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我一直在尝试使用 scikit-learn 中的 LR 来获得标准误差和 p 值。但没有成功。
我最终找到了这个 article :但 std 错误和 p 值与来自 statsmodel.api OLS 方法的不匹配
import numpy as np
from sklearn import datasets
from sklearn import linear_model
import regressor
import statsmodels.api as sm
boston = datasets.load_boston()
which_betas = np.ones(13, dtype=bool)
which_betas[3] = False
X = boston.data[:,which_betas]
y = boston.target
#scikit + regressor stats
ols = linear_model.LinearRegression()
ols.fit(X,y)
xlables = boston.feature_names[which_betas]
regressor.summary(ols, X, y, xlables)
# statsmodel
x2 = sm.add_constant(X)
models = sm.OLS(y,x2)
result = models.fit()
print result.summary()
Residuals:
Min 1Q Median 3Q Max
-26.3743 -1.9207 0.6648 2.8112 13.3794
Coefficients:
Estimate Std. Error t value p value
_intercept 36.925033 4.915647 7.5117 0.000000
CRIM -0.112227 0.031583 -3.5534 0.000416
ZN 0.047025 0.010705 4.3927 0.000014
INDUS 0.040644 0.055844 0.7278 0.467065
NOX -17.396989 3.591927 -4.8434 0.000002
RM 3.845179 0.272990 14.0854 0.000000
AGE 0.002847 0.009629 0.2957 0.767610
DIS -1.485557 0.180530 -8.2289 0.000000
RAD 0.327895 0.061569 5.3257 0.000000
TAX -0.013751 0.001055 -13.0395 0.000000
PTRATIO -0.991733 0.088994 -11.1438 0.000000
B 0.009827 0.001126 8.7256 0.000000
LSTAT -0.534914 0.042128 -12.6973 0.000000
---
R-squared: 0.73547, Adjusted R-squared: 0.72904
F-statistic: 114.23 on 12 features
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.735
Model: OLS Adj. R-squared: 0.729
Method: Least Squares F-statistic: 114.2
Date: Sun, 21 Aug 2016 Prob (F-statistic): 7.59e-134
Time: 21:56:26 Log-Likelihood: -1503.8
No. Observations: 506 AIC: 3034.
Df Residuals: 493 BIC: 3089.
Df Model: 12
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [95.0% Conf. Int.]
------------------------------------------------------------------------------
const 36.9250 5.148 7.173 0.000 26.811 47.039
x1 -0.1122 0.033 -3.405 0.001 -0.177 -0.047
x2 0.0470 0.014 3.396 0.001 0.020 0.074
x3 0.0406 0.062 0.659 0.510 -0.081 0.162
x4 -17.3970 3.852 -4.516 0.000 -24.966 -9.828
x5 3.8452 0.421 9.123 0.000 3.017 4.673
x6 0.0028 0.013 0.214 0.831 -0.023 0.029
x7 -1.4856 0.201 -7.383 0.000 -1.881 -1.090
x8 0.3279 0.067 4.928 0.000 0.197 0.459
x9 -0.0138 0.004 -3.651 0.000 -0.021 -0.006
x10 -0.9917 0.131 -7.547 0.000 -1.250 -0.734
x11 0.0098 0.003 3.635 0.000 0.005 0.015
x12 -0.5349 0.051 -10.479 0.000 -0.635 -0.435
==============================================================================
Omnibus: 190.837 Durbin-Watson: 1.015
Prob(Omnibus): 0.000 Jarque-Bera (JB): 897.143
Skew: 1.619 Prob(JB): 1.54e-195
Kurtosis: 8.663 Cond. No. 1.51e+04
==============================================================================
Warnings:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 1.51e+04. This might indicate that there are
strong multicollinearity or other numerical problems.
import pandas as pd
import statsmodels.api as sm
import numpy as np
import scipy
from sklearn.linear_model import LinearRegression
from sklearn import metrics
def readFile(filename, sheetname):
xlsx = pd.ExcelFile(filename)
data = xlsx.parse(sheetname, skiprows=1)
return data
def lr_statsmodel(X,y):
X = sm.add_constant(X)
model = sm.OLS(y,X)
results = model.fit()
print (results.summary())
def lr_scikit(X,y,featureCols):
model = LinearRegression()
results = model.fit(X,y)
predictions = results.predict(X)
print 'Coefficients'
print 'Intercept\t' , results.intercept_
df = pd.DataFrame(zip(featureCols, results.coef_))
print df.to_string(index=False, header=False)
# Query:: The numbers matches with Excel OLS but skeptical about relating score as rsquared
rSquare = results.score(X,y)
print '\nR-Square::', rSquare
# This looks like a better option
# source: http://scikit-learn.org/stable/modules/generated/sklearn.metrics.r2_score.html#sklearn.metrics.r2_score
r2 = metrics.r2_score(y,results.predict(X))
print 'r2', r2
# Query: No clue at all! http://scikit-learn.org/stable/modules/model_evaluation.html#regression-metrics
print 'Rsquared?!' , metrics.explained_variance_score(y, results.predict(X))
# INFO:: All three of them are providing the same figures!
# Adj-Rsquare formula @ https://www.easycalculation.com/statistics/learn-adjustedr2.php
# In ML, we don't use all of the data for training, and hence its highly unusual to find AdjRsquared. Thus the need for manual calculation
N = X.shape[0]
p = X.shape[1]
adjRsquare = 1 - ((1 - rSquare ) * (N - 1) / (N - p - 1))
print "Adjusted R-Square::", adjRsquare
# calculate standard errors
# mean_absolute_error
# mean_squared_error
# median_absolute_error
# r2_score
# explained_variance_score
mse = metrics.mean_squared_error(y,results.predict(X))
print mse
print 'Residual Standard Error:', np.sqrt(mse)
# OLS in Matrix : https://github.com/nsh87/regressors/blob/master/regressors/stats.py
n = X.shape[0]
X1 = np.hstack((np.ones((n, 1)), np.matrix(X)))
se_matrix = scipy.linalg.sqrtm(
metrics.mean_squared_error(y, results.predict(X)) *
np.linalg.inv(X1.T * X1)
)
print 'se',np.diagonal(se_matrix)
# https://github.com/nsh87/regressors/blob/master/regressors/stats.py
# http://regressors.readthedocs.io/en/latest/usage.html
y_hat = results.predict(X)
sse = np.sum((y_hat - y) ** 2)
print 'Standard Square Error of the Model:', sse
if __name__ == '__main__':
# read file
fileData = readFile('Linear_regression.xlsx','Input Data')
# list of independent variables
feature_cols = ['Price per week','Population of city','Monthly income of riders','Average parking rates per month']
# build dependent & independent data set
X = fileData[feature_cols]
y = fileData['Number of weekly riders']
# Statsmodel - OLS
# lr_statsmodel(X,y)
# ScikitLearn - OLS
lr_scikit(X,y,feature_cols)
Y X1 X2 X3 X4
City Number of weekly riders Price per week Population of city Monthly income of riders Average parking rates per month
1 1,92,000 $15 18,00,000 $5,800 $50
2 1,90,400 $15 17,90,000 $6,200 $50
3 1,91,200 $15 17,80,000 $6,400 $60
4 1,77,600 $25 17,78,000 $6,500 $60
5 1,76,800 $25 17,50,000 $6,550 $60
6 1,78,400 $25 17,40,000 $6,580 $70
7 1,80,800 $25 17,25,000 $8,200 $75
8 1,75,200 $30 17,25,000 $8,600 $75
9 1,74,400 $30 17,20,000 $8,800 $75
10 1,73,920 $30 17,05,000 $9,200 $80
11 1,72,800 $30 17,10,000 $9,630 $80
12 1,63,200 $40 17,00,000 $10,570 $80
13 1,61,600 $40 16,95,000 $11,330 $85
14 1,61,600 $40 16,95,000 $11,600 $100
15 1,60,800 $40 16,90,000 $11,800 $105
16 1,59,200 $40 16,30,000 $11,830 $105
17 1,48,800 $65 16,40,000 $12,650 $105
18 1,15,696 $102 16,35,000 $13,000 $110
19 1,47,200 $75 16,30,000 $13,224 $125
20 1,50,400 $75 16,20,000 $13,766 $130
21 1,52,000 $75 16,15,000 $14,010 $150
22 1,36,000 $80 16,05,000 $14,468 $155
23 1,26,240 $86 15,90,000 $15,000 $165
24 1,23,888 $98 15,95,000 $15,200 $175
25 1,26,080 $87 15,90,000 $15,600 $175
26 1,51,680 $77 16,00,000 $16,000 $190
27 1,52,800 $63 16,10,000 $16,200 $200
最佳答案
这是 reg 是 sklearn 的 lin 回归拟合方法的输出
计算调整后的 r2
def adjustedR2(x,y reg):
r2 = reg.score(x,y)
n = x.shape[0]
p = x.shape[1]
adjusted_r2 = 1-(1-r2)*(n-1)/(n-p-1)
return adjusted_r2
对于 p 值
from sklearn.feature_selection import f_regression
freg=f_regression(x,y)
p=freg[1]
print(p.round(3))
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