Scipy优化CurveFit计算错误值

Question

我很想知道两种正弦波类型之间的相移。为此，我试图用 scipy.cuve_fit 来拟合每个波浪。我一直在关注 this post。然而，我获得了负振幅，相移有时看起来像转发的 pi 弧度。

我使用的代码如下：

def fit_sin_LD(t_LD, y_LD):

'''Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc"'''

ff = np.fft.fftfreq(len(t_LD), (t_LD[1]-t_LD[0]))   # assume uniform spacing
Fyy = abs(np.fft.fft(y_LD))
guess_freq = abs(ff[np.argmax(Fyy[1:])+1])   # excluding the zero frequency "peak", which is related to offset
guess_amp = np.std(y_LD) * 2.**0.5
guess_offset = np.mean(y_LD)
guess = np.array([guess_amp, 2.*np.pi*guess_freq, 0., guess_offset])

def sinfunc_LD(t_LD, A, w, p, c):
    return A * np.sin(w*t_LD + p) + c
#boundary=([0,-np.inf,-np.pi, 1.5],[0.8, +np.inf, np.pi, 2.5])
popt, pcov = scipy.optimize.curve_fit(sinfunc_LD, t_LD, y_LD, p0=guess, maxfev=3000) # with maxfev= number I can increase the number of iterations
A, w, p, c = popt
f          = w/(2.*np.pi)
fitfunc_LD = lambda t_LD: A*np.sin(w*t_LD + p) + c
fitted_LD  = fitfunc_LD(t_LD)
dic_LD = {"amp_LD": A, "omega_LD": w, "phase_LD": p, "offset_LD": c, "freq_LD": f, "period_LD": 1./f, "fitfunc_LD": fitted_LD, "maxcov_LD": np.max(pcov), "rawres_LD": (guess, popt, pcov)}
return dic_LD


def fit_sin_APD(t_APD, y_APD):

''' Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc" '''

ff = np.fft.fftfreq(len(t_APD), (t_APD[1]-t_APD[0]))   # assume uniform spacing
Fyy = abs(np.fft.fft(y_APD))
guess_freq = abs(ff[np.argmax(Fyy[1:])+1])   # excluding the zero frequency "peak", which is related to offset
guess_amp = np.std(y_APD) * 2.**0.5
guess_offset = np.mean(y_APD)
guess = np.array([guess_amp, 2.*np.pi*guess_freq, 0., guess_offset])

def sinfunc_APD(t_APD, A, w, p, c):
    return A * np.sin(w*t_APD + p) + c
#boundary=([0,0,-np.pi, 0.0],[np.inf, np.inf, np.pi, 0.7])
popt, pcov  = scipy.optimize.curve_fit(sinfunc_APD, t_APD, y_APD, p0=guess, maxfev=5000) # with maxfev= number I can increase the number of iterations
A, w, p, c  = popt
f           = w/(2.*np.pi)
fitfunc_APD = lambda t_APD: A*np.sin(w*t_APD + p) + c
fitted_APD  = fitfunc_APD(t_APD)
dic_APD     = {"amp_APD": A, "omega_APD": w, "phase_APD": p, "offset_APD": c, "freq_APD": f, "period_APD": 1./f, "fitfunc_APD": fitted_APD, "maxcov_APD": np.max(pcov), "rawres_APD": (guess, popt, pcov)}       
return dic_APD

我不明白为什么 curve_fit 会返回一个负振幅（这在物理学上没有意义）。我也尝试将边界条件设置为 **kwargs* ：

bounds=([0.0, -np.inf,-np.pi, 0.0],[+np.inf, +np.inf,-np.pi, +np.inf])

但它会产生更奇怪的结果。

我添加了一张显示这种差异的图片：

有人知道如何解决这个相位和振幅问题吗？

提前致谢

Answer 1

这里有几个问题我不明白：

不需要在"fit function"
如果唯一不同的是字典的命名，则无需定义两次。（虽然我不明白为什么一开始就必须以不同的方式命名）
可以直接拟合频率而不是omega
预计算拟合值时，直接使用给定的拟合函数

总的来说，我不明白为什么第二次拟合会失败，并且在这里使用一些通用数据，但事实并非如此。考虑到在物理学中振幅可能很复杂这一事实，我对负面结果没有任何问题。不过，我理解 OP 中的重点。当然，拟合算法不了解物理学，从数学上讲，振幅为负没有问题。这只是给出了 pi 的额外相移。因此，在处理所需的相移时，可以很容易地强制正振幅。我在这里介绍了这个作为可能的关键字参数。此外，我将其简化为一个拟合函数，可能 "renaming" 的输出字典键作为关键字参数。

import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit


def sinfunc( t, A, f, p, c ):
    return A * np.sin( 2.0 * np.pi * f * t + p) + c

def fit_sin(t_APD, y_APD, addName="", posamp=False):

    ''' Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc" '''

    ff = np.fft.fftfreq( len( t_APD ), t_APD[1] - t_APD[0] ) # assume uniform spacing
    Fyy = abs( np.fft.fft( y_APD ) )
    guess_freq = abs( ff[np.argmax( Fyy[1:] ) + 1] )   # excluding the zero frequency "peak", which is related to offset
    guess_amp = np.std( y_APD ) * 2.**0.5
    guess_offset = np.mean( y_APD )
    guess = np.array( [ guess_amp, guess_freq, 0., guess_offset ] )

    popt, pcov  = curve_fit(sinfunc, t_APD, y_APD, p0=guess, maxfev=500) # with maxfev= number I can increase the number of iterations
    if popt[0] < 0 and posamp:
        popt[0] = -popt[0]
        popt[2] += np.pi 
        popt[2] = popt[2] % ( 2 * np.pi )
    A, f, p, c  = popt 
    fitted_APD  = sinfunc( t_APD, *popt )
    dic_APD     = {
            "amp{}".format(addName): A, 
            "omega{}".format(addName): 2.0 * np.pi * f, 
            "phase{}".format(addName): p, 
            "offset{}".format(addName): c, 
            "freq{}".format(addName): f, 
            "period{}".format(addName): 1.0 / f, 
            "fitfunc{}".format(addName): fitted_APD, 
            "maxcov{}".format(addName): np.max( pcov ), 
            "rawres{}".format(addName): ( guess, popt, pcov ) }
    return dic_APD

tl = np.linspace(0,1e-6, 150 )
sl1 = np.fromiter( (sinfunc(t, .18, 4998735, 3.6, 2.0 ) + .01 *( 1 - 2 * np.random.random() ) for t in tl ), np.float )
sl2 = np.fromiter( (sinfunc(t, .06, 4998735, 2.1, 0.4 ) + .01 *( 1 - 2 * np.random.random() ) for t in tl ), np.float )

ld = fit_sin(tl, sl1, addName="_ld" )
print ld["amp_ld"]
ld = fit_sin(tl, sl1, addName="_ld", posamp=True )
print ld["amp_ld"]
apd = fit_sin(tl, sl2 )

fig = plt.figure("1")
ax = fig.add_subplot( 1, 1, 1 )

ax.plot( tl, sl1, color="r" )
ax.plot( tl, ld["fitfunc_ld"], color="k", ls="--" )
ax.plot( tl, sl2, color="#50FF80" )
ax.plot( tl, apd["fitfunc"], color="k", ls="--" )

ax.grid()
plt.show()

这给了我：

-0.180108427200549
0.180108427200549

即在第一次尝试中，尽管对幅度的猜测很好，但结果却是负数。这可能是由于大相。由于该猜测为零，因此算法更容易先切换幅度符号然后调整相位。如上所述，这很容易纠正，甚至不需要错误传播。

Scipy优化CurveFit计算错误值

Scipy Optimize CurveFit calculates wrong values

python

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python-3.x

scipy-optimize