Question

我正在尝试找到一个python软件包，该软件包将提供一个选项以使自然平滑样条线具有用户可选的平滑因子。有没有实现的方法？如果没有，您将如何使用可用的工具自己实现？

根据自然样条曲线，我的意思是应该满足以下条件：拟合函数在端点处的二阶导数为零（线性）。
通过平滑样条线，我的意思是，样条线不应“插值”（穿过所有数据点）。我想自行决定正确的平滑系数lambda（有关平滑样条曲线，请参见Wikipedia page）。

我发现的东西

scipy.interpolate.CubicSpline [link]：执行自然（三次）样条拟合。进行插值，并且无法平滑数据。
scipy.interpolate.UnivariateSpline [link]：样条曲线是否适合用户可选的平滑因子。但是，没有选项可以使样条曲线自然。

Answer 1

的After hours of investigation, I did not find any pip installable packages which could fit a natural cubic spline with user-controllable smoothness. However, after deciding to write one myself, while reading about the topic I stumbled upon a

blog post madrury下载属性问题。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here一起使用a BSD-licence（NaturalCubicSpline）。他还在IPython notebook中写了一些示例。

但是，由于这是Internet，并且链接容易消失，因此我将在此处复制源代码的相关部分+我编写的帮助函数（get_natural_cubic_spline_model），并显示如何使用它的示例。配合的平滑度可以通过使用不同的结数来控制。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

Answer 2

python包patsy具有生成样条曲线基的函数，包括自然三次样条曲线基。在documentation中进行了描述。然后可以使用任何库来拟合模型，例如scikit学习或统计模型。

df的{{1}}参数可用于控制“平滑度”
请注意，cr()太低可能会导致欠拟合（请参见下文）。

使用scikit-learn的简单示例。

df

Answer 3

对于我的一个项目，我需要为时间序列建模创建间隔，并提高程序效率，我创建了tsmoothie：用于以矢量化方式进行时间序列平滑和离群值检测的python库

它提供了不同的平滑算法以及计算间隔的可能性。

对于import multiprocessing def example_func(): print("This is a targeted function for multiprocessing") if __name__ == "__main__": print("This is the main session, starting multiprocessing") multiprocessing.Process(target=example_func).start()是自然立方类型：

app/Http/Controllers/Auth/LoginController.php

我还指出tsmoothie可以向量化方式对多个时间序列进行平滑处理

Answer 4

您可以使用this numpy/scipy implementation的自然三次平滑样条进行单变量/多变量数据平滑。平滑参数应在[0.0，1.0]范围内。如果我们使用等于1.0的平滑参数，我们将获得自然三次样条插值而不进行数据平滑。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 5

编程语言R提供了很好的自然三次平滑样条曲线的实现。您可以在Python rpy2中使用R函数：

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)

_{如果您想直接设置lambda：spline1 = r_smooth_spline(x=r_x, y=r_y, lambda=42)不起作用，因为lambda在Python中已经具有另一种含义，但是有一个解决方案：{{3 }}。}

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

Python自然平滑样条线

我发现的东西

5 个答案:

示例

`get_natural_cubic_spline_model`