从 CSV 文件中分离字符串

Question

目前我让用户导入从Excel 或Numbers 创建的CSV-file。工作表有多个字段，我可以毫无问题地导入这些字段。

但是，其中一个字段如下所示：

"[Title1: (100: S11), (100: S12), (100: 10), (100: 2), (100: 3)], [Title2: (300: 50), (300: 90), (300: 100a), (300: D), (300: E)], [Title3: (500: 2), (500: 112), (500: 6), (500: 110), (500: 113)]"

我应该如何将此字段分成如下所示的字典：

//[String: [[Int: String]]]

[
    "Title1": [[100: "S11"], [100: "S12"], [100:   "10"], [100:   "2"], [100:   "3"]],
    "Title2": [[300:  "50"], [300:  "90"], [300: "100a"], [300:   "D"], [300:   "E"]],
    "Title3": [[500:   "2"], [500: "112"], [500:    "6"], [500: "110"], [500: "113"]]
]

这个字段是由用户自己创建的，这似乎有点太复杂了，容易出错。也许更好的问题是如何更好地格式化而不是将该字段拆分到我想要的字典中？

---

也许是简化版：

//[String: [Int: [String]]]

[
    "Title1": [100: ["S11", "S12",   "10",   "2",   "3"]],
    "Title2": [300:  ["50",  "90", "100a",   "D",   "E"]],
    "Title3": [500:   ["2", "112",    "6", "110", "113"]]
]

---

关于这个字段的一些解释：

标题：可以是任何东西，它是嵌套字典的标题

嵌套字典中的key可以是100, 200, 300, 400, 500 or 600

嵌套字典中的值是String，与它们在嵌套字典中的键一起构成唯一键。

这意味着：

100: "S11" and 200: "S11" // valid
100: "S11" and 100: "S11" // not valid

---

extension String {

    // Separate alphanumeric characters
    func split(_ number: Int = 1) -> [String] {
        let number = max(number, 1)

        return self.components(separatedBy: CharacterSet.alphanumerics.inverted).filter { $0.characters.count > number - 1 }
    }
}

let string = "Appetizer: (100: S11), (100: S12), (100: 10), (100: 2), (100: 3) * Main: (300: 50), (300: 90), (300: 100a), (300: D), (300: E) * Dessert: (500: 2), (500: 112), (500: 6), (500: 110), (500: 113)"

let values = string.components(separatedBy: "*").map { $0.split() }

var result: [String: [Int: [String]]] = [:]

values.forEach {
    guard let key = $0.first else { return }

    result[key] = result[key] ?? [:]

    let values = $0.dropFirst()

    let name = values.enumerated().filter { $0.0 % 2 == 0 }.map { $0.1 }
    let code = values.enumerated().filter { $0.0 % 2 != 0 }.map { $0.1 }

    zip(name, code).forEach {
        guard let category = Int($0.0) else { return }

        result[key]?[category] = (result[key]?[category] ?? []) + [$0.1]
    }
}

这段代码产生了我想要的结果，但它看起来很丑陋，而且过于依赖值的位置（而不是检查值）。

关于如何使用更简单、更优雅的方式来做到这一点有什么建议吗？

Answer 1

以下代码应该将您提供的数据解析为您想要的字典：

    let data = "[Title1: (100: S11), (100: S12), (100: 10), (100: 2), (100: 3)], [Title2: (300: 50), (300: 90), (300: 100a), (300: D), (300: E)], [Title3: (500: 2), (500: 112), (500: 6), (500: 110), (500: 113)]"

    let arr1 = data.components(separatedBy:"],")
    var dic = [String:[[String:Any]]]()
    for row1 in arr1 {
        let arr2 = row1.components(separatedBy:",")
        var key = ""
        for row2 in arr2 {
            var txt = row2.replacingOccurrences(of:" ", with:"")
            if txt.hasPrefix("[Title") {
                // First row
                txt = txt.replacingOccurrences(of:"[", with:"")
                let arr3 = txt.components(separatedBy:":")
                key = arr3[0]
                let key1 = arr3[1].replacingOccurrences(of:"(", with:"")
                let val1 = arr3[2].replacingOccurrences(of:")", with:"").replacingOccurrences(of:" ", with:"")
                dic[key] = [[key1:val1]]
            } else {
                let arr3 = row2.components(separatedBy:":")
                let key1 = arr3[0].replacingOccurrences(of:"(", with:"").replacingOccurrences(of:" ", with:"")
                let val1 = arr3[1].replacingOccurrences(of:")", with:"").replacingOccurrences(of:"]", with:"").replacingOccurrences(of:" ", with:"")
                dic[key]?.append([key1:val1])
            }
        }
    }
    print(dic)

以上回答了您的第一个问题，您如何将您提供的数据解析为您指定的字典。第二个问题有点复杂，因为我对数据的了解不够，无法提供有用的答案。但也许现在没有必要了，因为可以根据需要将数据拆分成字典？ :)

Answer 2

我假设您选择了第一种导出格式，即

[Title1: (100: S11), (100: S12), (100: 10), (100: 2), (100: 3)], [Title2: (300: 50), (300: 90), (300: 100a), (300: D), (300: E)], [Title3: (500: 2), (500: 112), (500: 6), (500: 110), (500: 113)]

您可以使用 RegEx 来解析该字符串。下面的答案不处理错误。如果您的标题包含分号（:），它也将不起作用

extension String {
    subscript(_ range: NSRange) -> String {
        let startIndex = self.utf16.index(self.utf16.startIndex, offsetBy: range.location)
        let endIndex = self.utf16.index(startIndex, offsetBy: range.length)

        return String(describing: self.utf16[startIndex..<endIndex])
    }
}

let str = "[Title1: (100: S11), (100: S12), (100: 10), (100: 2), (100: 3)], [Title2: (300: 50), (300: 90), (300: 100a), (300: D), (300: E)], [Title3: (500: 2), (500: 112), (500: 6), (500: 110), (500: 113)]"
let regex1 = try! NSRegularExpression(pattern: "\\[(.+?): (.+?)\\]", options: [])
let regex2 = try! NSRegularExpression(pattern: "\\((\\d+?): (.+?)\\)", options: [])

var result = [String: [Int: [String]]]()

regex1.enumerateMatches(in: str, options: [], range: NSMakeRange(0, str.utf16.count)) { match1, _, _ in
    guard let match1 = match1 else { return }

    let title = str[match1.rangeAt(1)]
    let categoriesStr = str[match1.rangeAt(2)]
    var categories = [Int: [String]]()

    regex2.enumerateMatches(in: categoriesStr, options: [], range: NSMakeRange(0, categoriesStr.utf16.count)) { match2, _, _ in
        guard let match2 = match2 else { return }

        let number = Int(categoriesStr[match2.rangeAt(1)])!
        let identifier = categoriesStr[match2.rangeAt(2)]

        if categories[number] == nil {
            categories[number] = [String]()
        }
        categories[number]!.append(identifier)
    }

    result[title] = categories
}

print(result)