英文:
Python Sklearn multi-linear Regression for probabilities - normalize coefficients to 1
问题
以下是您提供的内容的翻译：
问题很简单。我有三个方程式：OmegaMIX = beta1 * Omega1 + beta2 * Omega2，我想通过线性回归找到最佳系数，其中截距为0。我已经得到了代码，但beta1和beta2是概率，所以必须满足beta1 + beta2 = 1，但实际情况并非如此。应该如何设置？
这是现有的代码：
from sklearn import linear_modelinputfilename = 'JO.csv' #输入df = pd.read_csv(inputfilename)x = df.drop('OmegaMIX', axis=1) #参考温度y = df['OmegaMIX'] #多参数LIRsregr = linear_model.LinearRegression(positive=True, fit_intercept=False)regr.fit(x, y)print('β1 = ', regr.coef_[0])print('β2 = ', regr.coef_[1])print("质量 = ", regr.score(x, y, sample_weight=None))
输出结果为：
β1 = 0.33995522604783796β2 = 0.5794270911721245质量 = 0.9995968335914979
输入文件为：
OmegaMIX,Omega1,Omega22.70,4.43,2.091.84,3.00,1.370.50,1.19,0.17
英文:
The question is simple. I have three equations: OmegaMIX = beta1Omega1+beta2Omega2 and I want to find the best coefficients by the linear regression, where intercept is 0. I have obtained the code, but the beta1 and beta2 are probabilities, so it must be beta1+beta2 = 1, which is not the case. How can this be set?
This is the existing code:
from sklearn import linear_modelinputfilename = 'JO.csv' #inputdf = pd.read_csv(inputfilename)x = df.drop('OmegaMIX',axis=1) #Reference temperaturey = df['OmegaMIX'] #Multiparametric LIRsregr = linear_model.LinearRegression(positive=True,fit_intercept=False)regr.fit(x, y)print('\u03B21 = ', regr.coef_[0])print('\u03B22 = ', regr.coef_[1])print("Quality = ", regr.score(x, y, sample_weight=None))
and the output is:
β1 = 0.33995522604783796β2 = 0.5794270911721245Quality = 0.9995968335914979
The input file is:
OmegaMIX,Omega1,Omega22.70,4.43,2.091.84,3.00,1.370.50,1.19,0.17
答案1
得分: 1
我不认为这可以在sklearn上完成。我更愿意使用CVXPY，因为你可以控制约束条件。
这是一个示例：
import pandas as pdimport cvxpy as cpdf = pd.read_csv('JO.csv')# 设置矩阵X和向量y，cvxpy需要它们是numpy数组X = df.drop('OmegaMIX', axis=1).valuesy = df['OmegaMIX'].values# 定义问题的变量beta1 = cp.Variable()beta2 = cp.Variable()# 设置约束条件constraints = [beta1 >= 0, beta2 >= 0, beta1 + beta2 == 1]# 定义线性回归问题objective = cp.Minimize(cp.sum_squares(X[:, 0] * beta1 + X[:, 1] * beta2 - y))# 解决问题problem = cp.Problem(objective, constraints)problem.solve()# 打印beta值print('β1 =', beta1.value)print('β2 =', beta2.value)print('β1 + β2 =', (beta1.value + beta2.value))print("质量 =", problem.value)
请注意，这是原文的翻译。
英文:
I don't think this can be done on sklearn. I would rather use CVXPY as you can control the constraints.
Here's an example :
import pandas as pdimport cvxpy as cpdf = pd.read_csv('JO.csv')# Setting out matrix X and our vector y, cvxpy needs them to be numpy arraysX = df.drop('OmegaMIX', axis=1).valuesy = df['OmegaMIX'].values# Defining the variables of the problembeta1 = cp.Variable()beta2 = cp.Variable()# setting the constraintsconstraints = [beta1 >= 0, beta2 >= 0, beta1 + beta2 == 1]# Defining the linear regression problemobjective = cp.Minimize(cp.sum_squares(X[:, 0] * beta1 + X[:, 1] * beta2 - y))# Solving the probelemproblem = cp.Problem(objective, constraints)problem.solve()# printing the betasprint('\u03B21 =', beta1.value)print('\u03B22 =', beta2.value)print('\u03B21 + \u03B22 =', (beta1.value + beta2.value))print("Quality =", problem.value)
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