scipy.interpolate.UnivariateSpline 不平滑，无论参数如何

Question

我在插值时无法让 scipy.interpolate.UnivariateSpline 使用任何平滑。基于函数页面< /a> 以及一些 以前的帖子，我相信它应该提供平滑s 参数。

这是我的代码：

# Imports
import scipy
import pylab

# Set up and plot actual data
x = [0, 5024.2059124920379, 7933.1645067836089, 7990.4664106277542, 9879.9717114947653, 13738.60563208926, 15113.277958924193]
y = [0.0, 3072.5653360000988, 5477.2689107965398, 5851.6866463790966, 6056.3852496014106, 7895.2332350173638, 9154.2956175610598]
pylab.plot(x, y, "o", label="Actual")

# Plot estimates using splines with a range of degrees
for k in range(1, 4):
    mySpline = scipy.interpolate.UnivariateSpline(x=x, y=y, k=k, s=2)
    xi = range(0, 15100, 20)
    yi = mySpline(xi)
    pylab.plot(xi, yi, label="Predicted k=%d" % k)

# Show the plot
pylab.grid(True)
pylab.xticks(rotation=45)
pylab.legend( loc="lower right" )
pylab.show()

这是结果：

我已经尝试过一系列 s< /code> 值（0.01、0.1、1、2、5、50）以及显式权重，设置为相同的值（1.0）或随机。我仍然无法得到任何平滑，并且结的数量始终与数据点的数量相同。特别是，我正在寻找像第四点（7990.4664106277542、5851.6866463790966）这样的异常值进行平滑。

是因为我没有足够的数据吗？如果是这样，是否有类似的样条函数或聚类技术可以应用来实现这几个数据点的平滑？

Answer 1

简短的回答：您需要更仔细地选择 s 的值。

的文档指出：

Positive smoothing factor used to choose the number of knots. Number of 
knots will be increased until the     smoothing condition is satisfied:
sum((w[i]*(y[i]-s(x[i])))**2,axis=0) <= s

UnivariateSpline 如果您不传入显式权重，可以推断出平滑的“合理”值约为 s = m * v，其中 m 是数字数据点的数量和v 数据的方差。在本例中，s_good ~ 5e7。

编辑：s的合理值当然还取决于数据中的噪声水平。文档似乎建议在 (m - sqrt(2*m)) * std**2 <= s <= (m + sqrt(2*m)) 范围内选择 s )) * std**2 其中 std 是与要平滑的“噪声”相关的标准差。

Answer 2

@Zhenya 在数据点之间手动设置结的答案太粗糙，无法在噪声数据中提供良好的结果，而不选择如何应用该技术。然而，受到他/她建议的启发，我成功地使用了 Mean-Shift 聚类。它执行簇计数的自动确定，并且似乎做了相当好的平滑工作（实际上非​​常平滑）。

# Imports
import numpy
import pylab
import scipy
import sklearn.cluster

# Set up original data - note that it's monotonically increasing by X value!
data = {}
data['original'] = {}
data['original']['x'] = [0, 5024.2059124920379, 7933.1645067836089, 7990.4664106277542, 9879.9717114947653, 13738.60563208926, 15113.277958924193]
data['original']['y'] = [0.0, 3072.5653360000988, 5477.2689107965398, 5851.6866463790966, 6056.3852496014106, 7895.2332350173638, 9154.2956175610598]

# Cluster data, sort it and and save
inputNumpy = numpy.array([[data['original']['x'][i], data['original']['y'][i]] for i in range(0, len(data['original']['x']))])
meanShift = sklearn.cluster.MeanShift()
meanShift.fit(inputNumpy)
clusteredData = [[pair[0], pair[1]] for pair in meanShift.cluster_centers_]
clusteredData.sort(lambda pair1, pair2: cmp(pair1[0],pair2[0]))
data['clustered'] = {}
data['clustered']['x'] = [pair[0] for pair in clusteredData]
data['clustered']['y'] = [pair[1] for pair in clusteredData]

# Build a spline using the clustered data and predict
mySpline = scipy.interpolate.UnivariateSpline(x=data['clustered']['x'], y=data['clustered']['y'], k=1)
xi = range(0, round(max(data['original']['x']), -3) + 3000, 20)
yi = mySpline(xi)

# Plot the datapoints
pylab.plot(data['clustered']['x'], data['clustered']['y'], "D", label="Datapoints (%s)" % 'clustered')
pylab.plot(xi, yi, label="Predicted (%s)" %  'clustered')
pylab.plot(data['original']['x'], data['original']['y'], "o", label="Datapoints (%s)" % 'original')

# Show the plot
pylab.grid(True)
pylab.xticks(rotation=45)
pylab.legend( loc="lower right" )
pylab.show()

Answer 3

虽然我不知道有哪个库可以立即为你做这件事，但我会尝试更多的 DIY 方法：我会从在原始数据点之间制作一个带有结的样条线开始，在两个 x 和 y。在您的特定示例中，在第 4 点和第 5 点之间有一个结应该可以解决问题，因为它会消除 x=8000 处的巨大导数。

Answer 4

我在运行 BigChef 的答案时遇到了麻烦，这是一个适用于 python 3.6 的变体：

# Imports
import pylab
import scipy
import sklearn.cluster

# Set up original data - note that it's monotonically increasing by X value!
data = {}
data['original'] = {}
data['original']['x'] = [0, 5024.2059124920379, 7933.1645067836089, 7990.4664106277542, 9879.9717114947653, 13738.60563208926, 15113.277958924193]
data['original']['y'] = [0.0, 3072.5653360000988, 5477.2689107965398, 5851.6866463790966, 6056.3852496014106, 7895.2332350173638, 9154.2956175610598]

# Cluster data, sort it and and save
import numpy
inputNumpy = numpy.array([[data['original']['x'][i], data['original']['y'][i]] for i in range(0, len(data['original']['x']))])
meanShift = sklearn.cluster.MeanShift()
meanShift.fit(inputNumpy)
clusteredData = [[pair[0], pair[1]] for pair in meanShift.cluster_centers_]

clusteredData.sort(key=lambda li: li[0])
data['clustered'] = {}
data['clustered']['x'] = [pair[0] for pair in clusteredData]
data['clustered']['y'] = [pair[1] for pair in clusteredData]

# Build a spline using the clustered data and predict
mySpline = scipy.interpolate.UnivariateSpline(x=data['clustered']['x'], y=data['clustered']['y'], k=1)
xi = range(0, int(round(max(data['original']['x']), -3)) + 3000, 20)
yi = mySpline(xi)

# Plot the datapoints
pylab.plot(data['clustered']['x'], data['clustered']['y'], "D", label="Datapoints (%s)" % 'clustered')
pylab.plot(xi, yi, label="Predicted (%s)" %  'clustered')
pylab.plot(data['original']['x'], data['original']['y'], "o", label="Datapoints (%s)" % 'original')

# Show the plot
pylab.grid(True)
pylab.xticks(rotation=45)
pylab.show()