Swift 3 Array,使用.remove(at:i)一次删除多个项目

Question

是否可以从数组中删除多个项目,同时使用索引位置按照.remove(at:i)类似:

伪代码:

myArray.remove(at: 3, 5, 8, 12)

如果是这样,这样做的语法是什么？

---

更新:

我正在尝试这个,它有效,但下面答案中的扩展更具可读性和合理性,并且实现了与伪代码完全相同的目标.

创建一系列"位置":[3,5,8,12]

let sorted = positions.sorted(by: { $1 < $0 })
for index in sorted
{
    myArray.remove(at: index)
}

Answer 1

如果索引是连续使用removeSubrange方法,则可能.例如,如果要删除索引3到5处的项目:

myArray.removeSubrange(ClosedRange(uncheckedBounds: (lower: 3, upper: 5)))

对于非连续索引,我建议删除索引较大的项目到较小的索引.除了代码可能更短之外,我无法想到在同一行中"同时"删除项目.您可以使用扩展方法执行此操作:

extension Array {
  mutating func remove(at indexes: [Int]) {
    for index in indexes.sorted(by: >) {
      remove(at: index)
    }
  }
}

然后:

myArray.remove(at: [3, 5, 8, 12])

更新:使用上面的解决方案,您需要确保索引数组不包含重复索引.或者您可以避免重复,如下所示:

extension Array {
    mutating func remove(at indexes: [Int]) {
        var lastIndex: Int? = nil
        for index in indexes.sorted(by: >) {
            guard lastIndex != index else {
                continue
            }
            remove(at: index)
            lastIndex = index
        }
    }
}


var myArray = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
myArray.remove(at: [5, 3, 5, 12]) // duplicated index 5
// result: [0, 1, 2, 4, 6, 7, 8, 9, 10, 11, 13] only 3 elements are removed

Answer 2

使用数组元素的索引删除元素:

1. 字符串和索引的数组

   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"]
   

2. 整数和索引的数组

   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

3. 整数数组

   var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
   let elementsTobeRemoved = [3, 5, 8, 12]
   let arrayResult = numbers.filter { element in
       return !elementsTobeRemoved.contains(element)
   }
   print(arrayResult)
   
   //result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
   

4. 字符串数组

   let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
   let arrayRemoveLetters = ["a", "e", "g", "h"]
   let arrayRemainingLetters = arrayLetters.filter {
       !arrayRemoveLetters.contains($0)
   }
   
   print(arrayRemainingLetters)
   
   //result - ["b", "c", "d", "f", "i"]
   

Answer 3

斯威夫特4

extension Array {

    mutating func remove(at indexs: [Int]) {
        guard !isEmpty else { return }
        let newIndexs = Set(indexs).sorted(by: >)
        newIndexs.forEach {
            guard $0 < count, $0 >= 0 else { return }
            remove(at: $0)  
        }
    }

}

var arr = ["a", "b", "c", "d", "e", "f"]

arr.remove(at: [2, 3, 1, 4])

result: ["a", "f"]

Answer 4

简单明了的解决方案，只需Array扩展即可：

extension Array {

    mutating func remove(at indices: [Int]) {
        Set(indices)
            .sorted(by: >)
            .forEach { rmIndex in
                self.remove(at: rmIndex)
            }
    }
}

• Set(indices) -确保唯一性
• .sorted(by: >) -函数从最后到第一删除元素，因此在删除过程中，我们确保索引正确