我已經看到了一些這樣的示例，但是所有這些似乎都依賴于知道要計算發生次數的元素。我的數組是動態生成的，所以我無法知道要計算哪個元素的出現（我想計算所有元素的出現）。有人可以建議嗎？提前致謝編輯：也許我應該更清楚一點，數組將包含多個不同的字符串（例如 ["FOO", "FOO", "BAR", "FOOBAR"]我如何在不知道它們是什么的情況下計算foo，bar和foobar的出現？
3 回答
阿晨1998
TA貢獻2037條經驗 獲得超6個贊
使用Swift 5時，您可以根據需要選擇以下7個Playground示例代碼之一來計算數組中可哈希項的出現次數。
＃1。使用Array的reduce(into:_:)和Dictionary的subscript(_:default:)標
let array = [4, 23, 97, 97, 97, 23]
let dictionary = array.reduce(into: [:]) { counts, number in
counts[number, default: 0] += 1
}
print(dictionary) // [4: 1, 23: 2, 97: 3]
＃2。使用repeatElement(_:count:)函數，zip(_:_:)函數和Dictionary的init(_:uniquingKeysWith:)初始值設定項
let array = [4, 23, 97, 97, 97, 23]
let repeated = repeatElement(1, count: array.count)
//let repeated = Array(repeating: 1, count: array.count) // also works
let zipSequence = zip(array, repeated)
let dictionary = Dictionary(zipSequence, uniquingKeysWith: { (current, new) in
return current + new
})
//let dictionary = Dictionary(zipSequence, uniquingKeysWith: +) // also works
print(dictionary) // prints [4: 1, 23: 2, 97: 3]
＃3。使用Dictionary的init(grouping:by:)初始值設定項和mapValues(_:)方法
let array = [4, 23, 97, 97, 97, 23]
let dictionary = Dictionary(grouping: array, by: { $0 })
let newDictionary = dictionary.mapValues { (value: [Int]) in
return value.count
}
print(newDictionary) // prints: [97: 3, 23: 2, 4: 1]
＃4。使用Dictionary的init(grouping:by:)初始值設定項和map(_:)方法
let array = [4, 23, 97, 97, 97, 23]
let dictionary = Dictionary(grouping: array, by: { $0 })
let newArray = dictionary.map { (key: Int, value: [Int]) in
return (key, value.count)
}
print(newArray) // prints: [(4, 1), (23, 2), (97, 3)]
＃5。使用for循環和Dictionary的subscript(_:)下標
extension Array where Element: Hashable {
func countForElements() -> [Element: Int] {
var counts = [Element: Int]()
for element in self {
counts[element] = (counts[element] ?? 0) + 1
}
return counts
}
}
let array = [4, 23, 97, 97, 97, 23]
print(array.countForElements()) // prints [4: 1, 23: 2, 97: 3]
＃6。使用NSCountedSet和NSEnumerator的map(_:)方法（需要Foundation）
import Foundation
extension Array where Element: Hashable {
func countForElements() -> [(Element, Int)] {
let countedSet = NSCountedSet(array: self)
let res = countedSet.objectEnumerator().map { (object: Any) -> (Element, Int) in
return (object as! Element, countedSet.count(for: object))
}
return res
}
}
let array = [4, 23, 97, 97, 97, 23]
print(array.countForElements()) // prints [(97, 3), (4, 1), (23, 2)]
＃7。使用NSCountedSet和AnyIterator（需要Foundation）
import Foundation
extension Array where Element: Hashable {
func counForElements() -> Array<(Element, Int)> {
let countedSet = NSCountedSet(array: self)
var countedSetIterator = countedSet.objectEnumerator().makeIterator()
let anyIterator = AnyIterator<(Element, Int)> {
guard let element = countedSetIterator.next() as? Element else { return nil }
return (element, countedSet.count(for: element))
}
return Array<(Element, Int)>(anyIterator)
}
}
let array = [4, 23, 97, 97, 97, 23]
print(array.counForElements()) // [(97, 3), (4, 1), (23, 2)]
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