scipy.interpolate.UnivariateSpline无论参数如何都不平滑

Question

我无法获得scipy.interpolate.UnivariateSpline在插值时使用任何平滑.根据功能的页面以及之前的一些帖子,我相信它应该提供s参数的平滑.

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()

结果如下:

我尝试了一系列s值(0.01,0.1,1,2,5,50),以及显式权重,设置为相同的事物(1.0)或随机化.我仍然无法进行任何平滑,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4点(7990.4664106277542,5851.6866463790966)那样的异常值进行平滑处理.

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用于实现这几个数据点的平滑？

Answer 1

简短回答:你需要s更仔细地选择价值.

UnivariateSpline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

由此可以推断,"合理"值,平滑,如果你不明确的权重传递,周围s = m * v哪里m是数据点的数量和v数据的变化.在这种情况下,s_good ~ 5e7.

编辑:合理的值s当然也取决于数据中的噪音水平.文档似乎建议s在与要平滑的"噪声"相关的标准偏差范围内(m - sqrt(2*m)) * std**2 <= s <= (m + sqrt(2*m)) * std**2进行选择std.

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，我通过 scikit-learn 包中的Mean-Shift 聚类取得了成功。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()