在没有numpy polyfit的情况下在python中拟合二次函数 [英] Fitting a quadratic function in python without numpy polyfit

问题描述

我正在尝试将二次函数拟合到某些数据，并且在不使用numpy的polyfit函数的情况下尝试这样做.

I am trying to fit a quadratic function to some data, and I'm trying to do this without using numpy's polyfit function.

数学上，我试图关注此网站 https://neutrium .net/mathematics/least-squares-fitting-of-a-polynomial/，但是我不知为何我做对了.如果有人可以帮助我，那就太好了；或者，如果您可以建议另一种方法，那也就很棒.

Mathematically I tried to follow this website https://neutrium.net/mathematics/least-squares-fitting-of-a-polynomial/ but somehow I don't think that I'm doing it right. If anyone could assist me that would be great, or If you could suggest another way to do it that would also be awesome.

到目前为止我已经尝试过的:

What I've tried so far:

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

ones = np.ones(3)
A = np.array( ((0,1),(1,1),(2,1)))
xfeature = A.T[0]
squaredfeature = A.T[0] ** 2
b = np.array( (1,2,0), ndmin=2 ).T
b = b.reshape(3)

features = np.concatenate((np.vstack(ones), np.vstack(xfeature), np.vstack(squaredfeature)), axis = 1)
featuresc = features.copy()
print(features)
m_det = np.linalg.det(features)
print(m_det)
determinants = []
for i in range(3):
    featuresc.T[i] = b
    print(featuresc)
    det = np.linalg.det(featuresc)
    determinants.append(det)
    print(det)
    featuresc = features.copy()

determinants = determinants / m_det
print(determinants)
plt.scatter(A.T[0],b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
p2 = np.polyfit(A.T[0],b,2)
plt.plot(u, np.polyval(p2,u), 'b--')
plt.show()

如您所见，我的曲线与nnumpy的polyfit曲线相比效果不佳.

As you can see my curve doesn't compare well to nnumpy's polyfit curve.

更新: 我遍历了代码，并删除了所有愚蠢的错误，现在，当我尝试使其超过3分时，它可以工作，但是我不知道如何超过3分.

Update: I went through my code and removed all the stupid mistakes and now it works, when I try to fit it over 3 points, but I have no idea how to fit over more than three points.

这是新代码:

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

ones = np.ones(3)
A = np.array( ((0,1),(1,1),(2,1)))
xfeature = A.T[0]
squaredfeature = A.T[0] ** 2
b = np.array( (1,2,0), ndmin=2 ).T
b = b.reshape(3)

features = np.concatenate((np.vstack(ones), np.vstack(xfeature), np.vstack(squaredfeature)), axis = 1)
featuresc = features.copy()
print(features)
m_det = np.linalg.det(features)
print(m_det)
determinants = []
for i in range(3):
    featuresc.T[i] = b
    print(featuresc)
    det = np.linalg.det(featuresc)
    determinants.append(det)
    print(det)
    featuresc = features.copy()

determinants = determinants / m_det
print(determinants)
plt.scatter(A.T[0],b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
p2 = np.polyfit(A.T[0],b,2)
plt.plot(u, np.polyval(p2,u), 'r--')
plt.show()

推荐答案

实际上使用最小二乘法来求解系统，而不是使用Cramer规则.请记住，只有当您拥有的点总数等于多项式的期望阶数加1时，克莱默法则才有效. 如果您没有这个，那么Cramer规则将不起作用，因为您正试图找到问题的确切解决方案.如果您有更多点，则该方法不合适，因为我们将创建一个超定方程组.

Instead using Cramer's Rule, actually solve the system using least squares. Remember that Cramer's Rule will only work if the total number of points you have equals the desired order of polynomial plus 1. If you don't have this, then Cramer's Rule will not work as you're trying to find an exact solution to the problem. If you have more points, the method is unsuitable as we will create an overdetermined system of equations.

要将其调整为更多点，请

To adapt this to more points, numpy.linalg.lstsq would be a better fit as it solves the solution to the Ax = b by computing the vector x that minimizes the Euclidean norm using the matrix A. Therefore, remove the y values from the last column of the features matrix and solve for the coefficients and use numpy.linalg.lstsq to solve for the coefficients:

import numpy as np
import matplotlib.pyplot as plt


ones = np.ones(4)
xfeature = np.asarray([0,1,2,3])
squaredfeature = xfeature ** 2
b = np.asarray([1,2,0,3])

features = np.concatenate((np.vstack(ones),np.vstack(xfeature),np.vstack(squaredfeature)), axis = 1) # Change - remove the y values

determinants = np.linalg.lstsq(features, b)[0] # Change - use least squares
plt.scatter(xfeature,b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
plt.show()

我现在得到了这个图，它与图中的虚线曲线相匹配，也与numpy.polyfit给您的内容相匹配:

I get this plot now, which matches what the dashed curve is in your graph, also matching what numpy.polyfit gives you:

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