在Swift中有没有对应的Scala、Xtend、Groovy、Ruby等等?
var aofa = [[1,2,3],[4],[5,6,7,8,9]]
aofa.flatten() // shall deliver [1,2,3,4,5,6,7,8,9]
当然我可以用reduce来做，但那有点糟糕
var flattened = aofa.reduce(Int[]()){
a,i in var b : Int[] = a
b.extend(i)
return b
}
当前回答
struct Group {
var members: [String]?
}
let groups = [Group]()
let outputMembers: [String] = Array(groups.compactMap({ $0.members }).joined())
描述
如果要制作单数组的数组对象模型。示例:我们从所有组中获取outputMembers单个数组。
其他回答
Apple Swift 5.1.2版本(swiflang -1100.0.278) 目标:x86_64-apple-darwin19.2.0
let optionalNumbers = [[1, 2, 3, nil], nil, [4], [5, 6, 7, 8, 9]]
print(optionalNumbers.compactMap { $0 }) // [[Optional(1), Optional(2), Optional(3), nil], [Optional(4)], [Optional(5), Optional(6), Optional(7), Optional(8), Optional(9)]]
print(optionalNumbers.compactMap { $0 }.reduce([], +).map { $0 as? Int ?? nil }.compactMap{ $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(optionalNumbers.compactMap { $0 }.flatMap { $0 }.map { $0 as? Int ?? nil }.compactMap{ $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(Array(optionalNumbers.compactMap { $0 }.joined()).map { $0 as? Int ?? nil }.compactMap{ $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
let nonOptionalNumbers = [[1, 2, 3], [4], [5, 6, 7, 8, 9]]
print(nonOptionalNumbers.compactMap { $0 }) // [[1, 2, 3], [4], [5, 6, 7, 8, 9]]
print(nonOptionalNumbers.reduce([], +)) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(nonOptionalNumbers.flatMap { $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(Array(nonOptionalNumbers.joined())) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
flatten()在Swift 3中根据SE-0133重命名为joined():
https://github.com/apple/swift-evolution/blob/master/proposals/0133-rename-flatten-to-joined.md
矩阵是[[myDTO]]?
在swift 5中，你可以使用this = Array(self.matrix!.joined())
struct Group {
var members: [String]?
}
let groups = [Group]()
let outputMembers: [String] = Array(groups.compactMap({ $0.members }).joined())
描述
如果要制作单数组的数组对象模型。示例:我们从所有组中获取outputMembers单个数组。
func convert(){
let arr = [[1,2,3],[4],[5,6,7,8,9]]
print("Old Arr = ",arr)
var newArr = [Int]()
for i in arr{
for j in i{
newArr.append(j)
}
}
print("New Arr = ",newArr)
}
