Ale*_*ica 1 generics ambiguous swift swift2
我已经写了这个扩展SequenceType来模仿Python的collections.Counter.
let input = [
"a", "a", "a", "a", "a",
"b", "b", "b", "b",
"c", "c", "c",
"d", "d",
"e"
]
let counts = input.countRepetitions()
print(counts) //expected result: ["a": 5 , "b" : 4, "c" : 3, "d" : 2, "e" : 1]
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这是代码:
extension SequenceType where Self.Generator.Element : Hashable {
func countRepetitions() -> [Self.Generator.Element : Int] {
return self.reduce([Self.Generator.Element : Int]()) { dict, element in
dict[key: element] = (dict[element] ?? 0) + 1
}
}
}
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我收到以下错误:
Playground execution failed: OS X.playground:26:22: error: type of expression is ambiguous without more context
return self.reduce([Self.Generator.Element : Int]()) { dict, element in
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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在使用嵌套类型定义变量时,这似乎是一个限制.而
let foo1: [Self.Generator.Element : Int] = [:]
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在你的方法中编译,这不会:
let foo2 = [Self.Generator.Element : Int]()
// error: type of expression is ambiguous without more context
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作为解决方法,您可以定义一个类型:
typealias E = Self.Generator.Element
let foo3 = [E : Int]()
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适用于您的案例:
extension SequenceType where Self.Generator.Element : Hashable {
func countRepetitions() -> [Self.Generator.Element : Int] {
typealias E = Self.Generator.Element
return self.reduce([E : Int]()) { (dict, element) -> [E : Int] in
var dict = dict
dict[element] = (dict[element] ?? 0) + 1
return dict
}
}
}
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(请注意,闭包参数是常量,因此您必须首先创建一个可变副本.此外,闭包必须返回一个值.)
但实际上你可以避免这个问题并让编译器推断出类型:
extension SequenceType where Self.Generator.Element : Hashable {
func countRepetitions() -> [Self.Generator.Element : Int] {
return self.reduce([:]) { (dict, element) in
var dict = dict
dict[element] = (dict[element] ?? 0) + 1
return dict
}
}
}
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另请注意,reduce在每个迭代步骤中创建一个新字典.一个更有效的解决方案是
extension SequenceType where Generator.Element : Hashable {
func countRepetitions() -> [Generator.Element : Int] {
var dict: [Generator.Element: Int] = [:]
self.forEach {
dict[$0] = (dict[$0] ?? 0) + 1
}
return dict
}
}
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我在哪里也省略了(冗余)Self..
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