使用Swift 5,您可以使用以下4个Playground代码之一来解决您的问题。
在最简单的情况下,您可以zip(_:_:)
用来创建初始数组中元素对的新序列(元组)。
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
let sequence = zip(strArr1, strArr2)
for (el1, el2) in sequence {
print("\(el1) - \(el2)")
}
使用简单的while循环和迭代器,同时循环两个数组也很容易:
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
var iter1 = strArr1.makeIterator()
var iter2 = strArr2.makeIterator()
while let el1 = iter1.next(), let el2 = iter2.next() {
print("\(el1) - \(el2)")
}
在某些情况下,您可能需要创建自己的类型,以将姓名缩写数组的元素配对。可以通过使您的类型符合来实现IteratorProtocol
。请注意,通过使您的类型也符合Sequence
协议,您可以在for循环中直接使用它的实例:
struct TupleIterator: Sequence, IteratorProtocol {
private var firstIterator: IndexingIterator<[String]>
private var secondIterator: IndexingIterator<[String]>
init(firstArray: [String], secondArray: [String]) {
self.firstIterator = firstArray.makeIterator()
self.secondIterator = secondArray.makeIterator()
}
mutating func next() -> (String, String)? {
guard let el1 = firstIterator.next(), let el2 = secondIterator.next() else { return nil }
return (el1, el2)
}
}
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
let tupleSequence = TupleIterator(firstArray: strArr1, secondArray: strArr2)
for (el1, el2) in tupleSequence {
print("\(el1) - \(el2)")
}
作为上一个示例的替代方法,您可以使用AnyIterator
。以下代码显示了在Array
扩展方法中可能的实现:
extension Array {
func pairWithElements(of array: Array) -> AnyIterator<(Element, Element)> {
var iter1 = self.makeIterator()
var iter2 = array.makeIterator()
return AnyIterator({
guard let el1 = iter1.next(), let el2 = iter2.next() else { return nil }
return (el1, el2)
})
}
}
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
let iterator = strArr1.pairWithElements(of: strArr2)
for (el1, el2) in iterator {
print("\(el1) - \(el2)")
}