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?
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
我提出了以下扩展,负责从数组中删除元素,假设数组中的元素实现了Equatable:
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 += 1
currentItem = self[index]
}
self.remove(at: index)
}
}
用法:
var test1 = [1, 2, 1, 2]
test1.removeEqualItems(2) // [1, 1]
var test2 = [1, 2, 1, 2]
test2.removeFirstEqualItem(2) // [1, 1, 2]
其他回答
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
guard let index = orders.firstIndex(of: videoID) else { return }
orders.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]
Swift中很少涉及数组操作
创建数组
var stringArray = ["One", "Two", "Three", "Four"]
在数组中添加对象
stringArray = stringArray + ["Five"]
从索引对象中获取值
let x = stringArray[1]
添加对象
stringArray.append("At last position")
在索引处插入对象
stringArray.insert("Going", at: 1)
删除对象
stringArray.remove(at: 3)
Concat对象值
var string = "Concate Two object of Array \(stringArray[1]) + \(stringArray[2])"
上面的答案似乎假定您知道要删除的元素的索引。
通常,您知道对数组中要删除的对象的引用。在这种情况下,直接使用对象引用可能会更容易,而不必到处传递它的索引。因此,我建议这个解决方案。它使用标识符!==,用于测试两个对象引用是否都引用同一个对象实例。
func delete(element: String) {
list = list.filter { $0 != element }
}
当然,这不仅仅适用于字符串。