根据匹配数对数组进行排序
根据匹配数对数组进行排序
提问于 2018-10-13 02:25:40
我正在尝试查找多个测试数组和一个控制数组之间匹配的数组项的数量。找到匹配数组后，我想将测试数组附加到另一个数组中，该数组按控制数组和测试数组之间的匹配数排序。例如，具有3个匹配项的测试数组将位于索引0处，2个匹配项位于索引1处，依此类推。
let controlArray = ["milk", "honey"]
let test1 = ["honey", "water"]
let test2 = ["milk", "honey", "eggs"]
var sortedArrayBasedOnMatches = [[String]]()
/*I want to append test1 and test2 to sortedArrayBasedOnMatches based on how many items
test1 and test2 have in common with controlArray*/
/*in my example above, I would want sortedArrayBasedOnMatches to equal
[test2, test1] since test 2 has two matches and test 1 only has one*/回答 3
Stack Overflow用户
发布于 2018-10-13 02:43:19
这可以通过编写一个流水线来处理输入数组，以一种非常实用和快速的方式来完成：
let sortedArrayBasedOnMatches = [test1, test2] // initial unsorted array
.map { arr in (arr, arr.filter { controlArray.contains($0) }.count) } // making pairs of (array, numberOfMatches)
.sorted { $0.1 > $1.1 } // sorting by the number of matches
.map { $0.0 } // getting rid of the match count, if not needed更新正如@Carpsen90指出的那样，Switf5附带了对count(where:)的支持，这减少了第一次map()调用所需的代码量。一个利用这一点的解决方案可以这样写：
// Swift 5 already has this, let's add it for current versions too
#if !swift(>=5)
extension Sequence {
// taken from the SE proposal
// https://github.com/apple/swift-evolution/blob/master/proposals/0220-count-where.md#detailed-design
func count(where predicate: (Element) throws -> Bool) rethrows -> Int {
var count = 0
for element in self {
if try predicate(element) {
count += 1
}
}
return count
}
}
#endif
let sortedArrayBasedOnMatches = [test1, test2] // initial unsorted array
.map { (arr: $0, matchCount: $0.count(where: controlArray.contains)) } // making pairs of (array, numberOfMatches)
.sorted { $0.matchCount > $1.matchCount } // sorting by the number of matches
.map { $0.arr } // getting rid of the match count, if not needed与原始解决方案相比，样式的另一个变化是对元组组件使用标签，这使得代码更清晰，但也更冗长。
Stack Overflow用户
发布于 2018-10-13 02:35:01
一种选择是将每个数组转换为一个集合，并使用controlArray计算交叉点中的元素数。
let controlArray = ["milk", "honey"]
let test1 = ["honey", "water"]
let test2 = ["milk", "honey", "eggs"]
var sortedArrayBasedOnMatches = [ test1, test2 ].sorted { (arr1, arr2) -> Bool in
return Set(arr1).intersection(controlArray).count > Set(arr2).intersection(controlArray).count
}
print(sortedArrayBasedOnMatches)Stack Overflow用户
发布于 2018-10-16 02:53:43
这将涵盖在控件数组中元素不是唯一的情况(例如牛奶、牛奶、蜂蜜...)并且具有任意数量的测试阵列。
func sortedArrayBasedOnMatches(testArrays:[[String]], control: [String]) -> [[String]]{
var final = [[String]].init()
var controlDict:[String: Int] = [:]
var orderDict:[Int: [[String]]] = [:] // the value is a array of arrays because there could be arrays with the same amount of matches.
for el in control{
if controlDict[el] == nil{
controlDict[el] = 1
}
else{
controlDict[el] = controlDict[el]! + 1
}
}
for tArr in testArrays{
var totalMatches = 0
var tDict = controlDict
for el in tArr{
if tDict[el] != nil && tDict[el] != 0 {
totalMatches += 1
tDict[el] = tDict[el]! - 1
}
}
if orderDict[totalMatches] == nil{
orderDict[totalMatches] = [[String]].init()
}
orderDict[totalMatches]?.append(tArr)
}
for key in Array(orderDict.keys).sorted(by: >) {
for arr in orderDict[key]! {
final.append(arr)
}
}
return final
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52785168
复制相关文章
相似问题
腾讯云开发者
