问题是:
两个变量(x1, x2)
和两个输出(y1, y2)
,数据可用于计算输入和输出之间的关系
最后,我想知道最小可能的x1
来给我一个特定的y1
和y2
值
到目前为止,我考虑的方法是:
为了使用线性回归模型对x1,x2
和y1,y2
之间的关系进行曲线拟合,我只对示例代码做了一些小的修改。x1
和x2
之间的关系被扩展到适合三次函数,因为这给了我一个较低的均方误差
因此,输入数据所拟合的方程的形式如下
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
很抱歉阅读起来有多困难,但我不允许将它们作为图片发布(底部是LaTeX代码)
其中,方程式的系数(a, b)
已通过以下代码中的线性回归模型coeffs1
和coeffs2
求解
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([0,0,0,0,0,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,60,60,60,60,60]).reshape(n, 1)
x2 = array([100,200,300,400,500,100,200,300,400,500,100,200,300,400,500,100,200,300,400,500,100,200,300,400,500,100,200,300,400,500,100,200,300,400,500]).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], ypred[:, 1]))
对于给定的y1
和y2
和已知系数,我显然剩下两个联立方程,但我不确定最佳输入(Symphy?scipy Optimize?)的最佳解决方法,特别是考虑到我希望x1,其中一个输入是最小值,而不是输出
这个问题的实际固定输出是y1 = 500
和y2 = 9
方程的乳胶代码
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}
目前没有回答
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