def simple_regession(self):
''' The answer of exercise03-08:
(a)
(i) Yes, from F-stat
(ii) Explain it from RSE and R^2 stat
(iii)negative
(iv) Code, no prediction interval
(b) Code
(c) Residual/fitted: non-linearity
'''
# model = smf.ols(formula="mpg ~ horsepower", data=self.df)
y = self.df['mpg']
X = self.df[['horsepower']]
X = sm.add_constant(X)
print X
res = sm.OLS(y, X).fit()
# res = model.fit()
print res.summary()
print "The prediction is: ", res.predict(exog=[[1, 98]])
print "The prediction interval is: "
'''
self.df.plot(kind="scatter", x='horsepower', y='mpg', c='w')
graph_x = np.linspace(min(self.df['horsepower']), 200)
graph_y = res.predict(sm.add_constant(graph_x))
plt.plot(graph_x, graph_y)
'''
fig = rp.abline_plot(model_results=res)
ax = fig.axes[0]
ax.scatter(X['horsepower'], y, c='w')
plt.show()
lrplot.plot_R_graphs(res)
The answer of exercise-03-09:
(f) From the correlation matrix in 9a., displacement, horsepower and weight show a similar nonlinear pattern against our response mpg.
This nonlinear pattern is very close to a log form.
So in the next attempt, we use log(mpg) as our response variable.
"""
# Why choose these predictors? By brute force and choose the least p-value?
mod = smf.ols(
formula="np.log(mpg) ~ cylinders+displacement+horsepower+weight+acceleration+year+origin", data=self.df
)
res = mod.fit()
print res.summary()
return res
if __name__ == "__main__":
ex09 = Exec09()
# ex09.plot_scatter_matrix()
# ex09.show_covariance()
""" RegressionResults class """
res = ex09.multi_variate_regression()
# lrp.plot_scale_location(res)
# lrp.plot_qq(res)
# lrp.plot_fitted_student_residual(ex09.df, res)
# ex09.get_leverages_resid(res)
# ex09.get_vifs(ex09.X)
# ex09.regress_with_interaction()
# res = ex09.regress_with_poly_2()
lrp.plot_R_graphs(res)
