从Xcode 10+开始，根据WWDC 2018会议223“包含算法”，一个好的方法将是mutmutingfunc removeAll(where predicate: (Element) throws -> Bool)重新抛出

苹果的例子:

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."

请参阅Apple的文档

所以在OP的例子中，移除动物[2]，“黑猩猩”:

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

这种方法可能是首选的，因为它的伸缩性很好(线性vs二次)，可读和干净。请记住，它只能在Xcode 10+中工作，并且在写这篇文章时是测试版。

鉴于

var animals = ["cats", "dogs", "chimps", "moose"]

删除第一个元素

animals.removeFirst() // "cats"
print(animals)        // ["dogs", "chimps", "moose"]

删除最后一个元素

animals.removeLast() // "moose"
print(animals)       // ["cats", "dogs", "chimps"]

删除索引处的元素

animals.remove(at: 2) // "chimps"
print(animals)           // ["cats", "dogs", "moose"]

删除未知索引的元素

只针对一个元素

if let index = animals.firstIndex(of: "chimps") {
    animals.remove(at: index)
}
print(animals) // ["cats", "dogs", "moose"]

对于多个元素

var animals = ["cats", "dogs", "chimps", "moose", "chimps"]

animals = animals.filter(){$0 != "chimps"}
print(animals) // ["cats", "dogs", "moose"]

笔记

上述方法就地修改数组(过滤器除外)并返回被删除的元素。 快速指南地图滤镜减少 如果不想修改原始数组，可以使用dropFirst或dropLast创建一个新数组。

更新至Swift 5.2

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，你应该使用var代替，例如:

var animals = ["cats", "dogs", "chimps", "moose"]

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

一个保持原始集合不变的非突变替代方法是使用过滤器创建一个新的集合，而不删除你想要的元素，例如:

let pets = animals.filter { $0 != "chimps" }

如果你有一个自定义对象数组，你可以像这样通过特定的属性进行搜索:

if let index = doctorsInArea.firstIndex(where: {$0.id == doctor.id}){
    doctorsInArea.remove(at: index)
}

或者如果你想通过名字来搜索

if let index = doctorsInArea.firstIndex(where: {$0.name == doctor.name}){
    doctorsInArea.remove(at: index)
}

使用索引数组删除元素:

Array of Strings and indexes let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"] let indexAnimals = [0, 3, 4] let arrayRemainingAnimals = animals .enumerated() .filter { !indexAnimals.contains($0.offset) } .map { $0.element } print(arrayRemainingAnimals) //result - ["dogs", "chimps", "cow"] Array of Integers and indexes var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] let indexesToRemove = [3, 5, 8, 12] numbers = numbers .enumerated() .filter { !indexesToRemove.contains($0.offset) } .map { $0.element } print(numbers) //result - [0, 1, 2, 4, 6, 7, 9, 10, 11]

使用另一个数组的元素值删除元素

整数数组 let arrayResult = numbers。筛选器{元素 返回! indexesToRemove.contains(元素) ｝ 打印(arrayResult) //result - [0,1,2,4,6,7,9,10,11] 字符串数组 让arrayLetters =(“a”、“b”、“c”,“d”,“e”,“f”,“g”,“h”,“我”) let arrayRemoveLetters = ["a"， "e"， "g"， "h"] let arrayRemainingLetters = arrayLetters。过滤器{ ! arrayRemoveLetters.contains (0) ｝ 打印(arrayRemainingLetters) //result - ["b"， "c"， "d"， "f"， "i"]

斯威夫特5: 这是一个很酷的和简单的扩展来删除数组中的元素，而不需要过滤:

   extension Array where Element: Equatable {

    // Remove first collection element that is equal to the given `object`:
    mutating func remove(object: Element) {
        guard let index = firstIndex(of: object) else {return}
        remove(at: index)
    }

}

用法:

var myArray = ["cat", "barbecue", "pancake", "frog"]
let objectToRemove = "cat"

myArray.remove(object: objectToRemove) // ["barbecue", "pancake", "frog"]

也适用于其他类型，例如Int，因为Element是泛型类型:

var myArray = [4, 8, 17, 6, 2]
let objectToRemove = 17

myArray.remove(object: objectToRemove) // [4, 8, 6, 2]