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
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]
其他回答
在Swift标准库中，为所有符合Sequence协议的类型实现了join函数(或在Swift 3之前在SequenceType上flatten)，其中包括Array:
let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let flattened = Array(numbers.joined())
在某些情况下，使用joined()可能是有益的，因为它返回一个惰性集合而不是一个新数组，但当传递给array()初始化器时，总是可以转换为一个数组，就像上面的例子一样。
快4.倍
flatMap的这种用法并没有被弃用，它就是为此而生的。 https://developer.apple.com/documentation/swift/sequence/2905332-flatmap
var aofa = [[1,2,3],[4],[5,6,7,8,9]]
aofa.flatMap { $0 } //[1,2,3,4,5,6,7,8,9]
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]
Swift >= 3.0
减少:
let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let reduced = numbers.reduce([], +)
flatMap:
let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let flattened = numbers.flatMap { $0 }
加入:
let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let joined = Array(numbers.joined())
你可以用下面的方法来平嵌套数组:
var arrays = [1, 2, 3, 4, 5, [12, 22, 32], [[1, 2, 3], 1, 3, 4, [[[777, 888, 8999]]]]] as [Any]
func flatten(_ array: [Any]) -> [Any] {
return array.reduce([Any]()) { result, current in
switch current {
case(let arrayOfAny as [Any]):
return result + flatten(arrayOfAny)
default:
return result + [current]
}
}
}
let result = flatten(arrays)
print(result)
/// [1, 2, 3, 4, 5, 12, 22, 32, 1, 2, 3, 1, 3, 4, 777, 888, 8999]
