我遵循StatsModels示例here来绘制分位数回归线.只需对我的数据稍作修改,该示例效果很好,生成此绘图(请注意,我已修改代码以仅绘制0.05,0.25,0.5,0.75和0.95分位数)：

但是,我想绘制OLS拟合和相应分位数的二阶多项式拟合(而不是线性).例如,以下是相同数据的二阶OLS行：

如何修改链接示例中的代码以生成非线性分位数？

这是我从链接示例中修改的相关代码,以生成第一个图：

d = {'temp': x, 'dens': y}

df = pd.DataFrame(data=d)

# Least Absolute Deviation

#

# The LAD model is a special case of quantile regression where q=0.5

mod = smf.quantreg('dens ~ temp', df)

res = mod.fit(q=.5)

print(res.summary())

# Prepare data for plotting

#

# For convenience, we place the quantile regression results in a Pandas DataFrame, and the OLS results in a dictionary.

quantiles = [.05, .25, .50, .75, .95]

def fit_model(q):

res = mod.fit(q=q)

return [q, res.params['Intercept'], res.params['temp']] + res.conf_int().ix['temp'].tolist()

models = [fit_model(x) for x in quantiles]

models = pd.DataFrame(models, columns=['q', 'a', 'b','lb','ub'])

ols = smf.ols('dens ~ temp', df).fit()

ols_ci = ols.conf_int().ix['temp'].tolist()

ols = dict(a = ols.params['Intercept'],

b = ols.params['temp'],

lb = ols_ci[0],

ub = ols_ci[1])

print(models)

print(ols)

x = np.arange(df.temp.min(), df.temp.max(), 50)

get_y = lambda a, b: a + b * x

for i in range(models.shape[0]):

y = get_y(models.a[i], models.b[i])

plt.plot(x, y, linestyle='dotted', color='grey')

y = get_y(ols['a'], ols['b'])

plt.plot(x, y, color='red', label='OLS')

plt.scatter(df.temp, df.dens, alpha=.2)

plt.xlim((-10, 40))

plt.ylim((0, 0.4))

plt.legend()

plt.xlabel('temp')

plt.ylabel('dens')

plt.show()

解决方法:

经过一天的调查,想出了一个解决方案,所以发布我自己的答案.在StatsModels的Josef Perktold获得了很多帮助.

这是相关的代码和图：

d = {'temp': x, 'dens': y}

df = pd.DataFrame(data=d)

x1 = pd.DataFrame({'temp': np.linspace(df.temp.min(), df.temp.max(), 200)})

poly_2 = smf.ols(formula='dens ~ 1 + temp + I(temp ** 2.0)', data=df).fit()

plt.plot(x, y, 'o', alpha=0.2)

plt.plot(x1.temp, poly_2.predict(x1), 'r-',

label='2nd order poly fit, $R^2$=%.2f' % poly_2.rsquared,

alpha=0.9)

plt.xlim((-10, 50))

plt.ylim((0, 0.25))

plt.xlabel('mean air temp')

plt.ylabel('density')

plt.legend(loc="upper left")

# with quantile regression

# Least Absolute Deviation

# The LAD model is a special case of quantile regression where q=0.5

mod = smf.quantreg('dens ~ temp + I(temp ** 2.0)', df)

res = mod.fit(q=.5)

print(res.summary())

# Quantile regression for 5 quantiles

quantiles = [.05, .25, .50, .75, .95]

# get all result instances in a list

res_all = [mod.fit(q=q) for q in quantiles]

res_ols = smf.ols('dens ~ temp + I(temp ** 2.0)', df).fit()

plt.figure()

# create x for prediction

x_p = np.linspace(df.temp.min(), df.temp.max(), 50)

df_p = pd.DataFrame({'temp': x_p})

for qm, res in zip(quantiles, res_all):

# get prediction for the model and plot

# here we use a dict which works the same way as the df in ols

plt.plot(x_p, res.predict({'temp': x_p}), linestyle='--', lw=1,

color='k', label='q=%.2F' % qm, zorder=2)

y_ols_predicted = res_ols.predict(df_p)

plt.plot(x_p, y_ols_predicted, color='red', zorder=1)

#plt.scatter(df.temp, df.dens, alpha=.2)

plt.plot(df.temp, df.dens, 'o', alpha=.2, zorder=0)

plt.xlim((-10, 50))

plt.ylim((0, 0.25))

#plt.legend(loc="upper center")

plt.xlabel('mean air temp')

plt.ylabel('density')

plt.title('')

plt.show()

红线：二阶多项式拟合

黑色虚线：第5,第25,第50,第75,第95百分位数

标签：python,pandas,regression,statsmodels