Question

我有位数组：

var bits: [Bit]

如何将其转换为bytes array：

var bytes: [UInt8]

例如，我有280位，我应该有35个UInt8字节数组。我可以想到解决方案，我拿8位并检查第一个是否为真，如果第二个为真，那么总和结果并有价值。我会为我的位数组中的每8位做一次。但我认为这将是一个糟糕的解决方案（它可以工作，但有不必要的计算）。我认为可能有更快的解决方案，有一些转移，所以但我真的很糟糕，所以我正在寻求帮助。感谢

Answer 1

一种可能的解决方案是枚举数组中的所有位并为所有人＆＃34; One＆＃34;位设置UInt8中的相应位阵列：

func bitsToBytes(bits: [Bit]) -> [UInt8] {
    let numBits = bits.count
    let numBytes = (numBits + 7)/8
    var bytes = [UInt8](count : numBytes, repeatedValue : 0)

    for (index, bit) in enumerate(bits) {
        if bit == .One {
            bytes[index / 8] += 1 << (7 - index % 8)
        }
    }

    return bytes
}

主要思想是对于位数组中的给定index， index / 8是字节数组中的对应索引，并且index % 8是字节中的位位置。您可以使用index % 8或7 - index % 8作为班次金额，具体取决于所需的位顺序。

示例：

// 0110 0100  0000 1001
let bits : [Bit] = [.Zero, .One, .One, .Zero,  .Zero, .One, .Zero, .Zero,  .Zero, .Zero, .Zero, .Zero,  .One, .Zero, .Zero, .One]
let bytes = bitsToBytes(bits)
println(bytes) // [100, 9]

或者，你可以＆＃34;内联＆＃34;每组的计算 8位您必须检查哪种解决方案表现更好在你的情况下。

func bitsToBytes(bits: [Bit]) -> [UInt8] {
    let numBits = bits.count
    let numBytes = numBits/8
    var bytes = [UInt8](count : numBytes, repeatedValue : 0)
    for pos in 0 ..< numBytes {
        let val = 128 * bits[8 * pos].toIntMax() + 
            64 * bits[8 * pos + 1].toIntMax() +
            32 * bits[8 * pos + 2].toIntMax() +
            16 * bits[8 * pos + 3].toIntMax() +
            8 * bits[8 * pos + 4].toIntMax() +
            4 * bits[8 * pos + 5].toIntMax() +
            2 * bits[8 * pos + 6].toIntMax() +
            1 * bits[8 * pos + 7].toIntMax()
        bytes[pos] = UInt8(val)
    }
    return bytes
}

这里，为简单起见，如果位数不是8的倍数，则忽略任何多余的位。相同的代码也可以写一点＆＃34; Swiftier＆＃34;如

func bitsToBytes(bits: [Bit]) -> [UInt8] {
    return map(0 ..< bits.count/8) {
        pos in
        let val = 128 * bits[8 * pos].toIntMax() + 
            64 * bits[8 * pos + 1].toIntMax() +
            32 * bits[8 * pos + 2].toIntMax() +
            16 * bits[8 * pos + 3].toIntMax() +
            8 * bits[8 * pos + 4].toIntMax() +
            4 * bits[8 * pos + 5].toIntMax() +
            2 * bits[8 * pos + 6].toIntMax() +
            1 * bits[8 * pos + 7].toIntMax()
        return (UInt8(val))
    }
}

基准：现在这是一个快速而肮脏的基准测试应用程序（下面的代码），比较各种解决方案。它测量转换长度为256的10,000位数组的时间。测试是在MacBook Pro 2,3 GHz Intel Core i7上完成的，以及使用＆＃34; Release＆＃34;编译的代码配置。

Swift 1.1 / Xcode 6.2（6C131e）的结果：

Martin1: 0.0460730195045471
Martin2: 0.0280380249023438
Martin3: 0.0374950170516968
Antonio: 5.85363000631332
Nate   : 4.86936402320862

Swift 1.2 / Xcode 6.3（6D532l）的结果：

Martin1: 0.0228430032730103
Martin2: 0.00573796033859253
Martin3: 0.00732702016830444
Antonio: 0.515677988529205
Nate   : 0.634827971458435

代码：

protocol BitsToBytesConverter {
    var ident : String { get }
    func bitsToBytes(bits: [Bit]) -> [UInt8]
}

class MR1 : BitsToBytesConverter {

    let ident = "Martin1"
    func bitsToBytes(bits: [Bit]) -> [UInt8] {
        let numBits = bits.count
        let numBytes = (numBits + 7)/8
        var bytes = [UInt8](count : numBytes, repeatedValue : 0)

        for (index, bit) in enumerate(bits) {
            if bit == .One {
                bytes[index / 8] += UInt8(1 << (7 - index % 8))
            }
        }

        return bytes
    }
}

class MR2 : BitsToBytesConverter {

    let ident = "Martin2"

    func bitsToBytes(bits: [Bit]) -> [UInt8] {
        let numBits = bits.count
        let numBytes = numBits/8
        var bytes = [UInt8](count : numBytes, repeatedValue : 0)
        for pos in 0 ..< numBytes {
            let val = 128 * bits[8 * pos].toIntMax() + 
                64 * bits[8 * pos + 1].toIntMax() +
                32 * bits[8 * pos + 2].toIntMax() +
                16 * bits[8 * pos + 3].toIntMax() +
                8 * bits[8 * pos + 4].toIntMax() +
                4 * bits[8 * pos + 5].toIntMax() +
                2 * bits[8 * pos + 6].toIntMax() +
                1 * bits[8 * pos + 7].toIntMax()
            bytes[pos] = UInt8(val)
        }
        return bytes
    }
}

class MR3 : BitsToBytesConverter {

    let ident = "Martin3"

    func bitsToBytes(bits: [Bit]) -> [UInt8] {
        return map(0 ..< bits.count/8) {
            pos in
            let val = 128 * bits[8 * pos].toIntMax() + 
                64 * bits[8 * pos + 1].toIntMax() +
                32 * bits[8 * pos + 2].toIntMax() +
                16 * bits[8 * pos + 3].toIntMax() +
                8 * bits[8 * pos + 4].toIntMax() +
                4 * bits[8 * pos + 5].toIntMax() +
                2 * bits[8 * pos + 6].toIntMax() +
                1 * bits[8 * pos + 7].toIntMax()
            return (UInt8(val))
        }
    }
}

class AB : BitsToBytesConverter {

    let ident = "Antonio"

    typealias IntegerType = UInt8

    func bitsToBytes(bits: [Bit]) -> [UInt8] {

        let initial = [IntegerType]()

        return reduce(enumerate(bits), initial) { array, element in
            // The size in bits of a UInt8
            let size = sizeof(IntegerType) * 8

            // Create a mutable copy of the array returned at the previous iteration
            var next = array

            // If it's the first iteration, or an iteration divisible by the size of UInt8, 
            // append a new element to the array
            if element.index % size == 0 {
                next.append(0x00)
            }

            // Shift all bits of the last element to the left
            next[next.count - 1] <<= 1

            // If the current bit is one, add 1 to the rightmost bit
            // Using a logical OR
            if element.element == .One {
                next[next.count - 1] |= 0x01
            }

            return next
        }
    }
}

class NC : BitsToBytesConverter {

    let ident = "Nate   "

    func group<T>(array: [T], byCount groupCount: Int) -> [Slice<T>] {
        // get a list of the start indices
        let startIndices = stride(from: 0, to: array.count, by: groupCount)
        // add `groupCount` to each to get the end indices
        let endIndices = lazy(startIndices).map { advance($0, groupCount, array.count) }

        // zip those together & map onto an array of slices of the input array
        return map(Zip2(startIndices, endIndices)) {
            array[$0.0 ..< $0.1]
        }
    }

    func bitsToByte(bits: Slice<Bit>) -> UInt8 {
        return bits.reduce(0) { accumulated, current in
            accumulated << 1 | (current == .One ? 1 : 0)
        }
    }

    func bitsToBytes(bits: [Bit]) -> [UInt8] {
        return group(bits, byCount: 8).map(bitsToByte)
    }
}


let numBits = 256 // Bits per bit array
let numBitArrays = 10000 // Number of bit arrays

func randomBits() -> [Bit] {
    return map(0 ..< numBits) { _  in
        Bit(rawValue: Int(arc4random_uniform(2)))!
    }
}

func randomBitsArray() -> [[Bit]] {
    return map(0 ..< numBitArrays) { _ in
        randomBits()
    }
}

let bitsArray = randomBitsArray()

func test(conv : BitsToBytesConverter) {
    let x = conv.bitsToBytes([])
    let startTime = NSDate()
    for bits in bitsArray {
        let bytes = conv.bitsToBytes(bits)
    }
    let duration = -startTime.timeIntervalSinceNow
    println("\(conv.ident): \(duration)")
}

test(MR1())
test(MR2())
test(MR3())
test(AB())
test(NC())

Answer 2

这是一个有趣的问题。我将此视为两个较小的问题：（1）如何将Bit数组拆分为Bit数组的数组，其中每个较小的数组是一个字节的位，以及（2）如何将这些较小的数组转换为每个字节。

要解决第一个问题，我们可以编写一个函数，将数组分组为特定大小的片段：

func group<T>(array: [T], byCount groupCount: Int) -> [Slice<T>] {
    // get a list of the start indices
    let startIndices = stride(from: 0, to: s.count, by: groupCount)
    // add `groupCount` to each to get the end indices
    let endIndices = lazy(startIndices).map { advance($0, groupCount, array.count) }

    // zip those together & map onto an array of slices of the input array
    return map(zip(startIndices, endIndices)) {
        array[$0.0 ..< $0.1]
    }
}

要解决第二个问题，我们可以编写一个函数，从Slice<Bit>返回每个group(_:byCount:)并将其转换为UInt8。在每一步，它将值左移一位，然后设置1位，如果该元素为.One：

func bitsToByte(bits: Slice<Bit>) -> UInt8 {
    return bits.reduce(0) { accumulated, current in
        accumulated << 1 | (current == .One ? 1 : 0)
    }
}

最后，您可以依次调用其中的每一个，或者将它们组合起来以获得结果：

// 1111 1111  1000 0000  0000 0001  0101 0101
let bits : [Bit] = [.One, .One, .One, .One, .One, .One, .One, .One,
    .One, .Zero,  .Zero, .Zero, .Zero, .Zero, .Zero, .Zero,
    .Zero,  .Zero, .Zero, .Zero, .Zero, .Zero, .Zero, .One,
    .Zero, .One, .Zero, .One, .Zero, .One, .Zero, .One]
let bytes = group(bits, byCount: 8).map(bitsToByte)
// [255, 128, 1, 85]

Answer 3

如果您更喜欢功能性方法，而代价是更昂贵的计算，那么您可以将reduce与enumerate结合使用。

后者，给定一系列元素，创建一系列(index, element)元组。我们需要索引来知道位位置。

reduce用于将Bit数组缩减为UInt8

数组

typealias IntegerType = UInt8

let initial = [IntegerType]()

let result = reduce(enumerate(bits), initial) { array, element in
    // The size in bits of a UInt8
    let size = sizeof(IntegerType) * 8

    // Create a mutable copy of the array returned at the previous iteration
    var next = array

    // If it's the first iteration, or an iteration divisible by the size of UInt8, 
    // append a new element to the array
    if element.index % size == 0 {
        next.append(0x00)
    }

    // Shift all bits of the last element to the left
    next[next.count - 1] <<= 1

    // If the current bit is one, add 1 to the rightmost bit
    // Using a logical OR
    if element.element == .One {
        next[next.count - 1] |= 0x01
    }

    return next
}

返回的结果是UInt8。

的数组

<强>更新忘记提及如果要转换为其他整数类型，只需更改IntegerType别名。

Answer 4

这里有一些很酷的代码@ martin-r @antonio @ nate-cook，但是在将此代码转换为 Swift 3 供我使用时，我也发现了一些正确性问题。此修订代码段的功能：

Xcode 8，Swift 3.0，在Playground中运行
包括验证功能（已注释掉）
具有位枚举以表示位类型。（比特类型已从Swift 3中删除）
虽然这仍然有定时仪器，但我还没有尝试过真正测试不同方法的速度。似乎我们需要首先确保正确的结果。
只有Martin1和Martin2编码。其他方法留给读者练习;）。

Martin1: 0.112455010414124: hash 0
Martin2: 1.06640499830246: hash 0

无论如何，这是我的代码强烈基于他人的工作。希望有些人觉得这很有用。

import Foundation

enum Bit { case zero, one
    func asInt() -> Int {
        return (self == .one) ? 1 : 0
    }
}

protocol BitsToBytesConverter {
    var ident : String { get }
    func bitsToBytes(bits: [Bit]) -> [UInt8]
}

class MR1 : BitsToBytesConverter {

    let ident = "Martin1"

    func bitsToBytes(bits: [Bit]) -> [UInt8] {
        assert(bits.count % 8 == 0, "Bit array size must be multiple of 8")

        let numBytes = 1 + (bits.count - 1) / 8
        var bytes = [UInt8](repeating: 0, count: numBytes)

        for (index, bit) in bits.enumerated() {
            if bit == .one {
                bytes[index / 8] += UInt8(1 << (7 - index % 8))
            }
        }
        return bytes
    }
}

class MR2 : BitsToBytesConverter {

    let ident = "Martin2"

    func bitsToBytes(bits: [Bit]) -> [UInt8] {
        assert(bits.count % 8 == 0, "Bit array size must be multiple of 8")

        let numBytes = 1 + (bits.count - 1) / 8

        var bytes = [UInt8](repeating : 0, count : numBytes)
        for pos in 0 ..< numBytes {
            let val = 128 * bits[8 * pos].asInt() +
                64 * bits[8 * pos + 1].asInt() +
                32 * bits[8 * pos + 2].asInt() +
                16 * bits[8 * pos + 3].asInt() +
                8 * bits[8 * pos + 4].asInt() +
                4 * bits[8 * pos + 5].asInt() +
                2 * bits[8 * pos + 6].asInt() +
                1 * bits[8 * pos + 7].asInt()
            bytes[pos] = UInt8(val)
        }
        return bytes
    }
}

func format(bits: [Bit]) -> String {
    return bits.reduce("") { (current, next) in
        return current.appending((next == .one) ? "1" : "0")
    }
}
func randomBits(ofLength: Int) -> [Bit] {
    return (0 ..< ofLength).map() { _  in
        Int(arc4random_uniform(2)) == 1 ? .one : .zero
    }
}
func isValid(conv: BitsToBytesConverter, input: [Bit], expected: [UInt8]) -> Bool {
    let actual = conv.bitsToBytes(bits: input)
    var bIsValid = (actual.count == expected.count)
    if bIsValid {
        for index in 0 ..< expected.count {
            if actual[index] != expected[index] {
                bIsValid = false
                break
            }
        }
    }
    return bIsValid
}
func validateAll() {
    let input0: [Bit] = [.one, .zero, .one, .zero, .one, .zero, .one, .zero]
    let expected0: [UInt8] = [170]

    let input1: [Bit] = (0..<8).map { _ in .zero }
    let expected1: [UInt8] = [0]

    let input2: [Bit] = (0..<8).map { _ in .one }
    let expected2: [UInt8] = [255]

    let input3 = input1 + input2
    let expected3 = expected1 + expected2

    let input4 = input2 + input1
    let expected4 = expected2 + expected1

    let inputs = [input0, input1, input2, input3, input4]
    let expecteds = [expected0, expected1, expected2, expected3, expected4]

    let convs: [BitsToBytesConverter] = [MR1(), MR2()]

    for inIndex in 0 ..< inputs.count {
        let input = inputs[inIndex]
        let expected = expecteds[inIndex]

        let formattedBits = format(bits: input)
        print("Checking: \(formattedBits) -> \(expected)")

        for conv in convs {
            let bIsValid = isValid(conv: conv, input: input, expected: expected)
            print("\(conv.ident) valid = \(bIsValid)")
        }
    }
}
func timeTest(conv : BitsToBytesConverter, bitsArray: [[Bit]]) {
    // Prime compilation, caching, ...
    let _ = conv.bitsToBytes(bits: bitsArray[0])

    let startTime = NSDate()
    var hash = 0

    for bits in bitsArray {
        let _ = conv.bitsToBytes(bits: bits)

        // Hash to compare results across methods
        //let result = conv.bitsToBytes(bits: bits)
        //let bString = format(bits: bits)
        //print("Bits: \(bString): Bytes: \(result)")
        //hash = result.reduce(0) { (previous, next) in
        //    return previous + next.hashValue
        //}
    }
    let duration = -startTime.timeIntervalSinceNow
    print("\(conv.ident): \(duration): hash \(hash)")
}
func testAll() {
    let numBits = 128 // Bits per bit array
    let numBitArrays = 100 // Number of bit arrays

    let testValues = (0 ..< numBitArrays).map() { _ in
        randomBits(ofLength: numBits)
    }
    timeTest(conv: MR1(), bitsArray: testValues)
    timeTest(conv: MR2(), bitsArray: testValues)
}
//validateAll()
testAll()

Answer 5

快速 5

func convertBitsToBytes(bits : [Bit]) -> [UInt8] {
    let numBits = bits.count
    let numBytes = (numBits + 7)/8
    var bytes = [UInt8](unsafeUninitializedCapacity: numBytes) { (buffer, initializedCount) in
        
    }
    
    for (index, bit) in bits.enumerated() {
        if bit == .one {
            bytes[index / 8] += UInt8(1 << (7 - index % 8))
        }
    }
    return bytes
}

Swift位数组到字节数组（UInt8数组）

5 个答案: