Swift 4 中更简单的字符串切片

Question

Swift 4 改变了字符串的工作方式。但是，似乎变得更加复杂且可读性降低。谁能简化这个例子（简单地将字符串的第三个字母作为字符串）？（除了拆分线。）

let myString="abc"
let thirdLetter = String(myString[myString.index(myString.startIndex, offsetBy: 2)])

Answer 1

在 Swift 4 中，您可以使用 Array(myString) 将 String myString 转换为 [Character]。然后您可以使用 Int 索引该数组，然后将 [Character] 转换为 String.

let myString = "abc"
let thirdLetter = String(Array(myString)[2])    // "c"
let firstTwo = String(Array(myString)[0..<2])   // "ab"

如果您要对 String 进行大量操作，通常最好将其转换并保留为 [Character]。

注意：我修改了这一部分以尽量避免编译器可能进行的任何缓存优化。现在每种方法只测量一次，每种方法的总计运行。

转换为Array并用Int建立索引很容易写和读，但它的表现如何呢？为了回答这个问题，我在发布版本中测试了以下内容：

func time1(str: String, n: Int) -> (Double, String) {
    // Method 1: Index string with String.Index, convert to String

    let start = Date()
    let a = String(str[str.index(str.startIndex, offsetBy: n)])
    let interval = Date().timeIntervalSince(start)

    return (interval, a)
}

func time2(str: String, n: Int) -> (Double, String) {
    // Method 2: Convert string to array, index with Int, convert to String

    let start = Date()
    let a = String(Array(str)[n])
    let interval = Date().timeIntervalSince(start)

    return (interval, a)
}


func time3(str: String, n: Int) -> (Double, String) {
    // Method 3: Use prefix() and last(), convert to String

    let start = Date()
    let a = String(str.prefix(n + 1).last!)
    let interval = Date().timeIntervalSince(start)

    return (interval, a)
}

func time4(str: String, n: Int) -> (Double, String) {
    // Method 4: Use Leo Dabus' extensions
    // 

    let start = Date()
    let a = str[n]
    let interval = Date().timeIntervalSince(start)

    return (interval, a)
}

func time5(str: String, n: Int) -> (Double, String) {
    // Method 5: Same as 2 but don't measure Array conversion time

    let arr = Array(str)

    let start = Date()
    let a = String(arr[n])
    let interval = Date().timeIntervalSince(start)

    return (interval, a)
}

func test() {
    for repetitions in [1, 10, 100, 1000] {
        var input = ""
        for _ in 0 ..< repetitions {
            input.append("abcdefghijklmnopqrstuvwxyz")
        }

        var t = [0.0, 0.0, 0.0, 0.0, 0.0]
        let funcs = [time1, time2, time3, time4, time5]

        for i in 0 ..< input.count {
            for f in funcs.indices {
                let (interval, _) = funcs[f](input, i)
                t[f] += interval
            }
        }

        print("For string length \(input.count):")
        for i in 0 ..< 5 {
            print(String(format: "Method %d time: %.8f", i + 1, t[i]))
        }
        print("")
    }
}

结果：

For string length 26:
Method 1 time: 0.00108612
Method 2 time: 0.00085294
Method 3 time: 0.00005889
Method 4 time: 0.00002104
Method 5 time: 0.00000405

For string length 260:
Method 1 time: 0.00117570
Method 2 time: 0.00670648
Method 3 time: 0.00115579
Method 4 time: 0.00110406
Method 5 time: 0.00007111

For string length 2600:
Method 1 time: 0.09964919
Method 2 time: 0.57621503
Method 3 time: 0.09244329
Method 4 time: 0.09166771
Method 5 time: 0.00087011

For string length 26000:
Method 1 time: 9.78054154
Method 2 time: 56.92994779
Method 3 time: 9.02372885
Method 4 time: 9.01480001
Method 5 time: 0.03442019

分析：

转换为数组的成本很高，尤其是随着数组大小的增长。
如果可以保留转换后的 [Character]，对其进行索引操作会非常快。（见方法五）
方法1、3、4速度都差不多，根据个人喜好选择。

Answer 2

您似乎正在尝试提取第三个字符。我会说

let c = myString.prefix(3).last

当然，它是一个可选字符，但您可以随意解包并在需要时强制转换为字符串。

prefix 方法非常有价值，因为它需要一个 Int 而不是强迫您进入 String.Index 的狂野和古怪的世界。

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