何工作的示例:

val example = List((11, "That is great!"), (12, "Wow, impossible!"))
print(a(example)) 

结果

List((11, "that"),(11, "is"),(11, "great"),(12, "wow"),(12, "impossible"))

Answer 1

您可以为此使用 flatMap:

val example = List((11, "That is great!"), (12, "Wow, impossible!"))
example.flatMap { case (int, str) =>
  str
    .replaceAll("[^A-Za-z]+", " ")
    .toLowerCase()
    .split(' ')
    .map((int, _))
}

产量:

res0: List[(Int, String)] = List((11,that), (11,is), (11,great), (12,wow), (12,impossible))