arrays - 我如何在功能上将对象数组转换为不同的结构
问题描述
我有一个字体数组,每个字体都有一个familyName和一个fontName。
我想将它们转换为表单中的元组数组(familyName:String,fontNames:[String])。
我觉得应该有一种简单的功能性方法来做到这一点,但无法解决。我最接近的是两次调用reduce
:首先是字典,然后是数组。
let dictionary = fonts.reduce(into [String : [String]]() ) { result, font in
let array = result[font.fontFamily] ?? []
result[fontFamily] = array + [font.fontName]
}
let array = dictionary(into: [(String, [String])]() ) { result, element in
result.append( (element.key, element.value.sorted()) )
}.sorted { $0.0 < $1.0 }
我还在对元组数组和元组数组中的 fontNames 数组进行排序。
有没有办法可以避免中间字典?
非常感谢。
更新 我创建了一个游乐场来向 sanjaykmwt 展示他们的建议结果:
struct Font {
let family: String
let name: String
}
let fonts = [
Font(family: "ABC", name: "abc"),
Font(family: "ABC", name: "def"),
Font(family: "ABC", name: "ghi"),
Font(family: "XYZ", name: "xyz"),
Font(family: "XYZ", name: "uvw")
]
let sortedFamily = fonts.sorted(by: { (lhs, rhs) -> Bool in
return lhs.family < rhs.family
})
let dict = sortedFamily.map({["family":$0.family,
"fonts":$0.name]})
print("dict: \(dict)")
输出:
dict: [["family": "ABC", "fonts": "abc"], ["family": "ABC", "fonts": "def"], ["family": "ABC", "fonts": "ghi"], ["family": "XYZ", "fonts": "xyz"], ["family": "XYZ", "fonts": "uvw"]]
解决方案
Fonts
如果你有一个with 数组fontFamily
, fontName
grouping
那你可以做map
// Array Of Fonts Example
let array = [Font.init(fontFamily: "Cago", fontName: "AA"),
Font.init(fontFamily: "Cago", fontName: "CCCC"),
Font.init(fontFamily: "Mango", fontName: "AAsss"),
Font.init(fontFamily: "Mango", fontName: "mngoo")]
// Grouping
let groupedByFamilayName = Dictionary.init(grouping: array) {$0.fontFamily}
// Map
let arrayOfTuple = groupedByFamilayName.map { (key,array) -> (String,[String]) in
return (key,array.map({$0.fontName}))
}
print(arrayOfTuple)