同时迭代两个数组

Question

我是Swift的新手.我一直在做Java编程.我有一个在Swift中编码的场景.

以下代码使用Java.我需要在Swift中为以下场景编写代码

// With String array - strArr1
String strArr1[] = {"Some1","Some2"}

String strArr2[] = {"Somethingelse1","Somethingelse2"}

for( int i=0;i< strArr1.length;i++){
    System.out.println(strArr1[i] + " - "+ strArr2[i]);
}

我在swift中有几个数组

var strArr1: [String] = ["Some1","Some2"]
var strArr2: [String] = ["Somethingelse1","Somethingelse2"]

for data in strArr1{
    println(data)
}

for data in strArr2{
    println(data)
}
// I need to loop over in single for loop based on index.

您能否根据索引提供有关循环语法的帮助

Answer 1

您可以使用zip(),从两个给定序列创建一对配对:

let strArr1 = ["Some1", "Some2"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]

for (e1, e2) in zip(strArr1, strArr2) {
    print("\(e1) - \(e2)")
}

该序列仅列举给定序列/数组的"共同元素".如果它们具有不同的长度,则简单地忽略较长阵列/序列的附加元素.

Answer 2

试试这个:

zip([0,2,4,6], [1,3,5,7]).forEach {
    print($0,$1)
}

zip([0,2,4,6], [1,3,5,7]).forEach {
    print($0.0,$0.1)
}

Answer 3

您还可以enumerate使用一个数组并使用索引查看第二个数组:

Swift 1.2:

for (index, element) in enumerate(strArr1) {
    println(element)
    println(strArr2[index])
}

斯威夫特2:

for (index, element) in strArr1.enumerate() {
    print(element)
    print(strArr2[index])
}

斯威夫特3:

for (index, element) in strArr1.enumerated() {
    print(element)
    print(strArr2[index])
}

Answer 4

使用Swift 5，您可以使用以下4种Playground代码之一来解决您的问题。

＃1 使用zip(_:_:)功能

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)")
}

/*
 prints:
 Some1 - Somethingelse1
 Some2 - Somethingelse2
 */

＃2。使用Array的makeIterator()方法和while循环

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)")
}

/*
 prints:
 Some1 - Somethingelse1
 Some2 - Somethingelse2
 */

＃3。使用符合的自定义类型IteratorProtocol

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)")
}

/*
 prints:
 Some1 - Somethingelse1
 Some2 - Somethingelse2
 */

＃4。使用AnyIterator

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)")
}

/*
 prints:
 Some1 - Somethingelse1
 Some2 - Somethingelse2
 */

Answer 5

for(var i = 0; i < strArr1.count ; i++)
{
    println(strArr1[i] + strArr2[i])
}

应该可以做到这一点。以前从未使用过 swift，因此请务必进行测试。

更新到最新的 Swift 语法

for i in 0..< strArr1.count {
    print(strArr1[i] + strArr2[i])
}