blocks|key|109708|text|let关键字用于声明不能更改的常量。如果你想修改一个变量，你应该改用var，例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|109709|var+animals+=+["cats",+"dogs",+"chimps",+"moose"]

animals.remove(at:+2)++//["cats",+"dogs",+"moose"]|code-block|syntax|javascript|109710|保持原始集合不变的一种不变的替代方法是使用filter创建一个新的集合，而不删除您想要删除的元素，例如：|109711|let+pets+=+animals.filter+{+$0+!=+"chimps"+}|109712|entityMap^0|0|3|Y|3|0|0|L|6|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

The <code>let</code> keyword is for declaring constants that can't be changed. If you want to modify a variable you should use <code>var</code> instead, e.g:

<pre><code>var animals = ["cats", "dogs", "chimps", "moose"]

animals.remove(at: 2) //["cats", "dogs", "moose"]
</code></pre>

A non-mutating alternative that will keep the original collection unchanged is to use <code>filter</code> to create a new collection without the elements you want removed, e.g:

<pre><code>let pets = animals.filter { $0 != "chimps" }
</code></pre>

blocks|key|4616325|text|对于Swift4：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4616326|list+=+list.filter{$0+!=+"your+Value"}|code-block|syntax|javascript|4616327|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|$]]

For Swift4:

<pre><code>list = list.filter{$0 != "your Value"}
</code></pre>

blocks|key|4616374|text|从Xcode+10%2B开始，根据WWDC+2018+session+223,+"Embracing+Algorithms,"的说法，mutating+func+removeAll(where+predicate:+(Element)+throws+->+Bool)+rethrows将是未来的一个好方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4616375|苹果的例子：|4616376|var+phrase+=+"The+rain+in+Spain+stays+mainly+in+the+plain."
let+vowels:+Set<Character>+=+["a",+"e",+"i",+"o",+"u"]

phrase.removeAll(where:+{+vowels.contains($0)+})
//+phrase+==+"Th+rn+n+Spn+stys+mnly+n+th+pln."|code-block|syntax|javascript|4616377|请参阅Apple's+Documentation|4616378|因此，在OP的例子中，删除animals2，“黑猩猩”：|4616379|var+animals+=+["cats",+"dogs",+"chimps",+"moose"]
animals.removeAll(where:+{+$0+==+"chimps"+}+)
//+or+animals.removeAll+{+$0+==+"chimps"+}|4616380|这种方法可能是首选的，因为它具有很好的伸缩性(线性与二次)，可读且干净。请记住，它只能在Xcode中工作，而且在写这篇文章的时候，它还在测试版中。|4616381|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/videos/play/wwdc2018/223|1|https://developer.apple.com/documentation/swift/array/2994751-removeall^0|1T|23|F|1A|0|0|0|0|3|L|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|O|3|P|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1H|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|1I|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]]]

As of Xcode 10+, and according to the <a href="https://developer.apple.com/videos/play/wwdc2018/223" rel="noreferrer">WWDC 2018 session 223, "Embracing Algorithms,"</a> a good method going forward will be <code>mutating func removeAll(where predicate: (Element) throws -&gt; Bool) rethrows</code>

Apple's example:

<pre><code>var phrase = "The rain in Spain stays mainly in the plain."
let vowels: Set&lt;Character&gt; = ["a", "e", "i", "o", "u"]

phrase.removeAll(where: { vowels.contains($0) })
// phrase == "Th rn n Spn stys mnly n th pln."
</code></pre>

see <a href="https://developer.apple.com/documentation/swift/array/2994751-removeall" rel="noreferrer">Apple's Documentation</a>

So in the OP's example, removing animals[2], "chimps":

<pre><code>var animals = ["cats", "dogs", "chimps", "moose"]
animals.removeAll(where: { $0 == "chimps" } )
// or animals.removeAll { $0 == "chimps" }
</code></pre>

This method may be preferred because it scales well (linear vs quadratic), is readable and clean. Keep in mind that it only works in Xcode 10+, and as of writing this is in Beta.

blocks|key|109805|text|如果您不知道要删除的元素的索引，并且该元素符合Equatable协议，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|109806|animals.remove(at:+animals.firstIndex(of:+"dogs")!)|code-block|syntax|javascript|109807|请参阅可等同协议答案：How+do+I+do+indexOfObject+or+a+proper+containsObject|offset|length|109808|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/24010700/how-do-i-do-indexofobject-or-a-proper-containsobject^0|0|0|B|1G|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]]]]

If you don't know the index of the element that you want to remove, and the element is conform the Equatable protocol, you can do:
<pre><code>animals.remove(at: animals.firstIndex(of: &quot;dogs&quot;)!)
</code></pre>
See Equatable protocol answer:<a href="https://stackoverflow.com/questions/24010700/how-do-i-do-indexofobject-or-a-proper-containsobject">How do I do indexOfObject or a proper containsObject</a>

blocks|key|4616009|text|Swift中与数组相关的少数操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4616010|创建数组|4616011|var+stringArray+=+["One",+"Two",+"Three",+"Four"]|code-block|syntax|javascript|4616012|在数组中添加对象|4616013|stringArray+=+stringArray+%2B+["Five"]|4616014|从Index对象获取值|4616015|let+x+=+stringArray[1]|4616016|追加对象|4616017|stringArray.append("At+last+position")|4616018|在索引位置插入对象|4616019|stringArray.insert("Going",+at:+1)|4616020|删除对象|4616021|stringArray.remove(at:+3)|4616022|Concat对象值|4616023|var+string+=+"Concate+Two+object+of+Array+\(stringArray[1])+%2B+\(stringArray[2])"|4616024|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1B|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1D|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1F|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|6|7|1J|8|@]|9|@]|A|$]]|$1|10|3|11|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|12|3|13|5|6|7|1L|8|@]|9|@]|A|$]]|$1|14|3|15|5|F|7|1M|8|@]|9|@]|A|$G|H]]|$1|16|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|17|$]]

Few Operation relates to Array in Swift
Create Array
<pre><code>var stringArray = [&quot;One&quot;, &quot;Two&quot;, &quot;Three&quot;, &quot;Four&quot;]
</code></pre>
Add Object in Array
<pre><code>stringArray = stringArray + [&quot;Five&quot;]
</code></pre>
Get Value from Index object
<pre><code>let x = stringArray[1]
</code></pre>
Append Object
<pre><code>stringArray.append(&quot;At last position&quot;)
</code></pre>
Insert Object at Index
<pre><code>stringArray.insert(&quot;Going&quot;, at: 1)
</code></pre>
Remove Object
<pre><code>stringArray.remove(at: 3)
</code></pre>
Concat Object value
<pre><code>var string = &quot;Concate Two object of Array \(stringArray[1]) + \(stringArray[2])&quot;
</code></pre>

blocks|key|110005|text|如果您有自定义对象数组，则可以按特定属性进行搜索，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|110006|if+let+index+=+doctorsInArea.firstIndex(where:+{$0.id+==+doctor.id}){
++++doctorsInArea.remove(at:+index)
}|code-block|syntax|javascript|110007|或者如果您想要按名称进行搜索，例如|110008|if+let+index+=+doctorsInArea.firstIndex(where:+{$0.name+==+doctor.name}){
++++doctorsInArea.remove(at:+index)
}|110009|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you have array of custom Objects, you can search by specific property like this:
<pre><code>if let index = doctorsInArea.firstIndex(where: {$0.id == doctor.id}){
 doctorsInArea.remove(at: index)
}
</code></pre>
or if you want to search by name for example
<pre><code>if let index = doctorsInArea.firstIndex(where: {$0.name == doctor.name}){
 doctorsInArea.remove(at: index)
}
</code></pre>

blocks|key|109859|text|假设Array中的元素实现了Equatable，我想出了以下扩展来负责从Array中删除元素|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|109860|extension+Array+where+Element:+Equatable+{
++
++mutating+func+removeEqualItems(_+item:+Element)+{
++++self+=+self.filter+{+(currentItem:+Element)+->+Bool+in
++++++return+currentItem+!=+item
++++}
++}

++mutating+func+removeFirstEqualItem(_+item:+Element)+{
++++guard+var+currentItem+=+self.first+else+{+return+}
++++var+index+=+0
++++while+currentItem+!=+item+{
++++++index+%2B=+1
++++++currentItem+=+self[index]
++++}
++++self.remove(at:+index)
++}
++
}|code-block|syntax|javascript|109861|用法：|109862|var+test1+=+[1,+2,+1,+2]
test1.removeEqualItems(2)+//+[1,+1]

var+test2+=+[1,+2,+1,+2]
test2.removeFirstEqualItem(2)+//+[1,+1,+2]|109863|entityMap^0|2|5|E|9|10|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

I came up with the following extension that takes care of removing elements from an <code>Array</code>, assuming the elements in the <code>Array</code> implement <code>Equatable</code>:
<pre><code>extension Array where Element: Equatable {
 
 mutating func removeEqualItems(_ item: Element) {
 self = self.filter { (currentItem: Element) -&gt; Bool in
 return currentItem != item
 }
 }

 mutating func removeFirstEqualItem(_ item: Element) {
 guard var currentItem = self.first else { return }
 var index = 0
 while currentItem != item {
 index += 1
 currentItem = self[index]
 }
 self.remove(at: index)
 }
 
}
 
</code></pre>
<h3>Usage:</h3>
<pre><code>var test1 = [1, 2, 1, 2]
test1.removeEqualItems(2) // [1, 1]

var test2 = [1, 2, 1, 2]
test2.removeFirstEqualItem(2) // [1, 1, 2]
</code></pre>

blocks|key|109956|text|关于@Suragch对“移除未知索引的元素”的替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|109957|还有一个功能更强大的"indexOf(element)“版本，它将匹配谓词而不是对象本身。它使用相同的名称，但被myObjects.indexOf{$0.property+=+valueToMatch}调用。它返回在myObjects数组中找到的第一个匹配项的索引。|109958|如果元素是对象/结构，您可能希望根据元素的某个属性的值来删除该元素。例如，您有一个具有car.color属性的Car类，并且您希望从您的carsArray中删除“红色”汽车。|109959|if+let+validIndex+=+(carsArray.indexOf{$0.color+==+UIColor.redColor()})+{
++carsArray.removeAtIndex(validIndex)
}|code-block|syntax|javascript|109960|可以预见的是，通过将上面的if语句嵌入到一个repeat/while循环中，并附加一个else块来设置一个标志“中断”出循环，您可以修改这个代码来删除“所有”红色汽车。|109961|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Regarding @Suragch's Alternative to "Remove element of unknown index": 

There is a more powerful version of "indexOf(element)" that will match on a predicate instead of the object itself. It goes by the same name but it called by myObjects.indexOf{$0.property = valueToMatch}. It returns the index of the first matching item found in myObjects array.

If the element is an object/struct, you may want to remove that element based on a value of one of its properties. Eg, you have a Car class having car.color property, and you want to remove the "red" car from your carsArray.

<pre><code>if let validIndex = (carsArray.indexOf{$0.color == UIColor.redColor()}) {
 carsArray.removeAtIndex(validIndex)
}
</code></pre>

Foreseeably, you could rework this to remove "all" red cars by embedding the above if statement within a repeat/while loop, and attaching an else block to set a flag to "break" out of the loop.

blocks|key|109731|text|这应该可以做到这一点(未测试)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|109732|animals[2...3]+=+[]|code-block|syntax|javascript|109733|编辑:你需要使它成为一个var，而不是一个let，否则它是一个不可变的常量。|offset|length|style|CODE|109734|entityMap^0|0|0|C|3|L|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

This should do it (not tested):
<pre><code>animals[2...3] = []
</code></pre>
Edit: and you need to make it a <code>var</code>, not a <code>let</code>, otherwise it's an immutable constant.

blocks|key|110033|text|用于删除字符串对象的扩展|type|unstyled|depth|inlineStyleRanges|entityRanges|data|110034|extension+Array+{
++++mutating+func+delete(element:+String)+{
++++++++self+=+self.filter()+{+$0+as!+String+!=+element+}
++++}
}|code-block|syntax|javascript|110035|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|$]]

extension to remove String object

<pre><code>extension Array {
 mutating func delete(element: String) {
 self = self.filter() { $0 as! String != element }
 }
}
</code></pre>

blocks|key|4616315|text|我使用这个扩展，几乎和Varun的一样，但是这个(下面)是通用的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4616316|+extension+Array+where+Element:+Equatable++{
++++++++mutating+func+delete(element:+Iterator.Element)+{
++++++++++++++++self+=+self.filter{$0+!=+element+}
++++++++}
++++}|code-block|syntax|javascript|4616317|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|$]]

I use this extension, almost same as Varun's, but this one (below) is all-purpose:

<pre><code> extension Array where Element: Equatable {
 mutating func delete(element: Iterator.Element) {
 self = self.filter{$0 != element }
 }
 }
</code></pre>

blocks|key|110162|text|要从数组中删除元素，请使用remove(at:)、removeLast()和removeAll()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|110163|yourArray+=+[1,2,3,4]|code-block|syntax|javascript|110164|删除位于%252位置的值|110165|yourArray.remove(at:+2)|110166|从数组中删除最后一个值|110167|yourArray.removeLast()|110168|从集合中删除所有成员|110169|yourArray.removeAll()|110170|entityMap^0|D|B|P|C|12|B|0|0|0|0|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]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1A|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|X|$]]

To remove elements from an array, use the <code>remove(at:)</code>,
 <code>removeLast()</code> and <code>removeAll()</code>.

<pre><code>yourArray = [1,2,3,4]
</code></pre>

Remove the value at 2 position

<pre><code>yourArray.remove(at: 2)
</code></pre>

Remove the last value from array

<pre><code>yourArray.removeLast()
</code></pre>

Removes all members from the set

<pre><code>yourArray.removeAll()
</code></pre>

How can I unset/remove an element from an array in Apple's new language Swift?

Here's some code:

<pre><code>let animals = ["cats", "dogs", "chimps", "moose"]
</code></pre>

How could the element <code>animals[2]</code> be removed from the array?

How to remove an element from an array in Swift

Swift

Xcode

如何在Apple的新语言Swift中取消设置/删除数组中的元素？下面是一些代码：let animals = ["cats", "dogs", "chimps", "moose"]如何从数组中删除元素animals[2]？

问如何在Swift中从数组中删除元素
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