Python自然平滑样条线

Question

我试图找到一个python包，它提供了一个选项，可以用用户选择的平滑因子来适应自然平滑样条曲线。有没有实现的方法？如果没有，您将如何使用可用的工具来实现它？

• 通过自然样条，我的意思是应该有一个条件，即拟合函数在端点的二阶导数为零（线性）。

• 通过平滑样条曲线，我的意思是样条曲线不应该是“插值”（通过所有数据点）。我想自己决定正确的平滑因子lambda（请参见Wikipedia page中的平滑样条）。

我所发现的

• scipy.interpolate.cubic spline[link]：进行自然（三次）样条拟合。进行插值，并且无法平滑数据。

• scipy.interpolate.UnivariateSpline[link]：样条曲线是否符合用户可选择的平滑因子。但是，没有使样条线自然的选项。

Answer 1

可以使用自然三次平滑样条的this numpy/scipy implementation进行单变量/多变量数据平滑。平滑参数应在范围[0.0，1.0]内。如果我们使用等于1.0的平滑参数，我们就得到了没有数据平滑的自然三次样条插值。该实现还支持单变量数据的矢量化。

单变量示例：

import numpy as np
import matplotlib.pyplot as plt

import csaps

np.random.seed(1234)

x = np.linspace(-5., 5., 25)
y = np.exp(-(x/2.5)**2) + (np.random.rand(25) - 0.2) * 0.3

sp = csaps.UnivariateCubicSmoothingSpline(x, y, smooth=0.85)

xs = np.linspace(x[0], x[-1], 150)
ys = sp(xs)

plt.plot(x, y, 'o', xs, ys, '-')
plt.show()

双变量示例：

import numpy as np

import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

import csaps

xdata = [np.linspace(-3, 3, 61), np.linspace(-3.5, 3.5, 51)]
i, j = np.meshgrid(*xdata, indexing='ij')

ydata = (3 * (1 - j)**2. * np.exp(-(j**2) - (i + 1)**2)
         - 10 * (j / 5 - j**3 - i**5) * np.exp(-j**2 - i**2)
         - 1 / 3 * np.exp(-(j + 1)**2 - i**2))

np.random.seed(12345)
noisy = ydata + (np.random.randn(*ydata.shape) * 0.75)

sp = csaps.MultivariateCubicSmoothingSpline(xdata, noisy, smooth=0.988)
ysmth = sp(xdata)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

ax.plot_wireframe(j, i, noisy, linewidths=0.5, color='r')
ax.scatter(j, i, noisy, s=5, c='r')

ax.plot_surface(j, i, ysmth, linewidth=0, alpha=1.0)

plt.show()

Answer 2

程序设计语言R提供了一个很好的自然三次光滑样条的实现。您可以在Python中使用R函数和rpy2：

import rpy2.robjects as robjects
r_y = robjects.FloatVector(y_train)
r_x = robjects.FloatVector(x_train)

r_smooth_spline = robjects.r['smooth.spline'] #extract R function# run smoothing function
spline1 = r_smooth_spline(x=r_x, y=r_y, spar=0.7)
ySpline=np.array(robjects.r['predict'](spline1,robjects.FloatVector(x_smooth)).rx2('y'))
plt.plot(x_smooth,ySpline)

_{如果要直接设置lambda：spline1 = r_smooth_spline(x=r_x, y=r_y, lambda=42)不起作用，因为lambda在Python中已经有了另一个含义，但是有一个解决方案：How to use the lambda argument of smooth.spline in RPy WITHOUT Python interprating it as lambda。}

_{要运行代码，首先需要定义数据x_train和y_train，如果要以全高清分辨率在-3和5之间绘制，则可以定义x_smooth=np.array(np.linspace(-3,5,1920)). 。}

Answer 3

经过数小时的调查，我没有发现任何pip可安装的软件包，可以适应一个自然的三次样条与用户可控的平滑度。然而，在决定自己写一篇文章之后，在阅读这个主题时，我偶然发现了github用户madrury的blog post。他编写了能够生成自然三次样条模型的python代码。

模型代码可与a BSD-licence一起使用here（NaturalCubicSpline）。他还写了一些例子。

但由于这是Internet，链接往往会断开，我将在这里复制源代码的相关部分+一个由我编写的助手函数（get_natural_cubic_spline_model），并展示如何使用它的示例。通过使用不同的结数可以控制拟合的平滑度。结的位置也可以由用户指定。

示例

from matplotlib import pyplot as plt
import numpy as np

def func(x):
    return 1/(1+25*x**2)

# make example data
x = np.linspace(-1,1,300)
y = func(x) + np.random.normal(0, 0.2, len(x))

# The number of knots can be used to control the amount of smoothness
model_6 = get_natural_cubic_spline_model(x, y, minval=min(x), maxval=max(x), n_knots=6)
model_15 = get_natural_cubic_spline_model(x, y, minval=min(x), maxval=max(x), n_knots=15)
y_est_6 = model_6.predict(x)
y_est_15 = model_15.predict(x)


plt.plot(x, y, ls='', marker='.', label='originals')
plt.plot(x, y_est_6, marker='.', label='n_knots = 6')
plt.plot(x, y_est_15, marker='.', label='n_knots = 15')
plt.legend(); plt.show()

get_natural_cubic_spline_model的源代码

import numpy as np
import pandas as pd
from sklearn.base import BaseEstimator, TransformerMixin
from sklearn.linear_model import LinearRegression
from sklearn.pipeline import Pipeline


def get_natural_cubic_spline_model(x, y, minval=None, maxval=None, n_knots=None, knots=None):
    """
    Get a natural cubic spline model for the data.

    For the knots, give (a) `knots` (as an array) or (b) minval, maxval and n_knots.

    If the knots are not directly specified, the resulting knots are equally
    space within the *interior* of (max, min).  That is, the endpoints are
    *not* included as knots.

    Parameters
    ----------
    x: np.array of float
        The input data
    y: np.array of float
        The outpur data
    minval: float 
        Minimum of interval containing the knots.
    maxval: float 
        Maximum of the interval containing the knots.
    n_knots: positive integer 
        The number of knots to create.
    knots: array or list of floats 
        The knots.

    Returns
    --------
    model: a model object
        The returned model will have following method:
        - predict(x):
            x is a numpy array. This will return the predicted y-values.
    """

    if knots:
        spline = NaturalCubicSpline(knots=knots)
    else:
        spline = NaturalCubicSpline(max=maxval, min=minval, n_knots=n_knots)

    p = Pipeline([
        ('nat_cubic', spline),
        ('regression', LinearRegression(fit_intercept=True))
    ])

    p.fit(x, y)

    return p


class AbstractSpline(BaseEstimator, TransformerMixin):
    """Base class for all spline basis expansions."""

    def __init__(self, max=None, min=None, n_knots=None, n_params=None, knots=None):
        if knots is None:
            if not n_knots:
                n_knots = self._compute_n_knots(n_params)
            knots = np.linspace(min, max, num=(n_knots + 2))[1:-1]
            max, min = np.max(knots), np.min(knots)
        self.knots = np.asarray(knots)

    @property
    def n_knots(self):
        return len(self.knots)

    def fit(self, *args, **kwargs):
        return self


class NaturalCubicSpline(AbstractSpline):
    """Apply a natural cubic basis expansion to an array.
    The features created with this basis expansion can be used to fit a
    piecewise cubic function under the constraint that the fitted curve is
    linear *outside* the range of the knots..  The fitted curve is continuously
    differentiable to the second order at all of the knots.
    This transformer can be created in two ways:
      - By specifying the maximum, minimum, and number of knots.
      - By specifying the cutpoints directly.  

    If the knots are not directly specified, the resulting knots are equally
    space within the *interior* of (max, min).  That is, the endpoints are
    *not* included as knots.
    Parameters
    ----------
    min: float 
        Minimum of interval containing the knots.
    max: float 
        Maximum of the interval containing the knots.
    n_knots: positive integer 
        The number of knots to create.
    knots: array or list of floats 
        The knots.
    """

    def _compute_n_knots(self, n_params):
        return n_params

    @property
    def n_params(self):
        return self.n_knots - 1

    def transform(self, X, **transform_params):
        X_spl = self._transform_array(X)
        if isinstance(X, pd.Series):
            col_names = self._make_names(X)
            X_spl = pd.DataFrame(X_spl, columns=col_names, index=X.index)
        return X_spl

    def _make_names(self, X):
        first_name = "{}_spline_linear".format(X.name)
        rest_names = ["{}_spline_{}".format(X.name, idx)
                      for idx in range(self.n_knots - 2)]
        return [first_name] + rest_names

    def _transform_array(self, X, **transform_params):
        X = X.squeeze()
        try:
            X_spl = np.zeros((X.shape[0], self.n_knots - 1))
        except IndexError: # For arrays with only one element
            X_spl = np.zeros((1, self.n_knots - 1))
        X_spl[:, 0] = X.squeeze()

        def d(knot_idx, x):
            def ppart(t): return np.maximum(0, t)

            def cube(t): return t*t*t
            numerator = (cube(ppart(x - self.knots[knot_idx]))
                         - cube(ppart(x - self.knots[self.n_knots - 1])))
            denominator = self.knots[self.n_knots - 1] - self.knots[knot_idx]
            return numerator / denominator

        for i in range(0, self.n_knots - 2):
            X_spl[:, i+1] = (d(i, X) - d(self.n_knots - 2, X)).squeeze()
        return X_spl