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。

如果你不知道你想要删除的元素的索引，并且元素符合Equatable协议，你可以这样做:

animals.remove(at: animals.firstIndex(of: "dogs")!)

参见Equatable协议答案:我如何做indexOfObject或一个适当的containsObject

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

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]

你可以这么做。首先确保Dog确实存在于数组中，然后删除它。如果您认为Dog可能在数组中发生多次，则添加for语句。

var animals = ["Dog", "Cat", "Mouse", "Dog"]
let animalToRemove = "Dog"

for object in animals {
    if object == animalToRemove {
        animals.remove(at: animals.firstIndex(of: animalToRemove)!)
    }
}

如果你确定Dog在数组中退出并且只发生了一次，那么就这样做:

animals.remove(at: animals.firstIndex(of: animalToRemove)!)

如果两者都有，字符串和数字

var array = [12, 23, "Dog", 78, 23]
let numberToRemove = 23
let animalToRemove = "Dog"

for object in array {

    if object is Int {
        // this will deal with integer. You can change to Float, Bool, etc...
        if object == numberToRemove {
        array.remove(at: array.firstIndex(of: numberToRemove)!)
        }
    }
    if object is String {
        // this will deal with strings
        if object == animalToRemove {
        array.remove(at: array.firstIndex(of: animalToRemove)!)
        }
    }
}

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)
}