Question

使用Swift4，我想根据与给定searchTerm最接近的匹配对字符串数组进行排序。重要的是，如果searchTerm可以找到完全匹配，那么returnArray应该预先显示这个searchTerm！

示例：给定Dim vDbld As Variant 'for the date by building ReDim vDbld(1 To 13, 1 To 2) vDbld(1, 1) = "Irving Building" vDbld(2, 1) = "Memorial Building" vDbld(3, 1) = "West Tower" vDbld(4, 1) = "Witting Surgical Center" vDbld(5, 1) = "Madison Irving Surgical Center" vDbld(6, 1) = "Marley Education Center" vDbld(7, 1) = "410 South Crouse" vDbld(8, 1) = "Physicians Office Building" vDbld(9, 1) = "Crouse Business Center" vDbld(10, 1) = "Commonwealth Place" vDbld(11, 1) = "Irving - Memorial Connector" vDbld(12, 1) = "Crouse Garage" vDbld(13, 1) = "CNY Medical Center"

Array = ["Hello world", "Hello Jamaica", "Hello", "Family", "Hel"]，算法应返回：

searchTerm = "Hello"。

方法1：我尝试使用FuzzyMatching - 它以某种方式工作（即它确实根据给定的searchTerm对inputArray进行排序，但它没有提前确定匹配！即使用FuzzyMatching我根据子串实现了良好的排序 - 匹配和句法排序。但它并没有在returnArray中为我提供完全匹配。

方法2：然后我尝试了自己的算法 - （见下面的代码）。但是如果数组中有几个字符串都以我的searchTerm开头（即将searchTerm作为前缀），那么我的算法不太好用。

["Hello", "Hello world", "Hello Jamaica", "Hel", "Family"]

如何在Swift4中完成“最近匹配的字符串数组排序”？特别是在returnArray中为我提前精确匹配？任何帮助赞赏！

Answer 1

您可以使用Levenshtein distance分数将搜索词与数组中的每个字符串进行比较，得分最高的将是结果数组中的第一项，依此类推。结果将是一个字符串数组按得分的降顺序排序。

以下字符串的扩展名可以用来获取Levenshtein距离得分。在该算法中，值越高，平等性越好。

 extension String {
    func levenshteinDistanceScore(to string: String, ignoreCase: Bool = true, trimWhiteSpacesAndNewLines: Bool = true) -> Double {

        var firstString = self
        var secondString = string

        if ignoreCase {
            firstString = firstString.lowercased()
            secondString = secondString.lowercased()
        }
        if trimWhiteSpacesAndNewLines {
            firstString = firstString.trimmingCharacters(in: .whitespacesAndNewlines)
            secondString = secondString.trimmingCharacters(in: .whitespacesAndNewlines)
        }

        let empty = [Int](repeating:0, count: secondString.count)
        var last = [Int](0...secondString.count)

        for (i, tLett) in firstString.enumerated() {
            var cur = [i + 1] + empty
            for (j, sLett) in secondString.enumerated() {
                cur[j + 1] = tLett == sLett ? last[j] : Swift.min(last[j], last[j + 1], cur[j])+1
            }
            last = cur
        }

        // maximum string length between the two
        let lowestScore = max(firstString.count, secondString.count)

        if let validDistance = last.last {
            return  1 - (Double(validDistance) / Double(lowestScore))
        }

        return 0.0
    }
}

Swift中最接近的字符串数组排序

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