动态创建傅立叶级数的scipy曲线拟合定义

Question

我想使用 numpy 和 scipy 为x-y数据集实现傅立叶级数开发。

首先，我希望将我的数据与第一个 8 余弦相匹配，并仅绘制一次谐波。所以我写了以下两个函数定义：

# fourier series defintions
tau = 0.045
def fourier8(x, a1, a2, a3, a4, a5, a6, a7, a8):
    return a1 * np.cos(1 * np.pi / tau * x) + \
           a2 * np.cos(2 * np.pi / tau * x) + \
           a3 * np.cos(3 * np.pi / tau * x) + \
           a4 * np.cos(4 * np.pi / tau * x) + \
           a5 * np.cos(5 * np.pi / tau * x) + \
           a6 * np.cos(6 * np.pi / tau * x) + \
           a7 * np.cos(7 * np.pi / tau * x) + \
           a8 * np.cos(8 * np.pi / tau * x)


def fourier1(x, a1):
    return a1 * np.cos(1 * np.pi / tau * x)

然后我用它们来拟合我的数据：

# import and filename
filename = 'data.txt'
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

z, Ua = np.loadtxt(filename,delimiter=',', unpack=True)
tau = 0.045

# plot data
fig = plt.figure()
ax1 = fig.add_subplot(111)
p1, = plt.plot(z,Ua)

# fits
popt, pcov = curve_fit(fourier8, z, Ua)

# further plots
Ua_fit8 = fourier8(z,*popt)
Ua_fit1 = fourier1(z,popt[0])
p2, = plt.plot(z,Ua_fit8)
p3, = plt.plot(z,Ua_fit1)

plt.show()

按预期工作：

但是我知道我因为谐波的仲裁命令而变得通用，例如我想用第一个 15个谐波拟合我的数据并绘制第一个三个谐波。

如果不定义fourier1，fourier2，fourier3 ...，fourier15，我怎能实现这一目标？

以下是使用它的示例文本文件data.txt：

1.000000000000000021e-03,2.794682735905079767e+02
2.000000000000000042e-03,2.792294526290349950e+02
2.999999999999999629e-03,2.779794770690260179e+02
4.000000000000000083e-03,2.757183469104809888e+02
5.000000000000000104e-03,2.734572167519349932e+02
5.999999999999999258e-03,2.711960865933900209e+02
7.000000000000000146e-03,2.689349564348440254e+02
8.000000000000000167e-03,2.714324329982829909e+02
8.999999999999999320e-03,2.739299095617229796e+02
1.000000000000000021e-02,2.764273861251620019e+02
1.100000000000000110e-02,2.789248626886010243e+02
1.199999999999999852e-02,2.799669443683339978e+02
1.299999999999999940e-02,2.795536311643609793e+02
1.400000000000000029e-02,2.791403179603880176e+02
1.499999999999999944e-02,2.787270047564149991e+02
1.600000000000000033e-02,2.783136915524419805e+02
1.699999999999999775e-02,2.673604939531239779e+02
1.799999999999999864e-02,2.564072963538059753e+02
1.899999999999999953e-02,2.454540987544889958e+02
2.000000000000000042e-02,2.345009011551709932e+02
2.099999999999999784e-02,1.781413355804160119e+02
2.200000000000000219e-02,7.637540203022649621e+01
2.299999999999999961e-02,-2.539053151996269975e+01
2.399999999999999703e-02,-1.271564650701519952e+02
2.500000000000000139e-02,-2.289223986203409993e+02
2.599999999999999881e-02,-2.399383538664330047e+02
2.700000000000000316e-02,-2.509543091125239869e+02
2.800000000000000058e-02,-2.619702643586149975e+02
2.899999999999999800e-02,-2.729862196047059797e+02
2.999999999999999889e-02,-2.786861050144170235e+02
3.099999999999999978e-02,-2.790699205877460258e+02
3.200000000000000067e-02,-2.794537361610759945e+02
3.300000000000000155e-02,-2.798375517344049968e+02
3.399999999999999550e-02,-2.802213673077350222e+02
3.500000000000000333e-02,-2.776516459805940258e+02
3.599999999999999728e-02,-2.750819246534539957e+02
3.700000000000000511e-02,-2.725122033263129993e+02
3.799999999999999906e-02,-2.699424819991720028e+02
3.899999999999999994e-02,-2.698567311502329744e+02
4.000000000000000083e-02,-2.722549507794930150e+02
4.100000000000000172e-02,-2.746531704087540220e+02
4.199999999999999567e-02,-2.770513900380149721e+02
4.299999999999999656e-02,-2.794496096672759791e+02
4.400000000000000439e-02,-2.800761105821779893e+02
4.499999999999999833e-02,-2.800761105821779893e+02
4.599999999999999922e-02,-2.800761105821779893e+02
4.700000000000000011e-02,-2.800761105821779893e+02
4.799999999999999406e-02,-2.788333722531979788e+02
4.900000000000000189e-02,-2.763478955952380147e+02
5.000000000000000278e-02,-2.738624189372779938e+02
5.100000000000000366e-02,-2.713769422793179729e+02
5.199999999999999761e-02,-2.688914656213580088e+02
5.299999999999999850e-02,-2.715383673199499981e+02
5.400000000000000633e-02,-2.741852690185419874e+02
5.499999999999999334e-02,-2.768321707171339767e+02
5.600000000000000117e-02,-2.794790724157260229e+02
5.700000000000000205e-02,-2.804776351435970128e+02
5.799999999999999600e-02,-2.798278589007459800e+02
5.899999999999999689e-02,-2.791780826578950041e+02
5.999999999999999778e-02,-2.785283064150449945e+02
6.100000000000000561e-02,-2.778785301721940186e+02
6.199999999999999956e-02,-2.670252067497989970e+02
6.300000000000000044e-02,-2.561718833274049985e+02
6.400000000000000133e-02,-2.453185599050100052e+02
6.500000000000000222e-02,-2.344652364826150119e+02
6.600000000000000311e-02,-1.780224826854309867e+02
6.700000000000000400e-02,-7.599029851345700592e+01
6.799999999999999101e-02,2.604188565851649884e+01
6.900000000000000577e-02,1.280740698304900036e+02
7.000000000000000666e-02,2.301062540024639986e+02
7.100000000000000755e-02,2.404921248105050040e+02
7.199999999999999456e-02,2.508779956185460094e+02
7.299999999999999545e-02,2.612638664265870148e+02
7.400000000000001021e-02,2.716497372346279917e+02
7.499999999999999722e-02,2.773051723900500178e+02
7.599999999999999811e-02,2.782301718928520131e+02
7.699999999999999900e-02,2.791551713956549747e+02
7.799999999999999989e-02,2.800801708984579932e+02
7.900000000000001465e-02,2.810051704012610116e+02
8.000000000000000167e-02,2.785107135689390248e+02
8.099999999999998868e-02,2.760162567366169810e+02
8.200000000000000344e-02,2.735217999042949941e+02
8.300000000000000433e-02,2.710273430719730072e+02
8.399999999999999134e-02,2.706544464035359852e+02
8.500000000000000611e-02,2.724031098989830184e+02
8.599999999999999312e-02,2.741517733944299948e+02
8.699999999999999400e-02,2.759004368898779944e+02
8.800000000000000877e-02,2.776491003853250277e+02
8.899999999999999578e-02,2.783445666445250026e+02
8.999999999999999667e-02,2.790400329037249776e+02

Answer 1

我会通过定义一个接受可变数量参数的函数fourier来解决这个问题，使用循环遍历余弦来评估每个输入点的函数：

def fourier(x, *a):
    ret = a[0] * np.cos(np.pi / tau * x)
    for deg in range(1, len(a)):
        ret += a[deg] * np.cos((deg+1) * np.pi / tau * x)
    return ret

因为此函数采用可变数量的参数，所以需要提供正确维度的起始位置作为curve_fit函数的提示，以便它知道要使用的余弦数量。在示例中，您提供了8个余弦：

popt, pcov = curve_fit(fourier, z, Ua, [1.0] * 8)

现在，你要问的用例（符合15次谐波和绘制前三个）可以通过以下方式完成：

# Fit with 15 harmonics
popt, pcov = curve_fit(fourier, z, Ua, [1.0] * 15)

# Plot data, 15 harmonics, and first 3 harmonics
fig = plt.figure()
ax1 = fig.add_subplot(111)
p1, = plt.plot(z,Ua)
p2, = plt.plot(z, fourier(z, *popt))
p3, = plt.plot(z, fourier(z, popt[0], popt[1], popt[2]))
plt.show()

Answer 2

基于这个帖子。

Passing additional arguments using scipy.optimize.curve_fit?

# import and filename
filename = 'data.txt'
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

z, Ua = np.loadtxt(filename,delimiter=',', unpack=True)
tau = 0.045

def make_fourier(na, nb):
    def fourier(x, *a):
        ret = 0.0
        for deg in range(0, na):
            ret += a[deg] * np.cos((deg+1) * np.pi / tau * x)
        for deg in range(na, na+nb):
            ret += a[deg] * np.sin((deg+1) * np.pi / tau * x)
        return ret
    return fourier

popt, pcov = curve_fit(make_fourier(15,15), z, Ua, [0.0]*30)

# Plot data, 15 harmonics, and first 3 harmonics
fig = plt.figure()
ax1 = fig.add_subplot(111)
p1, = plt.plot(z,Ua)
p2, = plt.plot(z, (make_fourier(15,15))(z, *popt))
#p3, = plt.plot(z, (make_fourier(8,8))(z, popt[0], popt[1], popt[2]))
plt.show()