如何解决使用回归模型找到固定输出的最小输入组合
问题:
两个变量(x1,x2)
和两个输出(y1,y2)
,其数据可用于计算输入和输出之间的关系。
最后,我想知道为我提供特定的x1
和y1
值的最小y2
。
我到目前为止已考虑的方法:
为了对x1,x2
和y1,y2
之间的关系使用线性回归模型进行曲线拟合,我仅对示例代码进行了少量修改。 x1
和x2
之间的关系被扩展为适合三次函数,因为这给了我较低的均方误差。
因此输入数据适合的方程式为
y1 = a0 + a1*x1 + a2*x2 + a3*x1^2 + a4*x1*x2 + a5*x2^2 + a6 *x1^3 + a7*x1^2*x2 + a8*x1*x2^2 + a9*x2^3
y2 = b0 * intercept + b1*x1 + b2*x2 + b3*x1^2 + b4*x1*x2 + b5*x2^2 + b6*x1^3 + b7*x1^2*x2 + b8*x1*x2^2 + b9*x2^3
道歉,它们可能很难阅读,但不允许我将它们发布为图像(底部是LaTeX代码)。
以下代码中的线性回归模型(a,b)
和coeffs1
已经解决了方程的系数coeffs2
的问题。
from numpy import array,hstack,math
from sklearn.model_selection import train_test_split
from sklearn.metrics import mean_squared_error
from sklearn.linear_model import LinearRegression
from sklearn.multioutput import MultiOutputRegressor
from sklearn.preprocessing import PolynomialFeatures
import sympy as sym
def create_data(n):
# Input data
x1 = array([0,10,20,30,40,50,60,60]).reshape(n,1)
x2 = array([100,200,300,400,500,100,500]).reshape(n,1)
# Corresponding outputs
y1 = array([350.7214942,700.9404275,1049.29659,1392.818473,1727.293514,345.418542,690.4277426,1033.665635,1372.435114,1704.064163,329.6311055,658.9473636,986.6760855,1310.013523,1627.176216,303.8203576,607.4535149,909.770934,1207.960648,1499.905656,268.7786346,537.4766415,805.2109169,1069.428571,1327.209942,225.5578289,451.1324693,676.1020492,898.3732076,1114.408884,175.4807202,351.02104,526.3101109,699.6920184,867.7300064]).reshape(n,1)
y2 = array([12.06118197,13.2332737,14.93878735,16.94583691,19.09960095,11.52121175,12.23713054,13.62566473,15.44234451,17.5104543,10.97161994,11.18544616,12.21149801,13.81244743,15.77033639,10.42959162,10.09739159,10.709275,12.07256479,13.9161106,9.913894093,8.999435475,9.13551907,10.24129286,11.98680831,9.445995362,7.928597333,7.508774345,8.325111871,10.02131756,9.048387975,6.938567678,5.861061317,6.345192659,8.054102881]).reshape(n,1)
# Combine and form inputs into a third order polynomial
X = hstack((x1,x2))
poly = PolynomialFeatures(degree=3)
X = poly.fit_transform(X)
Y = hstack((y1,y2))
return X,Y
n = 35
X,Y = create_data(n)
xtrain,xtest,ytrain,ytest = train_test_split(X,Y,test_size=0.5)
print("xtrain:",xtrain.shape,"ytrain:",ytrain.shape)
print("xtest:",xtest.shape,"ytest:",ytest.shape)
lr = LinearRegression(fit_intercept=True)
model = MultiOutputRegressor(estimator=lr)
model.fit(xtrain,ytrain)
score = model.score(xtrain,ytrain)
print("Training score:",score)
coeffs1= model.estimators_[0].coef_
coeffs2= model.estimators_[1].coef_
intercept=model.estimators_[1].intercept_
ypred = model.predict(xtest)
print("y1 MSE:%.4f" % mean_squared_error(ytest[:,0],ypred[:,0]))
print("y2 MSE:%.4f" % mean_squared_error(ytest[:,1],1]))
对于给定的y1
和y2
以及已知的系数,我显然留下了两个联立方程,但是我不确定为最佳输入解决这些问题的最佳方法(sympy?scipy优化?) ,尤其是考虑到我想要x1时,输入之一应该是最小值而不是输出。
此问题的实际固定输出为y1 = 500
和y2 = 9
。
方程式的LaTeX代码
y_{2} = b_{0} + b_{1}x_{1} + b_{2}x_{2} + b_{3}x_{1}^{2} + b_{4}x_{1}x_{2} + b_{5}x_{2}^{2} + b_{6}x_{1}^{3} + b_{7}x_{1}^{2}x_{2} + b_{8}x_{1}x_{2}^{2} + b_{9}x_{2}^{3}
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