Python 如何平滑仅在特定部分有较大噪声的曲线？

我想平滑下图所示的散点图（点非常密集），数据是

曲线中间有较大的噪声，我想平滑曲线，y值也应<>强>单调增加p> 因为有很多这样的曲线，所以很难知道噪声在曲线中的位置

我尝试了

scipy.signal.savgol\u过滤器，但没有成功

我使用的代码是：

from scipy.signal import savgol_filter
from scipy import interpolate
import numpy as np
import matplotlib.pyplot as plt

s = np.loadtxt('data.csv', delimiter=',')
x = s[:, 0]
y = s[:, 1]
yhat = savgol_filter(y, 551, 3)

plt.plot(x, y, 'r')
plt.plot(x, yhat, 'b')
plt.show()

我们非常感谢您的建议。谢谢

--------------更新------------------

按照科林的方法，我得到了我想要的结果。代码如下：

from scipy.signal import savgol_filter
from scipy import interpolate
import numpy as np
import matplotlib.pyplot as plt

s = np.loadtxt('data.csv', delimiter=',')
x = s[:, 0]
y = s[:, 1]
yhat = savgol_filter(y, 551, 3)

tolerance = 0.2
increased_span = 150
filter_size = 11

first_pass = medfilt(y,filter_size)
diff = (y-first_pass)**2
first = np.argmax(diff>tolerance) - increased_span
last = len(y) - np.argmax(diff[::-1]>tolerance) + increased_span
print (first, last)
#interpolate between increased span
yhat[first:last] = np.interp(x[first:last], [x[first], x[last]],  [y[first], y[last]])


f = interpolate.interp1d(x, yhat, kind='slinear')
x_inter = np.linspace(x[0], x[-1], 1000)
y_inter = f(x_inter)
y_inter = savgol_filter(y_inter, 41, 3)

plt.plot(x, y, 'r')
plt.plot(x, yhat, 'b')
plt.show()

如果我们首先隔离故障区域，有很多方法可以消除它。以下是一个例子：

tolerance = 0.2
increased_span = 150
filter_size = 11

#find noise
first_pass = medfilt(y,filter_size)
diff = (yhat-first_pass)**2
first = np.argmax(diff>tolerance) - increased_span
last = len(y) - np.argmax(diff[::-1]>tolerance) + increased_span

#interpolate between increased span
yhat[first:last] = np.interp(x[first:last], [x[first], x[last]],  [y[first], y[last]])

这些参数来自哪里？我有一个预测，有时，特别是当它真的做得不好时，会产生一条比观察到的趋势线噪音大得多的线。如果能够自动更正，那将是一件非常棒的事情