Question

我一直在尝试将代码从Python重写为Swift，但我仍然坚持应该将最小二乘解法返回到线性矩阵方程的函数。有没有人知道用Swift编写的库，它有一个与 numpy.linalg.lstsq 等效的方法？我很感激你的帮助。

Python代码：

a = numpy.array([[p2.x-p1.x,p2.y-p1.y],[p4.x-p3.x,p4.y-p3.y],[p4.x-p2.x,p4.y-p2.y],[p3.x-p1.x,p3.y-p1.y]])
b = numpy.array([number1,number2,number3,number4])
res = numpy.linalg.lstsq(a,b) 
result = [float(res[0][0]),float(res[0][1])]
return result

到目前为止的Swift代码：

var matrix1 = [[p2.x-p1.x, p2.y-p1.y],[p4.x-p3.x, p4.y-p3.y], [p4.x-p2.x, p4.y-p2.y], [p3.x-p1.x, p3.y-p1.y]]
var matrix2 = [number1, number2, number3, number4]

Answer 1

Accelerate框架包含LAPACK线性代数包，它具有DGELS功能，可以解决欠线或超定线性系统。来自文档：

DGELS解决了超定或欠定的实际线性系统使用QR或LQ涉及M-by-N矩阵A或其转置 A的分解。假设A具有完全排名。

这是一个如何从Swift中使用该函数的示例。它本质上是this C sample code的翻译。

func solveLeastSquare(A A: [[Double]], B: [Double]) -> [Double]? {
    precondition(A.count == B.count, "Non-matching dimensions")

    var mode = Int8(bitPattern: UInt8(ascii: "N")) // "Normal" mode
    var nrows = CInt(A.count)
    var ncols = CInt(A[0].count)
    var nrhs = CInt(1)
    var ldb = max(nrows, ncols)

    // Flattened columns of matrix A
    var localA = (0 ..< nrows * ncols).map {
        A[Int($0 % nrows)][Int($0 / nrows)]
    }

    // Vector B, expanded by zeros if ncols > nrows
    var localB = B
    if ldb > nrows {
        localB.appendContentsOf([Double](count: ldb - nrows, repeatedValue: 0.0))
    }

    var wkopt = 0.0
    var lwork: CInt = -1
    var info: CInt = 0

    // First call to determine optimal workspace size
    dgels_(&mode, &nrows, &ncols, &nrhs, &localA, &nrows, &localB, &ldb, &wkopt, &lwork, &info)
    lwork = Int32(wkopt)

    // Allocate workspace and do actual calculation
    var work = [Double](count: Int(lwork), repeatedValue: 0.0)
    dgels_(&mode, &nrows, &ncols, &nrhs, &localA, &nrows, &localB, &ldb, &work, &lwork, &info)

    if info != 0 {
        print("A does not have full rank; the least squares solution could not be computed.")
        return nil
    }
    return Array(localB.prefix(Int(ncols)))
}

一些注意事项：

dgels_()修改传递的矩阵和矢量数据，并期望矩阵为＆＃34;扁平＆＃34;包含A列的数组。此外，右侧预计为长度为max(M, N)的数组。因此，首先将输入数据复制到局部变量。
所有参数必须通过引用dgels_()来传递，这就是原因它们都存储在var s。
C整数是一个32位整数，它可以进行一些转换必要Int和CInt。

示例1：来自http://www.seas.ucla.edu/~vandenbe/103/lectures/ls.pdf的超定系统。

let A = [[ 2.0, 0.0 ],
         [ -1.0, 1.0 ],
         [ 0.0, 2.0 ]]
let B = [ 1.0, 0.0, -1.0 ]
if let x = solveLeastSquare(A: A, B: B) {
    print(x) // [0.33333333333333326, -0.33333333333333343]
}

示例2：欠定系统，最小规范解决x_1 + x_2 + x_3 = 1.0。

let A = [[ 1.0, 1.0, 1.0 ]]
let B = [ 1.0 ]
if let x = solveLeastSquare(A: A, B: B) {
    print(x) // [0.33333333333333337, 0.33333333333333337, 0.33333333333333337]
}

更新 Swift 3 和 Swift 4：

func solveLeastSquare(A: [[Double]], B: [Double]) -> [Double]? {
    precondition(A.count == B.count, "Non-matching dimensions")

    var mode = Int8(bitPattern: UInt8(ascii: "N")) // "Normal" mode
    var nrows = CInt(A.count)
    var ncols = CInt(A[0].count)
    var nrhs = CInt(1)
    var ldb = max(nrows, ncols)

    // Flattened columns of matrix A
    var localA = (0 ..< nrows * ncols).map { (i) -> Double in
        A[Int(i % nrows)][Int(i / nrows)]
    }

    // Vector B, expanded by zeros if ncols > nrows
    var localB = B
    if ldb > nrows {
        localB.append(contentsOf: [Double](repeating: 0.0, count: Int(ldb - nrows)))
    }

    var wkopt = 0.0
    var lwork: CInt = -1
    var info: CInt = 0

    // First call to determine optimal workspace size
    var nrows_copy = nrows // Workaround for SE-0176
    dgels_(&mode, &nrows, &ncols, &nrhs, &localA, &nrows_copy, &localB, &ldb, &wkopt, &lwork, &info)
    lwork = Int32(wkopt)

    // Allocate workspace and do actual calculation
    var work = [Double](repeating: 0.0, count: Int(lwork))
    dgels_(&mode, &nrows, &ncols, &nrhs, &localA, &nrows_copy, &localB, &ldb, &work, &lwork, &info)

    if info != 0 {
        print("A does not have full rank; the least squares solution could not be computed.")
        return nil
    }
    return Array(localB.prefix(Int(ncols)))
}

将最小二乘解返回到线性矩阵方程

1 个答案: