最小二乘法适合python 3d曲面

Question

我想将我的表面方程拟合到一些数据。 我已经尝试过scipy.optimize.leastsq，但是由于无法指定范围，因此无法获得结果。 我也尝试了scipy.optimize.least_squares，但它给了我一个错误：

ValueError: too many values to unpack

我的方程式是：

f(x,y,z)=(x-A+y-B)/2+sqrt(((x-A-y+B)/2)^2+C*z^2)

应该找到参数A，B，C，以便当以下点用于x，y，z时，上述方程将尽可能接近零：

    [
   [-0.071, -0.85, 0.401],
   [-0.138, -1.111, 0.494],
   [-0.317, -0.317, -0.317],
   [-0.351, -2.048, 0.848]
   ]

范围为A> 0，B> 0，C> 1

我应该如何获得这样的身材？ python中执行此操作的最佳工具是什么。 我搜索了有关如何拟合3d曲面的示例，但是涉及函数拟合的大多数示例都是关于直线或平面拟合的。

Answer 1

我已经编辑了此答案，以提供一个更通用的示例，说明如何使用scipy的常规optimize.minimize方法以及scipy的optimize.least_squares方法来解决此问题。

---

首先让我们设置问题：

import numpy as np
import scipy.optimize

# ===============================================
# SETUP: define common compoments of the problem


def our_function(coeff, data):
    """
    The function we care to optimize.

    Args:
        coeff (np.ndarray): are the parameters that we care to optimize.
        data (np.ndarray): the input data
    """
    A, B, C = coeff
    x, y, z = data.T
    return (x - A + y - B) / 2 + np.sqrt(((x - A - y + B) / 2) ** 2 + C * z ** 2)


# Define some training data
data = np.array([
    [-0.071, -0.85, 0.401],
    [-0.138, -1.111, 0.494],
    [-0.317, -0.317, -0.317],
    [-0.351, -2.048, 0.848]
])
# Define training target
# This is what we want the target function to be equal to
target = 0

# Make an initial guess as to the parameters
# either a constant or random guess is typically fine
num_coeff = 3
coeff_0 = np.ones(num_coeff)
# coeff_0 = np.random.rand(num_coeff)

---

严格来说，这不是最小二乘，但类似的东西呢？ 此解决方案就像在问题上扔了一把大锤。 可能存在一种使用最小二乘法来使用SVD求解器更有效地求解的方法，但是如果您只是在寻找答案，则scipy.optimize.minimize会找到一个。

# ===============================================
# FORMULATION #1: a general minimization problem

# Here the bounds and error are all specified within the general objective function
def general_objective(coeff, data, target):
    """
    General function that simply returns a value to be minimized.
    The coeff will be modified to minimize whatever the output of this function
    may be.
    """
    # Constraints to keep coeff above 0
    if np.any(coeff < 0):
        # If any constraint is violated return infinity
        return np.inf
    # The function we care about
    prediction = our_function(coeff, data)
    # (optional) L2 regularization to keep coeff small
    # (optional) reg_amount = 0.0
    # (optional) reg = reg_amount * np.sqrt((coeff ** 2).sum())
    losses = (prediction - target) ** 2
    # (optional) losses += reg
    # Return the average squared error
    loss = losses.sum()
    return loss


general_result = scipy.optimize.minimize(general_objective, coeff_0,
                                         method='Nelder-Mead',
                                         args=(data, target))
# Test what the squared error of the returned result is
coeff = general_result.x
general_output = our_function(coeff, data)
print('====================')
print('general_result =\n%s' % (general_result,))
print('---------------------')
print('general_output = %r' % (general_output,))
print('====================')

输出看起来像这样：

====================
general_result =
 final_simplex: (array([[  2.45700466e-01,   7.93719271e-09,   1.71257109e+00],
       [  2.45692680e-01,   3.31991619e-08,   1.71255150e+00],
       [  2.45726858e-01,   6.52636219e-08,   1.71263360e+00],
       [  2.45713989e-01,   8.06971686e-08,   1.71260234e+00]]), array([ 0.00012404,  0.00012404,  0.00012404,  0.00012404]))
           fun: 0.00012404137498459109
       message: 'Optimization terminated successfully.'
          nfev: 431
           nit: 240
        status: 0
       success: True
             x: array([  2.45700466e-01,   7.93719271e-09,   1.71257109e+00])
---------------------
general_output = array([ 0.00527974, -0.00561568, -0.00719941,  0.00357748])
====================

---

我在文档中发现，将其调整为实际最小二乘法所需要做的就是指定计算残差的函数。

# ===============================================
# FORMULATION #2: a special least squares problem

# Here all that is needeed is a function that computes the vector of residuals
# the optimization function takes care of the rest
def least_squares_residuals(coeff, data, target):
    """
    Function that returns the vector of residuals between the predicted values
    and the target value. Here we want each predicted value to be close to zero
    """
    A, B, C = coeff
    x, y, z = data.T
    prediction = our_function(coeff, data)
    vector_of_residuals = (prediction - target)
    return vector_of_residuals


# Here the bounds are specified in the optimization call
bound_gt = np.full(shape=num_coeff, fill_value=0, dtype=np.float)
bound_lt = np.full(shape=num_coeff, fill_value=np.inf, dtype=np.float)
bounds = (bound_gt, bound_lt)

lst_sqrs_result = scipy.optimize.least_squares(least_squares_residuals, coeff_0,
                                               args=(data, target), bounds=bounds)
# Test what the squared error of the returned result is
coeff = lst_sqrs_result.x
lst_sqrs_output = our_function(coeff, data)
print('====================')
print('lst_sqrs_result =\n%s' % (lst_sqrs_result,))
print('---------------------')
print('lst_sqrs_output = %r' % (lst_sqrs_output,))
print('====================')

输出是：

====================
lst_sqrs_result =
 active_mask: array([ 0, -1,  0])
        cost: 6.197329866927735e-05
         fun: array([ 0.00518416, -0.00564099, -0.00710112,  0.00385024])
        grad: array([ -4.61826888e-09,   3.70771396e-03,   1.26659198e-09])
         jac: array([[-0.72611025, -0.27388975,  0.13653112],
       [-0.74479565, -0.25520435,  0.1644325 ],
       [-0.35777232, -0.64222767,  0.11601263],
       [-0.77338046, -0.22661953,  0.27104366]])
     message: '`gtol` termination condition is satisfied.'
        nfev: 13
        njev: 13
  optimality: 4.6182688779976278e-09
      status: 1
     success: True
           x: array([  2.46392438e-01,   5.39025298e-17,   1.71555150e+00])
---------------------
lst_sqrs_output = array([ 0.00518416, -0.00564099, -0.00710112,  0.00385024])
====================