如何使用pylab和numpy将正弦曲线拟合到我的数据中？

Question

对于一个学校项目，我试图表明经济遵循相对正弦的增长模式。除了它的经济学，这是不确定的，我正在构建一个python模拟，以表明，即使我们让某种程度的随机性保持不变，我们仍然可以产生一些相对正弦的东西。我对我正在制作的数据感到满意，但现在我想找到一些方法来获得与数据非常匹配的正弦图。我知道你可以做多项式拟合，但你能正确地适合吗？

提前感谢您的帮助。如果您想要查看代码的任何部分，请告诉我。

Answer 1

您可以使用scipy中的least-square optimization函数将任意函数与另一个函数相匹配。在拟合sin函数的情况下，拟合的3个参数是偏移（'a'），幅度（'b'）和相位（'c'）。

只要你对参数提供合理的初步猜测，优化就应该很好地收敛。幸运的是，对于正弦函数，首先估计其中2个很容易：可以通过取数据的平均值来估计偏移量。通过RMS的振幅（3 *标准差/ sqrt（2））。

注意：作为后期编辑，还添加了频率拟合。这不能很好地工作（可能导致极差的适应性）。因此，根据您的判断，我的建议是不使用频率拟合，除非频率误差小于百分之几。

这导致以下代码：

import numpy as np
from scipy.optimize import leastsq
import pylab as plt

N = 1000 # number of data points
t = np.linspace(0, 4*np.pi, N)
f = 1.15247 # Optional!! Advised not to use
data = 3.0*np.sin(f*t+0.001) + 0.5 + np.random.randn(N) # create artificial data with noise

guess_mean = np.mean(data)
guess_std = 3*np.std(data)/(2**0.5)/(2**0.5)
guess_phase = 0
guess_freq = 1
guess_amp = 1

# we'll use this to plot our first estimate. This might already be good enough for you
data_first_guess = guess_std*np.sin(t+guess_phase) + guess_mean

# Define the function to optimize, in this case, we want to minimize the difference
# between the actual data and our "guessed" parameters
optimize_func = lambda x: x[0]*np.sin(x[1]*t+x[2]) + x[3] - data
est_amp, est_freq, est_phase, est_mean = leastsq(optimize_func, [guess_amp, guess_freq, guess_phase, guess_mean])[0]

# recreate the fitted curve using the optimized parameters
data_fit = est_amp*np.sin(est_freq*t+est_phase) + est_mean

# recreate the fitted curve using the optimized parameters

fine_t = np.arange(0,max(t),0.1)
data_fit=est_amp*np.sin(est_freq*fine_t+est_phase)+est_mean

plt.plot(t, data, '.')
plt.plot(t, data_first_guess, label='first guess')
plt.plot(fine_t, data_fit, label='after fitting')
plt.legend()
plt.show()

编辑：我假设您知道正弦波中的周期数。如果你不这样做，那么适合它会有些棘手。您可以尝试通过手动绘图来猜测周期数，并尝试将其优化为第6个参数。

Answer 2

这是一个无参数拟合函数fit_sin()，不需要手动猜测频率：

import numpy, scipy.optimize

def fit_sin(tt, yy):
    '''Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc"'''
    tt = numpy.array(tt)
    yy = numpy.array(yy)
    ff = numpy.fft.fftfreq(len(tt), (tt[1]-tt[0]))   # assume uniform spacing
    Fyy = abs(numpy.fft.fft(yy))
    guess_freq = abs(ff[numpy.argmax(Fyy[1:])+1])   # excluding the zero frequency "peak", which is related to offset
    guess_amp = numpy.std(yy) * 2.**0.5
    guess_offset = numpy.mean(yy)
    guess = numpy.array([guess_amp, 2.*numpy.pi*guess_freq, 0., guess_offset])

    def sinfunc(t, A, w, p, c):  return A * numpy.sin(w*t + p) + c
    popt, pcov = scipy.optimize.curve_fit(sinfunc, tt, yy, p0=guess)
    A, w, p, c = popt
    f = w/(2.*numpy.pi)
    fitfunc = lambda t: A * numpy.sin(w*t + p) + c
    return {"amp": A, "omega": w, "phase": p, "offset": c, "freq": f, "period": 1./f, "fitfunc": fitfunc, "maxcov": numpy.max(pcov), "rawres": (guess,popt,pcov)}

初始频率猜测由频域中的峰值频率使用FFT给出。假设只有一个主导频率（零频率峰值除外），拟合结果几乎是完美的。

import pylab as plt

N, amp, omega, phase, offset, noise = 500, 1., 2., .5, 4., 3
#N, amp, omega, phase, offset, noise = 50, 1., .4, .5, 4., .2
#N, amp, omega, phase, offset, noise = 200, 1., 20, .5, 4., 1
tt = numpy.linspace(0, 10, N)
tt2 = numpy.linspace(0, 10, 10*N)
yy = amp*numpy.sin(omega*tt + phase) + offset
yynoise = yy + noise*(numpy.random.random(len(tt))-0.5)

res = fit_sin(tt, yynoise)
print( "Amplitude=%(amp)s, Angular freq.=%(omega)s, phase=%(phase)s, offset=%(offset)s, Max. Cov.=%(maxcov)s" % res )

plt.plot(tt, yy, "-k", label="y", linewidth=2)
plt.plot(tt, yynoise, "ok", label="y with noise")
plt.plot(tt2, res["fitfunc"](tt2), "r-", label="y fit curve", linewidth=2)
plt.legend(loc="best")
plt.show()

即使噪音很高，结果也很好：

  

幅度= 1.00660540618，角频率= 2.03370472482，相位= 0.360276844224，偏移= 3.95747467506，最大值。 COV。= 0.0122923578658

Answer 3

我们对用户更友好的是功能曲线拟合。这是一个例子：

import numpy as np
from scipy.optimize import curve_fit
import pylab as plt

N = 1000 # number of data points
t = np.linspace(0, 4*np.pi, N)
data = 3.0*np.sin(t+0.001) + 0.5 + np.random.randn(N) # create artificial data with noise

guess_freq = 1
guess_amplitude = 3*np.std(data)/(2**0.5)
guess_phase = 0
guess_offset = np.mean(data)

p0=[guess_freq, guess_amplitude,
    guess_phase, guess_offset]

# create the function we want to fit
def my_sin(x, freq, amplitude, phase, offset):
    return np.sin(x * freq + phase) * amplitude + offset

# now do the fit
fit = curve_fit(my_sin, t, data, p0=p0)

# we'll use this to plot our first estimate. This might already be good enough for you
data_first_guess = my_sin(t, *p0)

# recreate the fitted curve using the optimized parameters
data_fit = my_sin(t, *fit[0])

plt.plot(data, '.')
plt.plot(data_fit, label='after fitting')
plt.plot(data_first_guess, label='first guess')
plt.legend()
plt.show()

Answer 4

将sin曲线拟合到给定数据集的当前方法需要首先猜测参数，然后是交互过程。这是一个非线性回归问题。

另一种方法在于通过方便的积分方程将非线性回归转换为线性回归。然后，不需要初始猜测，也不需要迭代过程：直接获得拟合。

如果功能y = a + r*sin(w*x+phi)或y=a+b*sin(w*x)+c*cos(w*x)，请参阅Scribd上发表的论文"Régression sinusoidale"的第35-36页

如果函数y = a + p*x + r*sin(w*x+phi)：“混合线性和正弦回归”一章的第49-51页。

如果功能更复杂，一般过程在章节"Generalized sinusoidal regression"第54-61页中进行了解释，后面是数字示例y = r*sin(w*x+phi)+(b/x)+c*ln(x)，第62-63页

Answer 5

下面的Python脚本执行正弦曲线的最小二乘拟合，如Bloomfield（2000），“时间序列的傅立叶分析”，Wiley中所述。关键步骤如下：

1. 定义一系列可能的不同频率。
2. 对于上面第1点指定的每个频率，用普通最小二乘法（OLS）估算正弦曲线的所有参数（均值，幅度和相位）。
3. 在以上第2点估计的所有不同参数集中，选择一个最小化残差平方和（RSS）的参数。

import pandas as pd
import numpy as np
import statsmodels.api as sm
import matplotlib.pyplot as plt
######################################################################################
# (1) generate the data
######################################################################################
omega=4.5  # frequency
theta0=0.5 # mean
theta1=1.5 # amplitude
phi=-0.25  # phase

n=1000 # number of observations
sigma=1.25 # error standard deviation
e=np.random.normal(0,sigma,n) # Gaussian error

t=np.linspace(1,n,n)/n # time index
y=theta0+theta1*np.cos(2*np.pi*(omega*t+phi))+e # time series
######################################################################################
# (2) estimate the parameters
######################################################################################
# define the range of different possible frequencies
f=np.linspace(1,12,100)

# create a data frame to store the estimated parameters and the associated
# residual sum of squares for each of the considered frequencies
coefs=pd.DataFrame(data=np.zeros((len(f),5)), columns=["omega","theta0","theta1","phi","RSS"])

for i in range(len(f)): # iterate across the different considered frequencies

    x1=np.cos(2*np.pi*f[i]*t) # define the first regressor
    x2=np.sin(2*np.pi*f[i]*t) # define the second regressor

    X=sm.add_constant(np.transpose(np.vstack((x1,x2)))) # add the intercept
    reg=sm.OLS(y,X).fit() # fit the regression by OLS

    A=reg.params[1] # estimated coefficient of first regressor
    B=reg.params[2] # estimated coefficient of second regressor

    # estimated mean
    mean=reg.params[0]

    # estimated amplitude
    amplitude=np.sqrt(A**2+B**2)

    # estimated phase
    if A>0:          phase=np.arctan(-B/A)/(2*np.pi)
    if A<0 and B>0:  phase=(np.arctan(-B/A)-np.pi)/(2*np.pi)
    if A<0 and B<=0: phase=(np.arctan(-B/A)+np.pi)/(2*np.pi)
    if A==0 and B>0: phase=-1/4
    if A==0 and B<0: phase=1/4

    # fitted sinusoid
    s=mean+amplitude*np.cos(2*np.pi*(f[i]*t+phase))

    # residual sum of squares
    RSS=np.sum((y-s)**2)

    # save the estimation results
    coefs["omega"][i]=f[i]
    coefs["theta0"][i]=mean
    coefs["theta1"][i]=amplitude
    coefs["phi"][i]=phase
    coefs["RSS"][i]=RSS

    del x1, x2, X, reg, A, B, mean, amplitude, phase, s, RSS
######################################################################################
# (3) analyze the results
######################################################################################
# extract the set of parameters that minimizes the residual sum of squares
coefs=coefs.loc[coefs["RSS"]==coefs["RSS"].min(),]

# calculate the fitted sinusoid
s=coefs["theta0"].values+coefs["theta1"].values*np.cos(2*np.pi*(coefs["omega"].values*t+coefs["phi"].values))

# plot the fitted sinusoid
plt.plot(y,color="black",linewidth=1,label="actual")
plt.plot(s,color="lightgreen",linewidth=3,label="fitted")
plt.ylim(ymax=np.max(y)*1.3)
plt.legend(loc=1)
plt.show()