scala - 为什么这个 foldLeft scala 代码不起作用？

Question

以下代码片段旨在为我提供Map[String, (String, Int)].

def genList(xx: String) = {
    Seq("one", "two", "three", "four")
}

val oriwords = Set("hello", "how", "are", "you")
val newMap = (Map[String, (String, Int)]() /: oriwords) (
    (cmap, currentWord) => {
        val xv = 2

        genList(currentWord).map(ps => {
            val src = cmap get ps

            if(src == None) {
                cmap + (ps -> (w, xv))
            }
            else {
                if(src.get._2 < xv) {
                    cmap + (ps -> (w, xv))
                }
                else cmap
            }

        })
    }
)

但我收到以下错误：

error: too many arguments for method ->: (y: B)(String, B)
                       cmap + (ps -> (w, xv))
                                  ^

更新：通过答案中提到的建议更改，上述错误被删除。

val newMap = (Map[String, (String, Int)]() /: oriwords) (
    (cmap, currentWord) => {
        val xv = 2

        genList(currentWord).map(ps => {
            val src = cmap get ps

            if(src == None) {
                cmap + (ps -> ((currentWord, xv)))
            }
            else {
                if(src.get._2 < xv) {
                    cmap + (ps -> ((currentWord, xv)))
                }
                else cmap
            }

        })
    }
)

但是现在上面的代码出现了一个新的错误，如下所示：

error: type mismatch;
 found   : Seq[scala.collection.immutable.Map[String,(String, Int)]]
 required: scala.collection.immutable.Map[String,(String, Int)]
               genList(currentWord).map(ps => {
                                       ^

标签： scala

Answer 1

ps -> (w, xv)

被解释为

ps.->(w, xv)

即作为传递两个参数而不是你想要的，这是传递一个 2 元组作为单个参数：

ps.->((w, xv))

或在运算符语法中：

ps -> ((w, xv))