回答与疑问：
1. 非线性关系的数据拟合，可以先将自变量转为非线性。如转化为多项式（sklearn的PolynomialFeatures）。再用线性回归的方法去拟合。
2. 请问老师对于求解逆矩阵有没有什么高效的方法？
附上以前写的polyfit方法，请老师指点。谢谢
    def oneDPolynomiaTransform(self, x_origin):
        '''
        @description: generate polynomial for 1D input data. rule: [x0, x0^2, x0^3,...,x0^degreee]
        @param {type} x_origin- data before transformed[nX1]
        @return: x_transformed- data after transformed[nXdegree]
        '''
        len_features = len(x_origin)
        # polynomial feature data after transformed.
        x_transformed = np.array([])
        for i in range(len_features):
            for j in range(self.degree+1):
                x_transformed = np.append(
                    x_transformed, [(x_origin[i])**(j)])
        x_transformed = x_transformed.reshape(-1, self.degree+1)
        return x_transformed

from numpy import *

x=mat([[1,3,-7],[2,5,4],[-3,-7,-2],[1,4,-12]])

y=mat([[-7.5],[5.2],[-7.5],[-15]])

print("\n系数矩阵B:\n",(x.transpose().dot(x)).I.dot(x.transpose()).dot(y))

系数矩阵B:

 [[12.01208791]

 [-4.35934066]

 [ 0.82527473]]

"""
思考题同理
"""
x = np.mat([[1, 3, -7], [2, 5, 4], [-3, -7, -2], [1, 4, -12]])
y = np.mat([[-7.5], [5.2], [-7.5], [-15]])
print("\n 系数矩阵 B: \n", (x.transpose().dot(x)).I.dot(x.transpose()).dot(y))

 系数矩阵 B:
 [[12.01208791]
 [-4.35934066]
 [ 0.82527473]]