如何在苹果的新语言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"]

如何将元素animals[2]从数组中移除?


当前回答

斯威夫特5

guard let index = orders.firstIndex(of: videoID) else { return }
orders.remove(at: index)

其他回答

我使用这个扩展,几乎与Varun的一样,但这一个(下面)是万能的:

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

要从数组中删除元素,使用remove(at:), removeLast()和removeAll()。

yourArray = [1,2,3,4]

删除2位置的值

yourArray.remove(at: 2)

从数组中移除最后一个值

yourArray.removeLast()

从集合中移除所有成员

yourArray.removeAll()

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)
}

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]

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