如何轻松地将 Dictionary<K,V>.Subsequence 转换为 Dictionary<K,V>

如何解决如何轻松地将 Dictionary<K,V>.Subsequence 转换为 Dictionary<K,V>

[更新] 挖掘我发现的字典方法：

.split(whereSeparator: (key,value)) -> [Slice.Dictionary<K,V>]

返回由以下组成的子序列：

_startIndex,_endindex and _base (that contains the original Dict)

我是为了好玩而尝试的。使用结果来获得一个字典需要一个循环来从索引创建一个字典。

你知道一种轻松地将子序列转换为字典的方法吗？我们用子序列字符串做到了吗？：

String(subsequence)

解决方法

虽然 here 已经回答了您的问题，但对于 split 来说还不够，因为它涉及多个切片，它还需要平面映射或等效方法。

Dictionary(
  uniqueKeysWithValues:
    [1: 1,2: 2,3: 3]
    .split { $0.key > 2 } // filters out (key: 3,value: 3)
    .flatMap { $0 }
)

但是，我认为它实际上没有用，因为使用 filter 和反转条件会产生相同的结果。

[1: 1,3: 3].filter { $0.key <= 2 }

相反，如果您想要两个分割部分，作为字典，您可以使用它，它依赖于相同的 uniqueKeysWithValues 扩展初始值设定项。

// [false: [2: 2,1: 1],true: [3: 3]]
Dictionary(grouping: [1: 1,3: 3]) { $0.key > 2 }
  .mapValues(Dictionary.init)

extension Dictionary {
  /// Creates a new dictionary from the key-value pairs in the given sequence.
  ///
  /// - Parameter keysAndValues: A sequence of key-value pairs to use for
  ///   the new dictionary. Every key in `keysAndValues` must be unique.
  /// - Returns: A new dictionary initialized with the elements of `keysAndValues`.
  /// - Precondition: The sequence must not have duplicate keys.
  /// - Note: Differs from the initializer in the standard library,which doesn't allow labeled tuple elements.
  ///     This can't support *all* labels,but it does support `(key:value:)` specifically,///     which `Dictionary` and `KeyValuePairs` use for their elements.
  init<Elements: Sequence>(uniqueKeysWithValues keysAndValues: Elements)
  where Elements.Element == Element {
    self.init(
      uniqueKeysWithValues: keysAndValues.map { ($0.key,$0.value) }
    )
  }
}