返回线性矩阵方程的最小二乘解的函数

我一直在尝试将代码从 Python 重写为 Swift，但我被困在应该返回线性矩阵方程的最小二乘解的函数上。有谁知道用 Swift 编写的库，它具有与numpy.linalg.lstsq http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.linalg.lstsq.html？我将非常感谢你的帮助。

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]

加速框架包括LAPACK http://www.netlib.org/lapack/index.html线性代数包， 其中有一个DGELS http://www.netlib.org/lapack/explore-html/d7/d3b/group__double_g_esolve_ga1df516c81d3e902cca1fc79a7220b9cb.html#ga1df516c81d3e902cca1fc79a7220b9cb函数来求解欠定或超定线性系统。从文档中：

DGELS 求解超定或欠定实线性系统 涉及 M×N 矩阵 A 或其转置，使用 QR 或 LQ A 的因式分解。假设 A 具有满秩。

以下是如何从 Swift 使用该函数的示例。 它本质上是一个翻译这个C示例代码 https://software.intel.com/sites/products/documentation/doclib/mkl_sa/11/mkl_lapack_examples/dgels_ex.c.htm.

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_()修改传递的矩阵和向量数据，并期望 矩阵作为“平面”数组，包含以下列A。 此外，右侧预计是一个具有长度的数组max(M, N)。 因此，输入数据首先被复制到局部变量中。
• 所有参数必须通过引用传递dgels_()， 这就是为什么 它们都存储在vars.
• C 整数是一个 32 位整数，它在Int and CInt必要的。

示例1：超定系统，由http://www.seas.ucla.edu/~vandenbe/103/lectures/ls.pdf http://www.seas.ucla.edu/~vandenbe/103/lectures/ls.pdf.

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，最低范数 解决方案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]
}

更新为Swift 3 and 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)))
}