在带有附加参数的numpy数组上使用scipy.optimize.root

Question

给定优化问题(1) (如下所示)，其中p_i，p'_i和w_ji用于i=0,...,6889，我想使用Levenberg-Marquardt方法来使用scipy.optimize.root为R_j和v_j找到最佳解决方案(我对任何其他建议都持开放态度)。

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但是，我不知道如何设置需要传递给root的可调用函数。到目前为止，我所拥有的就是这个，这显然是错误的。

def fun(x, (old_points, new_points, weights, n_joints)):
    """
    :param x: variable to optimize. It is supposed to encapsulate R and v from (1)
    :param old_points: original vertex positions, (6890,3) numpy array
    :param new_points: transformed vertex positions, (6890,3) numpy array
    :param weights: weight matrix obtained from spectral clustering, (n_joints, 6890) numpy array
    :param n_joints: number of joints
    :return: non-linear cost function to find the root of
    """
    # Extract rotations and offsets
    R = np.array([(np.array(x[j * 15:j * 15 + 9]).reshape(3, 3)) for j in range(n_joints)])
    v = np.array([(np.array(x[j * 15 + 9:j * 15 + 12])) for j in range(n_joints)])

    # Use equation (1) for the non-linear pass.
    # R_j p_i
    Rp = np.einsum('jkl,il', x, old_points) # x shall replace R
    # w_ji (Rp_ij + v_j)
    wRpv = np.einsum('ji,ijk->ik', weights, Rp + x) # x shall replace v

    # Set up a non-linear cost function, then compute the squared norm.
    d = new_points - wRpv
    result = np.einsum('ik,ik', d, d)

    return result

编辑:现在这是正确的结果。

Answer 1

使用你原来的fun (但给它一个更好的名字)

def fun(x, (old_points, new_points, weights, n_joints)):
    """
    :param x: variable to optimize. It is supposed to encapsulate R and v from (1)
    :param old_points: original vertex positions, (6890,3) numpy array
    :param new_points: transformed vertex positions, (6890,3) numpy array
    :param weights: weight matrix obtained from spectral clustering, (n_joints, 6890) numpy array
    :param n_joints: number of joints
    :return: non-linear cost function to find the root of
    """
    # Extract rotations and offsets
    R = np.array([(np.array(x[j * 15:j * 15 + 9]).reshape(3, 3)) for j in range(n_joints)])
    v = np.array([(np.array(x[j * 15 + 9:j * 15 + 12])) for j in range(n_joints)])

    # Use equation (1) for the non-linear pass.
    # R_j p_i
    Rp = np.einsum('jkl,il', x, old_points) # x shall replace R
    # w_ji (Rp_ij + v_j)
    wRpv = np.einsum('ji,ijk->ik', weights, Rp + x) # x shall replace v

    # Set up a non-linear cost function, then compute the squared norm.
    d = new_points - wRpv
    result = np.einsum('ik,ik', d, d)

    return result

对它做一个闭包，这样它就只有一个输入(您正在求解的变量)：

old_points = ...
new_points = ...
weights = ...
rv = ...
n_joints = ...
def cont_function(x):
    return fun(x, old_points, new_points, weights, rv, n_joints)

现在尝试在roots中使用cost_function