如何从多项式拟合中提取导数？

Question

我有几个样本点的数据集共享相同的x坐标，并且多项式拟合考虑了所有这些样本点。这很好，如本图所示：

使用以下代码：

import numpy as np
import matplotlib.pyplot as plt

from sklearn.linear_model import Ridge
from sklearn.preprocessing import PolynomialFeatures
from sklearn.pipeline import make_pipeline

x = np.array([0., 4., 9., 12., 16., 20., 24., 27.])
y = np.array([[3620000.,26000000.,187000000.,348000000.,475000000.,483000000.,456000000.,384000000.],
              [3750000.,25900000.,187000000.,362000000.,449000000.,465000000.,488000000.,408000000.],
              [3720000.,26100000.,184000000.,341000000.,455000000.,458000000.,446000000.,430000000.]])


x_all = np.ravel(x + np.zeros_like(y))
y_all = np.ravel(y)

plt.scatter(x, y[0], label="training points 1", c='r')
plt.scatter(x, y[1], label="training points 2", c='b')
plt.scatter(x, y[2], label="training points 3", c='g')

x_plot = np.linspace(0, max(x), 100)

for degree in np.arange(5, 6, 1):
    model = make_pipeline(PolynomialFeatures(degree), Ridge(alpha=50, fit_intercept=False))
    model.fit(x_all[:, None], y_all)
    y_plot = model.predict(x_plot[:, None])
    plt.plot(x_plot, y_plot, label="degree %d" % degree)

    ridge = model.named_steps['ridge']
    print(degree, ridge.coef_)

plt.legend(loc='best')

plt.show()

我真正感兴趣的不是拟合多项式的等式，而是实际的导数。

有没有办法直接访问拟合函数的导数？上面代码中的对象model具有以下属性：

model.decision_function  model.fit_transform      model.inverse_transform  model.predict            model.predict_proba      model.set_params         model.transform          
model.fit                model.get_params         model.named_steps        model.predict_log_proba  model.score              model.steps 

所以在理想的情况下，我希望有类似（伪代码）的东西：

myDerivative = model.derivative(x_plot)

编辑：

我也非常乐意使用另一个完成工作的模块/库，所以我也愿意接受建议。

Answer 1

既然你知道拟合的多项式系数会得到你想要的吗？

deriv = np.polyder(ridge.coef_[::-1])
yd_plot = np.polyval(deriv,x_plot)