python - SciPy leastsq 适合正弦波失败

Question

我想弄清楚我在这里不明白的是什么。

我正在关注 http://www.scipy.org/Cookbook/FittingData并试图拟合正弦波。真正的问题是卫星磁力计数据，它在旋转的航天器上产生了一个很好的正弦波。我创建了一个数据集，然后尝试对其进行拟合以恢复输入。

这是我的代码:

import numpy as np
from scipy import optimize

from scipy.optimize import curve_fit, leastsq

import matplotlib.pyplot as plt


class Parameter:
    def __init__(self, value):
            self.value = value

    def set(self, value):
            self.value = value

    def __call__(self):
            return self.value

def fit(function, parameters, y, x = None):
    def f(params):
        i = 0
        for p in parameters:
            p.set(params[i])
            i += 1
        return y - function(x)

    if x is None: x = np.arange(y.shape[0])
    p = [param() for param in parameters]
    return optimize.leastsq(f, p, full_output=True, ftol=1e-6, xtol=1e-6)

# generate a perfect data set (my real data have tiny error)
def mysine(x, a1, a2, a3):
    return a1 * np.sin(a2 * x + a3)

xReal = np.arange(500)/10.
a1 = 200.
a2 = 2*np.pi/10.5  # omega, 10.5 is the period
a3 = np.deg2rad(10.) # 10 degree phase offset
yReal = mysine(xReal, a1, a2, a3)

# plot the real data
plt.figure(figsize=(15,5))
plt.plot(xReal, yReal, 'r', label='Real Values')

# giving initial parameters
amplitude = Parameter(175.)
frequency = Parameter(2*np.pi/8.)
phase = Parameter(0.0)

# define your function:
def f(x): return amplitude() * np.sin(frequency() * x + phase())

# fit! (given that data is an array with the data to fit)
out = fit(f, [amplitude, frequency, phase], yReal, xReal)
period = 2*np.pi/frequency()
print amplitude(), period, np.rad2deg(phase())

xx = np.linspace(0, np.max(xReal), 50)
plt.plot( xx, f(xx) , label='fit')
plt.legend(shadow=True, fancybox=True)

这使得这个情节:

[44.2434221897 8.094832581 -61.6204033699] 的恢复拟合参数与我开始时的参数完全不同。

对我不理解或做错的事情有什么想法吗？

scipy.__version__
'0.10.1'

---

编辑: 建议固定一个参数。在上面的示例中，将振幅固定为 np.histogram(yReal)[1][-1] 仍然会产生 Not Acceptable 输出。拟合:[175.0 8.31681375217 6.0] 我应该尝试不同的拟合方法吗？有什么建议？

Answer 1

这里是一些实现 Zhenya 想法的代码。 它使用

yhat = fftpack.rfft(yReal)
idx = (yhat**2).argmax()
freqs = fftpack.rfftfreq(N, d = (xReal[1]-xReal[0])/(2*pi))
frequency = freqs[idx]

猜测数据的主要频率，以及

amplitude = yReal.max()

猜测振幅。

---

import numpy as np
import scipy.optimize as optimize
import scipy.fftpack as fftpack
import matplotlib.pyplot as plt
pi = np.pi
plt.figure(figsize = (15, 5))

# generate a perfect data set (my real data have tiny error)
def mysine(x, a1, a2, a3):
    return a1 * np.sin(a2 * x + a3)

N = 5000
xmax = 10
xReal = np.linspace(0, xmax, N)
a1 = 200.
a2 = 2*pi/10.5  # omega, 10.5 is the period
a3 = np.deg2rad(10.) # 10 degree phase offset
print(a1, a2, a3)
yReal = mysine(xReal, a1, a2, a3) + 0.2*np.random.normal(size=len(xReal))

yhat = fftpack.rfft(yReal)
idx = (yhat**2).argmax()
freqs = fftpack.rfftfreq(N, d = (xReal[1]-xReal[0])/(2*pi))
frequency = freqs[idx]

amplitude = yReal.max()
guess = [amplitude, frequency, 0.]
print(guess)
(amplitude, frequency, phase), pcov = optimize.curve_fit(
    mysine, xReal, yReal, guess)

period = 2*pi/frequency
print(amplitude, frequency, phase)

xx = xReal
yy = mysine(xx, amplitude, frequency, phase)
# plot the real data
plt.plot(xReal, yReal, 'r', label = 'Real Values')
plt.plot(xx, yy , label = 'fit')
plt.legend(shadow = True, fancybox = True)
plt.show()

产量

(200.0, 0.5983986006837702, 0.17453292519943295)   # (a1, a2, a3)
[199.61981404516041, 0.61575216010359946, 0.0]     # guess
(200.06145097308041, 0.59841420869261097, 0.17487141943703263) # fitted parameters

请注意，通过使用 fft，对频率的猜测已经非常接近最终拟合参数。

看来您不需要修复任何参数。 通过使频率猜测更接近实际值，optimize.curve_fit 能够收敛到一个合理的答案。