在没有 numpy polyfit 的情况下在 python 中拟合二次函数

我正在尝试将二次函数拟合到某些数据，并且我尝试在不使用 numpy 的 polyfit 函数的情况下执行此操作。

从数学上讲我试图关注这个网站https://neutrium.net/mathematics/least-squares-fitting-of-a-polynomial/ https://neutrium.net/mathematics/least-squares-fitting-of-a-polynomial/但不知怎的，我不认为我做得对。如果有人可以帮助我，那就太好了，或者如果你能建议另一种方法来做到这一点，那就太棒了。

到目前为止我尝试过的：

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()

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

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()

实际上，使用最小二乘法求解系统，而不是使用克莱默法则。请记住，只有当您拥有的总点数等于所需的多项式阶数加 1 时，克莱默法则才有效。 如果你没有这个，那么当你试图找到问题的精确解决方案时，克莱默规则将不起作用。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。

为了适应更多的点，numpy.linalg.lstsq https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.linalg.lstsq.html会更合适，因为它解决了Ax = b通过计算向量x使用矩阵最小化欧几里得范数A。因此，删除y特征矩阵最后一列的值并求解系数并使用numpy.linalg.lstsq求解系数：

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，它与图中的虚线相匹配，也与numpy.polyfit给你：