让我们假设我们有一个数据集，它大概是
import numpy as np
x = np.linspace(0,2*np.pi,100)
y = np.sin(x) + np.random.random(100) * 0.2
Therefore we have a variation of 20% of the dataset. My first idea was to use the UnivariateSpline function of scipy, but the problem is that this does not consider the small noise in a good way. If you consider the frequencies, the background is much smaller than the signal, so a spline only of the cutoff might be an idea, but that would involve a back and forth fourier transformation, which might result in bad behaviour.
Another way would be a moving average, but this would also need the right choice of the delay.
有什么提示/书籍或链接可以解决这个问题吗?
如果你对周期信号的“平滑”版本感兴趣(就像你的例子)，那么FFT是正确的方法。进行傅里叶变换并减去低贡献频率:
import numpy as np
import scipy.fftpack
N = 100
x = np.linspace(0,2*np.pi,N)
y = np.sin(x) + np.random.random(N) * 0.2
w = scipy.fftpack.rfft(y)
f = scipy.fftpack.rfftfreq(N, x[1]-x[0])
spectrum = w**2
cutoff_idx = spectrum < (spectrum.max()/5)
w2 = w.copy()
w2[cutoff_idx] = 0
y2 = scipy.fftpack.irfft(w2)
即使你的信号不是完全周期性的，这也能很好地去除白噪声。有许多类型的过滤器可以使用(高通，低通，等等…)，合适的一个取决于你正在寻找什么。
另一个选择是在statmodel中使用KernelReg:
from statsmodels.nonparametric.kernel_regression import KernelReg
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0,2*np.pi,100)
y = np.sin(x) + np.random.random(100) * 0.2
# The third parameter specifies the type of the variable x;
# 'c' stands for continuous
kr = KernelReg(y,x,'c')
plt.plot(x, y, '+')
y_pred, y_std = kr.fit(x)
plt.plot(x, y_pred)
plt.show()
编辑:看看这个答案。使用np。Cumsum比np.卷积快得多
我使用了一种快速而肮脏的方法来平滑数据，基于移动平均盒(通过卷积):
x = np.linspace(0,2*np.pi,100)
y = np.sin(x) + np.random.random(100) * 0.8
def smooth(y, box_pts):
box = np.ones(box_pts)/box_pts
y_smooth = np.convolve(y, box, mode='same')
return y_smooth
plot(x, y,'o')
plot(x, smooth(y,3), 'r-', lw=2)
plot(x, smooth(y,19), 'g-', lw=2)
