我需要删除[Bool]数组中最后一次出现的特定元素。例如,在JavaScript中,这将是:
var k = [true, true, false];
k.splice(k.lastIndexOf(true), 1);
==> [true, false]
我如何在Swift中实现相同的行为?
答案 0 :(得分:7)
您可以通过反向枚举轻松找到最后一次出现的值。当您找到要查找的值时,只需将其删除并从循环中断开即可。使用reverse()
以相反的顺序枚举索引范围:
for i in array.indices.reversed() where array[i] == searchValue {
array.remove(at: i)
break
}
答案 1 :(得分:5)
Xcode 8.2•Swift 3.0.2
var k = [true, true, true, false, true, false]
if let index = k.reversed().index(of: true) {
k.remove(at: index.base - 1)
}
print(k) // "[true, true, true, false, false]"
如果您想创建一个扩展以将此功能添加到Array,您需要将其约束为等值元素:
extension Array where Element: Equatable {
/// Returns the last index where the specified value appears in the collection.
/// After using lastIndex(of:) to find the last position of a particular element in a collection, you can use it to access the element by subscripting.
/// - Parameter element: The element to find the last Index
func lastIndex(of element: Element) -> Index? {
if let index = reversed().index(of: element) {
return index.base - 1
}
return nil
}
/// Removes the last occurrence where the specified value appears in the collection.
/// - Returns: True if the last occurrence element was found and removed or false if not.
/// - Parameter element: The element to remove the last occurrence.
@discardableResult
mutating func removeLastOccurrence(of element: Element) -> Bool {
if let index = lastIndex(of: element) {
remove(at: index)
return true
}
return false
}
}
游乐场测试
var k = [true, true, true, false, true, false]
k.removeLastOccurrence(of: true)
print(k) // "[true, true, true, false, false]"
答案 2 :(得分:2)
我认为没有像lastIndexOf
那样的内置函数,所以它的工作量更多。
var removeIndex: Int?
for (index, item) in enumerate(arr) {
if item == search {
removeIndex = index
}
}
if let removeIndex = removeIndex {
arr.removeAtIndex(removeIndex)
}
arr
是您要搜索的数组(示例中为k
),search
是您要搜索的内容(示例中为true
)。