blocks|key|1951257|text|是。从SWIFT3.0开始，如果您需要每个元素的索引及其值，可以使用方法对数组进行迭代。它返回一个由索引和数组中每个项的值组成的对序列。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1951258|for+(index,+element)+in+list.enumerated()+{
++print("Item+\(index):+\(element)")
}|code-block|syntax|javascript|1951259|在SWIFT3.0之前和SWIFT2.0之后，这个函数被称为enumerate()|style|CODE|1951260|for+(index,+element)+in+list.enumerate()+{
++++print("Item+\(index):+\(element)")
}|1951261|在SWIFT2.0之前，enumerate是一个全局功能。|1951262|for+(index,+element)+in+enumerate(list)+{
++++println("Item+\(index):+\(element)")
}|1951263|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/ios/documentation/swift/conceptual/Swift_Programming_Language/CollectionTypes.html#//apple_ref/doc/uid/TP40014097-CH8-XID_177^0|Y|2|0|0|0|U|B|0|0|C|9|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|15|8|@$A|16|B|17|K|L]]|9|@]|C|$]]|$1|M|3|N|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|19|8|@$A|1A|B|1B|K|L]]|9|@]|C|$]]|$1|Q|3|R|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|S|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

Yes. As of Swift 3.0, if you need the index for each element along with its value, you can use the <a href="https://developer.apple.com/library/ios/documentation/swift/conceptual/Swift_Programming_Language/CollectionTypes.html#//apple_ref/doc/uid/TP40014097-CH8-XID_177" rel="noreferrer"><code>enumerated()</code> method</a> to iterate over the array. It returns a sequence of pairs composed of the index and the value for each item in the array. For example:

<pre><code>for (index, element) in list.enumerated() {
 print("Item \(index): \(element)")
}
</code></pre>

Before Swift 3.0 and after Swift 2.0, the function was called <code>enumerate()</code>:

<pre><code>for (index, element) in list.enumerate() {
 print("Item \(index): \(element)")
}
</code></pre>

Prior to Swift 2.0, <code>enumerate</code> was a global function.

<pre><code>for (index, element) in enumerate(list) {
 println("Item \(index): \(element)")
}
</code></pre>

blocks|key|345142|text|我在寻找使用字典的方法时找到了这个答案，结果发现它很容易适应，只需为元素传递一个元组即可。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|345143|//+Swift+2

var+list+=+["a":+1,+"b":+2]

for+(index,+(letter,+value))+in+list.enumerate()+{
++++print("Item+\(index):+\(letter)+\(value)")
}|code-block|syntax|javascript|345144|entityMap^0|6|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I found this answer while looking for a way to do that with a Dictionary, and it turns out it's quite easy to adapt it, just pass a tuple for the element.

<pre><code>// Swift 2

var list = ["a": 1, "b": 2]

for (index, (letter, value)) in list.enumerate() {
 print("Item \(index): \(letter) \(value)")
}
</code></pre>

blocks|key|7888|text|从Swift+2开始，需要对集合调用枚举函数，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7889|for+(index,+element)+in+list.enumerate()+{
++++print("Item+\(index):+\(element)")
}|code-block|syntax|javascript|7890|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Starting with Swift 2, the enumerate function needs to be called on the collection like so:

<pre><code>for (index, element) in list.enumerate() {
 print("Item \(index): \(element)")
}
</code></pre>

blocks|key|7904|text|Swift+5提供了一个名为enumerated()+for+Array的方法。enumerated()有以下声明：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|7905|func+enumerated()+->+EnumeratedSequence<Array<Element>>|code-block|syntax|javascript|7906|返回对序列(n，x)，其中n表示从零开始的连续整数，x表示序列的一个元素。|blockquote|7907|在最简单的情况下，您可以使用enumerated()和for循环。例如：|7908|let+list+=+["Car",+"Bike",+"Plane",+"Boat"]
for+(index,+element)+in+list.enumerated()+{
++++print(index,+":",+element)
}

/*
prints:
0+:+Car
1+:+Bike
2+:+Plane
3+:+Boat
*/|7909|但是，请注意，您并不局限于在for循环中使用enumerated()。实际上，如果您计划将enumerated()与类似于以下代码的for循环一起使用，那么您就错了：|7910|let+list+=+[Int](1...5)
var+arrayOfTuples+=+[(Int,+Int)]()

for+(index,+element)+in+list.enumerated()+{
++++arrayOfTuples+%2B=+[(index,+element)]
}

print(arrayOfTuples)+//+prints+[(0,+1),+(1,+2),+(2,+3),+(3,+4),+(4,+5)]|7911|一个更快捷的方法是：|7912|let+list+=+[Int](1...5)
let+arrayOfTuples+=+Array(list.enumerated())
print(arrayOfTuples)+//+prints+[(offset:+0,+element:+1),+(offset:+1,+element:+2),+(offset:+2,+element:+3),+(offset:+3,+element:+4),+(offset:+4,+element:+5)]|7913|作为另一种选择，您也可以将enumerated()与map结合使用。|7914|let+list+=+[Int](1...5)
let+arrayOfDictionaries+=+list.enumerated().map+{+(a,+b)+in+return+[a+:+b]+}
print(arrayOfDictionaries)+//+prints+[[0:+1],+[1:+2],+[2:+3],+[3:+4],+[4:+5]]|7915|此外，尽管有一些局限性，但forEach可以很好地替代for循环：|7916|let+list+=+[Int](1...5)
list.reversed().enumerated().forEach+{+print($0,+":",+$1)+}

/*
prints:
0+:+5
1+:+4
2+:+3
3+:+2
4+:+1
*/|7917|通过使用enumerated()和makeIterator()，您甚至可以在Array上手动迭代。例如：|7918|import+UIKit
import+PlaygroundSupport

class+ViewController:+UIViewController+{

++++var+generator+=+["Car",+"Bike",+"Plane",+"Boat"].enumerated().makeIterator()

++++override+func+viewDidLoad()+{
++++++++super.viewDidLoad()

++++++++let+button+=+UIButton(type:+.system)
++++++++button.setTitle("Tap",+for:+.normal)
++++++++button.frame+=+CGRect(x:+100,+y:+100,+width:+100,+height:+100)
++++++++button.addTarget(self,+action:+#selector(iterate(_:)),+for:+.touchUpInside)
++++++++view.addSubview(button)
++++}

++++@objc+func+iterate(_+sender:+UIButton)+{
++++++++let+tuple+=+generator.next()
++++++++print(String(describing:+tuple))
++++}

}

PlaygroundPage.current.liveView+=+ViewController()

/*
+Optional((offset:+0,+element:+"Car"))
+Optional((offset:+1,+element:+"Bike"))
+Optional((offset:+2,+element:+"Plane"))
+Optional((offset:+3,+element:+"Boat"))
+nil
+nil
+nil
+*/|7919|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/swift/array/1687832-enumerated|1|https://developer.apple.com/documentation/swift/sequence/3018369-foreach^0|E|C|V|5|14|C|E|C|0|0|0|0|E|C|0|0|M|C|19|C|0|0|0|0|D|C|Q|3|0|0|D|7|8|3|1|0|0|4|C|H|E|12|5|0|0^^$0|@$1|2|3|4|5|6|7|1L|8|@$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|C]]|D|@$9|1S|A|1T|1|1U]]|E|$]]|$1|F|3|G|5|H|7|1V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|M|7|1W|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1X|8|@$9|1Y|A|1Z|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|H|7|20|8|@]|D|@]|E|$I|J]]|$1|R|3|S|5|6|7|21|8|@$9|22|A|23|B|C]|$9|24|A|25|B|C]]|D|@]|E|$]]|$1|T|3|U|5|H|7|26|8|@]|D|@]|E|$I|J]]|$1|V|3|W|5|6|7|27|8|@]|D|@]|E|$]]|$1|X|3|Y|5|H|7|28|8|@]|D|@]|E|$I|J]]|$1|Z|3|10|5|6|7|29|8|@$9|2A|A|2B|B|C]|$9|2C|A|2D|B|C]]|D|@]|E|$]]|$1|11|3|12|5|H|7|2E|8|@]|D|@]|E|$I|J]]|$1|13|3|14|5|6|7|2F|8|@$9|2G|A|2H|B|C]]|D|@$9|2I|A|2J|1|2K]]|E|$]]|$1|15|3|16|5|H|7|2L|8|@]|D|@]|E|$I|J]]|$1|17|3|18|5|6|7|2M|8|@$9|2N|A|2O|B|C]|$9|2P|A|2Q|B|C]|$9|2R|A|2S|B|C]]|D|@]|E|$]]|$1|19|3|1A|5|H|7|2T|8|@]|D|@]|E|$I|J]]|$1|1B|3|-4|5|6|7|2U|8|@]|D|@]|E|$]]]|1C|$1D|$5|1E|1F|1G|E|$1H|1I]]|1J|$5|1E|1F|1G|E|$1H|1K]]]]

Swift 5 provides a method called <a href="https://developer.apple.com/documentation/swift/array/1687832-enumerated" rel="noreferrer"><code>enumerated()</code></a> for <code>Array</code>. <code>enumerated()</code> has the following declaration:

<pre><code>func enumerated() -&gt; EnumeratedSequence&lt;Array&lt;Element&gt;&gt;
</code></pre>

<blockquote>
 Returns a sequence of pairs (n, x), where n represents a consecutive integer starting at zero and x represents an element of the sequence.
</blockquote>

<hr>

In the simplest cases, you may use <code>enumerated()</code> with a for loop. For example:

<pre><code>let list = ["Car", "Bike", "Plane", "Boat"]
for (index, element) in list.enumerated() {
 print(index, ":", element)
}

/*
prints:
0 : Car
1 : Bike
2 : Plane
3 : Boat
*/
</code></pre>

<hr>

Note however that you're not limited to use <code>enumerated()</code> with a for loop. In fact, if you plan to use <code>enumerated()</code> with a for loop for something similar to the following code, you're doing it wrong:

<pre><code>let list = [Int](1...5)
var arrayOfTuples = [(Int, Int)]()

for (index, element) in list.enumerated() {
 arrayOfTuples += [(index, element)]
}

print(arrayOfTuples) // prints [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
</code></pre>

A swiftier way to do this is:

<pre><code>let list = [Int](1...5)
let arrayOfTuples = Array(list.enumerated())
print(arrayOfTuples) // prints [(offset: 0, element: 1), (offset: 1, element: 2), (offset: 2, element: 3), (offset: 3, element: 4), (offset: 4, element: 5)]
</code></pre>

<hr>

As an alternative, you may also use <code>enumerated()</code> with <code>map</code>:

<pre><code>let list = [Int](1...5)
let arrayOfDictionaries = list.enumerated().map { (a, b) in return [a : b] }
print(arrayOfDictionaries) // prints [[0: 1], [1: 2], [2: 3], [3: 4], [4: 5]]
</code></pre>

<hr>

Moreover, although it has some <a href="https://developer.apple.com/documentation/swift/sequence/3018369-foreach" rel="noreferrer">limitations</a>, <code>forEach</code> can be a good replacement to a for loop:

<pre><code>let list = [Int](1...5)
list.reversed().enumerated().forEach { print($0, ":", $1) }

/*
prints:
0 : 5
1 : 4
2 : 3
3 : 2
4 : 1
*/
</code></pre>

<hr>

By using <code>enumerated()</code> and <code>makeIterator()</code>, you can even iterate manually on your <code>Array</code>. For example:

<pre><code>import UIKit
import PlaygroundSupport

class ViewController: UIViewController {

 var generator = ["Car", "Bike", "Plane", "Boat"].enumerated().makeIterator()

 override func viewDidLoad() {
 super.viewDidLoad()

 let button = UIButton(type: .system)
 button.setTitle("Tap", for: .normal)
 button.frame = CGRect(x: 100, y: 100, width: 100, height: 100)
 button.addTarget(self, action: #selector(iterate(_:)), for: .touchUpInside)
 view.addSubview(button)
 }

 @objc func iterate(_ sender: UIButton) {
 let tuple = generator.next()
 print(String(describing: tuple))
 }

}

PlaygroundPage.current.liveView = ViewController()

/*
 Optional((offset: 0, element: "Car"))
 Optional((offset: 1, element: "Bike"))
 Optional((offset: 2, element: "Plane"))
 Optional((offset: 3, element: "Boat"))
 nil
 nil
 nil
 */
</code></pre>

blocks|key|7597|text|这是枚举循环的公式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7598|for+(index,+value)+in+shoppingList.enumerate()+{
print("Item+\(index+%2B+1):+\(value)")
}|code-block|syntax|javascript|7599|有关更多细节，您可以查看这里。|offset|length|7600|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/CollectionTypes.html#//apple_ref/doc/uid/TP40014097-CH8-ID105^0|0|0|C|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

This is the Formula of loop of Enumeration:

<pre><code>for (index, value) in shoppingList.enumerate() {
print("Item \(index + 1): \(value)")
}
</code></pre>

for more detail you can check <a href="https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/CollectionTypes.html#//apple_ref/doc/uid/TP40014097-CH8-ID105">Here</a>.

blocks|key|1951388|text|基本枚举|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951389|for+(index,+element)+in+arrayOfValues.enumerate()+{
//+do+something+useful
}|code-block|syntax|javascript|1951390|或者和斯威夫特3..。|1951391|for+(index,+element)+in+arrayOfValues.enumerated()+{
//+do+something+useful
}|1951392|枚举、筛选和映射|1951393|但是，我经常将枚举与地图或过滤器结合使用。例如，对两个数组进行操作。|1951394|在这个数组中，我希望过滤奇数或偶数索引元素，并将它们从Ints转换为Double。因此，enumerate()获取索引和元素，然后筛选检查索引，最后为了去掉结果元组，我只将它映射到元素。|offset|length|style|CODE|1951395|let+evens+=+arrayOfValues.enumerate().filter({
++++++++++++++++++++++++++++(index:+Int,+element:+Int)+->+Bool+in
++++++++++++++++++++++++++++return+index+%25+2+==+0
++++++++++++++++++++++++}).map({+(_:+Int,+element:+Int)+->+Double+in
++++++++++++++++++++++++++++return+Double(element)
++++++++++++++++++++++++})
let+odds+=+arrayOfValues.enumerate().filter({
++++++++++++++++++++++++++++(index:+Int,+element:+Int)+->+Bool+in
++++++++++++++++++++++++++++return+index+%25+2+!=+0
++++++++++++++++++++++++}).map({+(_:+Int,+element:+Int)+->+Double+in
++++++++++++++++++++++++++++return+Double(element)
++++++++++++++++++++++++})|1951396|entityMap^0|0|0|0|0|0|0|18|B|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

<h3>Basic enumerate</h3>

<pre><code>for (index, element) in arrayOfValues.enumerate() {
// do something useful
}
</code></pre>

or with Swift 3...

<pre><code>for (index, element) in arrayOfValues.enumerated() {
// do something useful
}
</code></pre>

<h3>Enumerate, Filter and Map</h3>

However, I most often use enumerate in combination with map or filter. For example with operating on a couple of arrays.

In this array I wanted to filter odd or even indexed elements and convert them from Ints to Doubles. So <code>enumerate()</code> gets you index and the element, then filter checks the index, and finally to get rid of the resulting tuple I map it to just the element.

<pre><code>let evens = arrayOfValues.enumerate().filter({
 (index: Int, element: Int) -&gt; Bool in
 return index % 2 == 0
 }).map({ (_: Int, element: Int) -&gt; Double in
 return Double(element)
 })
let odds = arrayOfValues.enumerate().filter({
 (index: Int, element: Int) -&gt; Bool in
 return index % 2 != 0
 }).map({ (_: Int, element: Int) -&gt; Double in
 return Double(element)
 })
</code></pre>

blocks|key|1951445|text|使用.enumerate()可以工作，但它没有提供元素的真正索引；它只提供以0开头的Int，并为每个连续的元素提供1的增量。这通常是不相关的，但是当与ArraySlice类型一起使用时，可能会出现意想不到的行为。请使用以下代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1951446|let+a+=+["a",+"b",+"c",+"d",+"e"]
a.indices+//=>+0..<5

let+aSlice+=+a[1..<4]+//=>+ArraySlice+with+["b",+"c",+"d"]
aSlice.indices+//=>+1..<4

var+test+=+[Int:+String]()
for+(index,+element)+in+aSlice.enumerate()+{
++++test[index]+=+element
}
test+//=>+[0:+"b",+1:+"c",+2:+"d"]+//+indices+presented+as+0..<3,+but+they+are+actually+1..<4
test[0]+==+aSlice[0]+//+ERROR:+out+of+bounds|code-block|syntax|javascript|1951447|这是一个有点人为的例子，这不是一个常见的问题在实践中，但我仍然认为值得知道这是可能发生的。|1951448|entityMap^0|2|C|23|A|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

Using <code>.enumerate()</code> works, but it does not provide the true index of the element; it only provides an Int beginning with 0 and incrementing by 1 for each successive element. This is usually irrelevant, but there is the potential for unexpected behavior when used with the <code>ArraySlice</code> type. Take the following code:

<pre><code>let a = ["a", "b", "c", "d", "e"]
a.indices //=&gt; 0..&lt;5

let aSlice = a[1..&lt;4] //=&gt; ArraySlice with ["b", "c", "d"]
aSlice.indices //=&gt; 1..&lt;4

var test = [Int: String]()
for (index, element) in aSlice.enumerate() {
 test[index] = element
}
test //=&gt; [0: "b", 1: "c", 2: "d"] // indices presented as 0..&lt;3, but they are actually 1..&lt;4
test[0] == aSlice[0] // ERROR: out of bounds
</code></pre>

It's a somewhat contrived example, and it's not a common issue in practice but still I think it's worth knowing this can happen.

blocks|key|7649|text|从Swift+3开始，它是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7650|for+(index,+element)+in+list.enumerated()+{
++print("Item+\(index):+\(element)")
}|code-block|syntax|javascript|7651|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Starting with Swift 3, it is 

<pre><code>for (index, element) in list.enumerated() {
 print("Item \(index): \(element)")
}
</code></pre>

blocks|key|7664|text|Xcode+8和Swift+3:数组可以使用tempArray.enumerated()枚举|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|7665|示例：|7666|var+someStrs+=+[String]()

someStrs.append("Apple")++
someStrs.append("Amazon")++
someStrs+%2B=+["Google"]++++


for+(index,+item)+in+someStrs.enumerated()++
{++
++++++++print("Value+at+index+=+\(index)+is+\(item)").++
}|code-block|syntax|javascript|7667|控制台：|7668|Value+at+index+=+0+is+Apple
Value+at+index+=+1+is+Amazon
Value+at+index+=+2+is+Google|7669|entityMap^0|M|M|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Xcode 8 and Swift 3: 
Array can be enumerated using <code>tempArray.enumerated()</code>

Example:

<pre><code>var someStrs = [String]()

someStrs.append("Apple") 
someStrs.append("Amazon") 
someStrs += ["Google"] 


for (index, item) in someStrs.enumerated() 
{ 
 print("Value at index = \(index) is \(item)"). 
}
</code></pre>

console:

<pre><code>Value at index = 0 is Apple
Value at index = 1 is Amazon
Value at index = 2 is Google
</code></pre>

blocks|key|1951520|text|Swift+5.x：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951521|我个人更喜欢使用forEach方法：|offset|length|style|CODE|1951522|list.enumerated().forEach+{+(index,+element)+in
++++...
}|code-block|syntax|javascript|1951523|您还可以使用简短的版本：|1951524|list.enumerated().forEach+{+print("index:+\($0.0),+value:+\($0.1)")+}|1951525|entityMap^0|0|8|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

<h2>Swift 5.x:</h2>
I personally prefer using the <code>forEach</code> method:
<pre><code>list.enumerated().forEach { (index, element) in
 ...
}
</code></pre>
You can also use the short version:
<pre><code>list.enumerated().forEach { print(&quot;index: \($0.0), value: \($0.1)&quot;) }
</code></pre>

blocks|key|1951549|text|您可以简单地使用枚举循环来获得所需的结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951550|Swift+2:|offset|length|style|BOLD|1951551|for+(index,+element)+in+elements.enumerate()+{
++++print("\(index):+\(element)")
}|code-block|syntax|javascript|1951552|Swift+3&+4:|1951553|for+(index,+element)+in+elements.enumerated()+{
++++print("\(index):+\(element)")
}|1951554|或者，您可以简单地通过for循环获得相同的结果：|1951555|for+index+in+0..<elements.count+{
++++let+element+=+elements[index]
++++print("\(index):+\(element)")
}|1951556|希望能帮上忙。|1951557|entityMap^0|0|0|8|0|0|0|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|13|8|@$D|14|E|15|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|X|$]]

You can simply use loop of enumeration to get your desired result:

Swift 2:

<pre><code>for (index, element) in elements.enumerate() {
 print("\(index): \(element)")
}
</code></pre>

Swift 3 &amp; 4:

<pre><code>for (index, element) in elements.enumerated() {
 print("\(index): \(element)")
}
</code></pre>

Or you can simply go through a for loop to get the same result:

<pre><code>for index in 0..&lt;elements.count {
 let element = elements[index]
 print("\(index): \(element)")
}
</code></pre>

Hope it helps.

blocks|key|7714|text|对于那些想要使用forEach的人。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|7715|Swift+4|7716|extension+Array+{
++func+forEachWithIndex(_+body:+(Int,+Element)+throws+->+Void)+rethrows+{
++++try+zip((startIndex+..<+endIndex),+self).forEach(body)
++}
}|code-block|syntax|javascript|7717|或|7718|array.enumerated().forEach+{+...+}|7719|entityMap^0|8|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

For those who want to use <code>forEach</code>.

Swift 4

<pre><code>extension Array {
 func forEachWithIndex(_ body: (Int, Element) throws -&gt; Void) rethrows {
 try zip((startIndex ..&lt; endIndex), self).forEach(body)
 }
}
</code></pre>

Or

<pre><code>array.enumerated().forEach { ... }
</code></pre>

blocks|key|1951586|text|为了完整起见，您可以简单地迭代数组索引，并使用下标在相应的索引处访问元素：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951587|let+list+=+[100,200,300,400,500]
for+index+in+list.indices+{
++++print("Element+at:",+index,+"+Value:",+list[index])
}|code-block|syntax|javascript|1951588|使用forEach|1951589|list.indices.forEach+{
++++print("Element+at:",+$0,+"+Value:",+list[$0])
}|1951590|使用集合enumerated()方法。注意，它返回带有offset和element的元组集合。|offset|length|style|CODE|1951591|for+item+in+list.enumerated()+{
++++print("Element+at:",+item.offset,+"+Value:",+item.element)
}|1951592|使用forEach：|1951593|list.enumerated().forEach+{
++++print("Element+at:",+$0.offset,+"+Value:",+$0.element)
}|1951594|那些会打印出来|1951595|元素:0值:+100
元素:1值:+200
元素:2值:+300
元素:3值:+400
元素:4值:+500|blockquote|1951596|如果需要数组索引(而不是偏移量)及其元素，则可以扩展Collection并创建自己的方法以获取索引元素：|1951597|extension+Collection+{
++++func+indexedElements(body:+((index:+Index,+element:+Element))+throws+->+Void)+rethrows+{
++++++++var+index+=+startIndex
++++++++for+element+in+self+{
++++++++++++try+body((index,element))
++++++++++++formIndex(after:+&index)
++++++++}
++++}
}|1951598|Alex建议的另一个可能的实现是压缩集合索引及其元素：|1951599|extension+Collection+{
++++func+indexedElements(body:+((index:+Index,+element:+Element))+throws+->+Void)+rethrows+{
++++++++for+element+in+zip(indices,+self)+{+try+body(element)+}
++++}
++++var+indexedElements:+Zip2Sequence<Indices,+Self>+{+zip(indices,+self)+}
}|1951600|测试：|1951601|let+list+=++["100","200","300","400","500"]

//+You+can+iterate+the+index+and+its+elements+using+a+closure
list.dropFirst(2).indexedElements+{
++++print("Index:",+$0.index,+"Element:",+$0.element)
}

//+or+using+a+for+loop
for+(index,+element)+in+list.indexedElements++{
++++print("Index:",+index,+"Element:",+element)
}|1951602|这将p[rint+]|1951603|指数:2要素:+300
指数:3要素:+400
指数:4要素:+500
索引:0元素:+100
指数:1元素:+200
指数:2要素:+300
指数:3要素:+400
指数:4要素:+500|1951604|entityMap^0|0|0|0|0|4|C|R|6|Y|7|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1L|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1M|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1N|8|@$M|1O|N|1P|O|P]|$M|1Q|N|1R|O|P]|$M|1S|N|1T|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1U|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1V|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1W|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1X|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|10|7|1Y|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|13|3|14|5|D|7|20|8|@]|9|@]|A|$E|F]]|$1|15|3|16|5|6|7|21|8|@]|9|@]|A|$]]|$1|17|3|18|5|D|7|22|8|@]|9|@]|A|$E|F]]|$1|19|3|1A|5|6|7|23|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|D|7|24|8|@]|9|@]|A|$E|F]]|$1|1D|3|1E|5|6|7|25|8|@]|9|@]|A|$]]|$1|1F|3|1G|5|10|7|26|8|@]|9|@]|A|$]]|$1|1H|3|-4|5|6|7|27|8|@]|9|@]|A|$]]]|1I|$]]

For completeness you can simply iterate over your array indices and use subscript to access the element at the corresponding index:
<pre><code>let list = [100,200,300,400,500]
for index in list.indices {
 print(&quot;Element at:&quot;, index, &quot; Value:&quot;, list[index])
}
</code></pre>
Using forEach
<pre><code>list.indices.forEach {
 print(&quot;Element at:&quot;, $0, &quot; Value:&quot;, list[$0])
}
</code></pre>
Using collection <code>enumerated()</code> method. Note that it returns a collection of tuples with the <code>offset</code> and the <code>element</code>:
<pre><code>for item in list.enumerated() {
 print(&quot;Element at:&quot;, item.offset, &quot; Value:&quot;, item.element)
}
</code></pre>
using forEach:
<pre><code>list.enumerated().forEach {
 print(&quot;Element at:&quot;, $0.offset, &quot; Value:&quot;, $0.element)
}
</code></pre>
<hr />
Those will print
<blockquote>
Element at: 0 Value: 100
Element at: 1 Value: 200
Element at: 2 Value: 300
Element at: 3 Value: 400
Element at: 4 Value: 500
</blockquote>
If you need the array index (not the offset) and its element you can extend Collection and create your own method to get the indexed elements:
<pre><code>extension Collection {
 func indexedElements(body: ((index: Index, element: Element)) throws -&gt; Void) rethrows {
 var index = startIndex
 for element in self {
 try body((index,element))
 formIndex(after: &amp;index)
 }
 }
}
</code></pre>
<hr />
Another possible implementation as suggested by Alex is to zip the collection indices with its elements:
<pre><code>extension Collection {
 func indexedElements(body: ((index: Index, element: Element)) throws -&gt; Void) rethrows {
 for element in zip(indices, self) { try body(element) }
 }
 var indexedElements: Zip2Sequence&lt;Indices, Self&gt; { zip(indices, self) }
}
</code></pre>
<hr />
Testing:
<pre><code>let list = [&quot;100&quot;,&quot;200&quot;,&quot;300&quot;,&quot;400&quot;,&quot;500&quot;]

// You can iterate the index and its elements using a closure
list.dropFirst(2).indexedElements {
 print(&quot;Index:&quot;, $0.index, &quot;Element:&quot;, $0.element)
}

// or using a for loop
for (index, element) in list.indexedElements {
 print(&quot;Index:&quot;, index, &quot;Element:&quot;, element)
}
</code></pre>
This will p[rint
<blockquote>
Index: 2 Element: 300
Index: 3 Element: 400
Index: 4 Element: 500
Index: 0 Element: 100
Index: 1 Element: 200
Index: 2 Element: 300
Index: 3 Element: 400
Index: 4 Element: 500
</blockquote>

blocks|key|1951621|text|我们调用枚举函数来实现这个功能。喜欢|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951622|++++for+(index,+element)+in+array.enumerate()+{
+++++index+is+indexposition+of+array
+++++element+is+element+of+array+
+++}|code-block|syntax|javascript|1951623|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

We called enumerate function to implements this. like
<pre><code> for (index, element) in array.enumerate() {
 index is indexposition of array
 element is element of array 
 }
</code></pre>

blocks|key|345493|text|对于您想要做的事情，应该在您的enumerated()阵列上使用方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|345494|for+(index,+element)+in+list.enumerated()+{
++++print("\(index)+-+\(element)")
}|code-block|syntax|javascript|345495|entityMap^0|F|C|W|2|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|S|8|@]|E|@]|F|$J|K]]|$1|L|3|-4|5|6|7|T|8|@]|E|@]|F|$]]]|M|$]]

For what you are wanting to do, you should use the <code>enumerated()</code> method on your Array:

<pre><code>for (index, element) in list.enumerated() {
 print("\(index) - \(element)")
}
</code></pre>

blocks|key|7754|text|在函数式编程中使用像这样的.enumerated()：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7755|list.enumerated().forEach+{+print($0.offset,+$0.element)+}+|code-block|syntax|javascript|7756|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use .enumerated() like this in functional programming:
<pre><code>list.enumerated().forEach { print($0.offset, $0.element) } 
</code></pre>

blocks|key|7761|text|Swift+5.x:|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|7762|让列表=+0，1，2，3，4，5|7763|list.enumerated().forEach+{+(index,+value)+in
++++print("index:+\(index),+value:+\(value)")
}|code-block|syntax|javascript|7764|或,|7765|list.enumerated().forEach+{+
++++print("index:+\($0.offset),+value:+\($0.element)")
}+|7766|7767|for+(index,+value)+in+list.enumerated()+{
++++print("index:+\(index),+value:+\(value)")
}|7768|entityMap^0|0|A|0|0|G|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|12|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|Q|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|R|3|S|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|T|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|U|$]]

Swift 5.x:
let list = [0, 1, 2, 3, 4, 5]
<pre><code>list.enumerated().forEach { (index, value) in
 print(&quot;index: \(index), value: \(value)&quot;)
}
</code></pre>
Or,
<pre><code>list.enumerated().forEach { 
 print(&quot;index: \($0.offset), value: \($0.element)&quot;)
} 
</code></pre>
Or,
<pre><code>for (index, value) in list.enumerated() {
 print(&quot;index: \(index), value: \(value)&quot;)
}
</code></pre>

blocks|key|1951718|text|在iOS+8.0中/SWIFT4.0%2B|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1951719|您可以按照forEach的苹果文档使用|CODE|1951720|返回对序列(n，x)，其中n表示从零开始的连续整数，x表示序列的一个元素。|blockquote|1951721|let+numberWords+=+["one",+"two",+"three"]

numberWords.enumerated().forEach+{+(key,+value)+in
+++print("Key:+\(key)+-+Value:+\(value)")
}|code-block|syntax|javascript|1951722|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/swift/array/1689783-foreach^0|1|8|A|8|0|5|7|D|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@$9|14|A|15|B|H]]|D|@$9|16|A|17|1|18]]|E|$]]|$1|I|3|J|5|K|7|19|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1A|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

In iOS 8.0/Swift 4.0+
You can use <code>forEach</code>
As per the <a href="https://developer.apple.com/documentation/swift/array/1689783-foreach" rel="nofollow noreferrer">Apple docs</a>:
<blockquote>
Returns a sequence of pairs (n, x), where n represents a consecutive integer starting at zero and x represents an element of the sequence.
</blockquote>
<pre><code>let numberWords = [&quot;one&quot;, &quot;two&quot;, &quot;three&quot;]

numberWords.enumerated().forEach { (key, value) in
 print(&quot;Key: \(key) - Value: \(value)&quot;)
}
</code></pre>

Is there a function that I can use to iterate over an array and have both index and element, like Python's <code>enumerate</code>?
<pre><code>for index, element in enumerate(list):
 ...
</code></pre>

How to iterate a loop with index and element in Swift

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

是否有一个函数可以用于遍历数组并同时具有索引和元素，比如Python的enumeratefor index, element in enumerate(list):    ...

问如何在Swift中使用索引和元素迭代循环
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Swift中使用索引和元素迭代循环EN