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ios swift无法使用类型为的参数列表调用映射_Swift_Dictionary - Fatal编程技术网

ios swift无法使用类型为的参数列表调用映射

ios swift无法使用类型为的参数列表调用映射,swift,dictionary,Swift,Dictionary,我是xCode(Swift)开发的新手,我正在尝试创建以下结构: struct artistSection : Comparable { var artist : String var vinyls : [Vinyl] var collapsed : Bool static func group(vinyls : [Vinyl]) -> [artistSection] { let groups = Dictionary(grouping : vinyls) {

我是xCode(Swift)开发的新手,我正在尝试创建以下结构:

struct artistSection : Comparable {
  var artist : String
  var vinyls : [Vinyl]
  var collapsed : Bool

  static func group(vinyls : [Vinyl]) -> [artistSection] {
     let groups = Dictionary(grouping : vinyls) { (vinyl) -> String in
       return (vinyl.artist!)
     }
     return groups.map(artistSection.init(artist: vinyls: collapsed:)).sorted
  }
}
return.map上出现以下错误:

无法使用类型为“((字符串,[乙烯基],Bool)->VinylTableViewController.artistSection)的参数列表调用“map”


您需要向
map
中的
ArtistSection
的初始值设定项提供输入参数

struct ArtistSection:Comparable {
    var artist : String
    var vinyls : [Vinyl]
    var collapsed : Bool

    static func group(vinyls : [Vinyl]) -> [ArtistSection] {
        let groups = Dictionary(grouping : vinyls) { (vinyl) -> String in
            return (vinyl.artist!)
        }
        return groups.map({ArtistSection(artist: $0.key, vinyls: $0.value, collapsed: false)}).sorted()
    }
}

与错误无关,但还要确保您遵守Swift命名约定,该约定是类型的大写(
ArtistSection
,而不是
ArtistSection
)。

非常感谢。成功了。感谢您对命名约定的建议。