Python Sklearn 多元线性回归用于概率 – 将系数归一化为1

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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_model
inputfilename = '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LIRs
regr = 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 =  0.33995522604783796
β2 =  0.5794270911721245
质量 =  0.9995968335914979

输入文件为:

OmegaMIX,Omega1,Omega2
2.70,4.43,2.09
1.84,3.00,1.37
0.50,1.19,0.17
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_model
inputfilename = 'JO.csv' #input
df = pd.read_csv(inputfilename)
x = df.drop('OmegaMIX',axis=1) #Reference temperature
y = df['OmegaMIX'] #Multiparametric LIRs
regr = 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 =  0.33995522604783796
β2 =  0.5794270911721245
Quality =  0.9995968335914979

The input file is:

OmegaMIX,Omega1,Omega2
2.70,4.43,2.09
1.84,3.00,1.37
0.50,1.19,0.17

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 pd
import cvxpy as cp

df = 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).values
y = 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 pd
import cvxpy as cp

df = pd.read_csv('JO.csv')

# Setting out matrix X and our vector y, cvxpy needs them to be numpy arrays
X = df.drop('OmegaMIX', axis=1).values
y = df['OmegaMIX'].values

# Defining the variables of the problem
beta1 = cp.Variable()
beta2 = cp.Variable()

# setting the constraints
constraints = [beta1 >= 0, beta2 >= 0, beta1 + beta2 == 1]

# Defining the linear regression problem
objective = cp.Minimize(cp.sum_squares(X[:, 0] * beta1 + X[:, 1] * beta2 - y))

# Solving the probelem
problem = cp.Problem(objective, constraints)
problem.solve()

# printing the betas
print('\u03B21 =', beta1.value)
print('\u03B22 =', beta2.value)
print('\u03B21 + \u03B22 =', (beta1.value + beta2.value))
print("Quality =", problem.value)

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  • 本文由 发表于 2023年5月21日 09:28:59
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