如何使用具有非线性约束的 LMFIT 解决最小二乘问题？

Question

我手头有一个最小二乘问题：Ax=b

没有任何约束，可以如下解决：

import numpy as np
from lmfit import Model, Parameters, Minimizer, report_fit
from numpy.linalg import lstsq

A = np.array([[ 143, -144, -343, 56, 98, 54]
,[-7632,  77, -277, 63, 999, 345 ]
,[ 55, -75,   74, 744, 765, 77]
,[-22 , 177, -28, 12, 54, 577]
,[-848 , -433 , 121, 54, 643, 88]
,[98, 75, 155, 87, 23, 54]
,[12, 56, 44, 44, 86, 46]
,[75, 22, 111, 965, 486, 345]])

b = [-114, -734, -270,  577,  676, 122, 245, 654]
b = np.reshape(b, (8, 1))

x, residuals, rnk, s = lstsq(A, b, rcond=-1)
print (" x =", x)

 x = [[-0.04886372]
 [-2.38958265]
 [ 1.44719216]
 [ 0.42944648]
 [-1.23440929]
 [ 1.98725466]]

我想在每对结果上添加一个重复约束，并通过使用 LMFIT 来解决这个问题，因为定义约束似乎更方便。我试过这个，但它不起作用。约束条件是，例如，对于第一对结果，x1 必须是正数，-x1 <= x2 <= x1 和下一对结果相同。

    def fcn2min(params, A, b):

        x1 = params['x1']
        x2 = params['x2']
        x3 = params['x3']
        x4 = params['x4']
        x5 = params['x5']
        x6 = params['x6']
        x = np.array([x1, x2, x3, x4, x5, x6])
        x = np.reshape(x, (6, 1))

        return np.sum((np.dot(A, x)-b)**2)


    params = Parameters()
    params.add('x1', min=0)
    params.add('xangle1', value=0.05, vary=True, min=(-np.pi/2)+(0.00000000001), max=(np.pi/2)-(0.00000000001))
    params.add('x2', expr='(x1)*(sin(xangle1))')
    params.add('x3', min=0)
    params.add('xangle2', value=0.05, vary=True, min=(-np.pi/2)+(0.00000000001), max=(np.pi/2)-(0.00000000001))
    params.add('x4', expr='(x3)*(sin(xangle2))')
    params.add('x5', min=0)
    params.add('xangle3', value=0.05, vary=True, min=(-np.pi/2)+(0.00000000001), max=(np.pi/2)-(0.00000000001))
    params.add('x6', expr='(x5)*(sin(xangle3))')


    mini = Minimizer(fcn2min, params, fcn_args=(A, b))
    out = mini.leastsq()

    print (" report_fit(result) =", report_fit(out.params))

首先，它不起作用，我很难定义和解决问题。其次，手动定义大规模问题的参数和约束非常困难，假设x 有 100 或 1000 个元素。 感谢您对处理此问题的任何帮助。谢谢你。

Answer 1

说出你实际得到了什么总是有帮助的。说“它不起作用”基本上是没有用的。您应该报告输出和收到的任何错误消息。包括对您认为应该获得的内容的描述有时也很有帮助。

几点：

• 您需要不返回np.sum((np.dot(A, x)-b)**2) 并改为返回np.dot(A, x)-b。拟合将进行平方和求和。
• 您应该为变量参数提供明确的初始值。

这样，您将获得一个可以完成并满足您的约束的合身：

def fcn2min(params, A, b):
    x1 = params['x1']
    x2 = params['x2']
    x3 = params['x3']
    x4 = params['x4']
    x5 = params['x5']
    x6 = params['x6']
    x = np.array([x1, x2, x3, x4, x5, x6]).reshape((6, 1))
    return  np.dot(A, x) - b

params = Parameters()
params.add('x1', 0.25, min=0)
params.add('xangle1', value=0.05, vary=True, min=(-np.pi/2)+(0.00000000001), max=(np.pi/2)-(0.00000000001))
params.add('x2', expr='(x1)*(sin(xangle1))')
params.add('x3', 0.25, min=0)
params.add('xangle2', value=0.05, vary=True, min=(-np.pi/2)+(0.00000000001), max=(np.pi/2)-(0.00000000001))
params.add('x4', expr='(x3)*(sin(xangle2))')
params.add('x5', 0.25, min=0)
params.add('xangle3', value=0.05, vary=True, min=(-np.pi/2)+(0.00000000001), max=(np.pi/2)-(0.00000000001))
params.add('x6', expr='(x5)*(sin(xangle3))')

mini = Minimizer(fcn2min, params, fcn_args=(A, b))
out = mini.leastsq()
report_fit(out, min_correl=0.5)

将打印出来：

[[Fit Statistics]]
    # fitting method   = leastsq
    # function evals   = 75
    # data points      = 8
    # variables        = 6
    chi-square         = 597869.757
    reduced chi-square = 298934.879
    Akaike info crit   = 101.773493
    Bayesian info crit = 102.250143
[[Variables]]
    x1:       0.21393627 +/- 0.19714856 (92.15%) (init = 0.25)
    xangle1: -0.01575861 +/- 9.59829799 (60908.27%) (init = 0.05)
    x2:      -0.00337120 +/- 2.05111043 (60842.16%) == '(x1)*(sin(xangle1))'
    x3:       0.89232913 +/- 1.36874096 (153.39%) (init = 0.25)
    xangle2: -0.86188487 +/- 2.50527077 (290.67%) (init = 0.05)
    x4:      -0.67734114 +/- 0.99201291 (146.46%) == '(x3)*(sin(xangle2))'
    x5:       0.89482649 +/- 1.32577015 (148.16%) (init = 0.25)
    xangle3:  1.57078737 +/- 398996.439 (25401047.11%) (init = 0.05)
    x6:       0.89482649 +/- 4.38420893 (489.95%) == '(x5)*(sin(xangle3))'
[[Correlations]] (unreported correlations are < 0.500)
    C(x1, x5)           =  0.878
    C(x5, xangle3)      =  0.853
    C(xangle1, xangle3) =  0.829
    C(xangle1, x5)      =  0.808
    C(x1, xangle2)      = -0.802
    C(x3, xangle2)      =  0.732
    C(xangle2, x5)      = -0.729
    C(x1, xangle1)      =  0.663
    C(x1, xangle3)      =  0.640
    C(xangle2, xangle3) = -0.548
    C(xangle1, xangle2) = -0.531

不确定性非常非常高，这表明您的模型可能不适合约束条件。