python - Scipy 非常奇怪地拟合并创建了多个拟合曲线，这不应该是这种情况，我知道曲线拟合不会返回多条曲线

Question

这是我的代码

yp = df.final_sulfur/df.start_sulfur
xp = df.mag/df.hm_weight
sns.set(font_scale=1.5, font='DejaVu Sans')
fig, ax = plt.subplots(1,1, figsize=(9, 9))
yp = df.final_sulfur/df.start_sulfur
xp = df.mag/df.hm_weight
p = ax.plot(xp, yp, marker='o', markersize=1, alpha=0.2, linestyle='none')
p = ax.set_xlabel('mag/hm_weight')
p = ax.set_ylabel('final/start sulfur')
p = ax.set_title('final S vs mag')
p = ax.set_xlim(0.08, 12)
p = ax.set_ylim(3.0e-3, 1.5)
p = ax.set_xscale('log')
p = ax.set_yscale('log')
leg = plt.legend()

试图拟合下面的曲线是我认为指数衰减应该做的方程，但我得到的是完全糟糕的结果

import scipy as sp
from scipy.optimize import curve_fit
def func(xp, a, b, c):
    return a*np.exp(-b*xp) + c  
popt2, pcov2 = curve_fit(func, xp, yp, p0=None)
a2, b2, c2 = popt2
print ('a2=', a2, 'b2=', b2, 'c2=', c2)
print ('func=', func(xp, a2, b2, c2))
ax.plot(xp, func(xp, *popt2), 'b-', label='Fit',linestyle='--',color='red')

注意：我需要使用对数刻度进行绘图，这是有意义的

我的示例数据看起来如何（不幸的是，很难将所有数据都放在这里）xp（示例）为了 SO 而转换为列表

[1.8535530937584435,
 0.3220338983050847,
 1.184544992374174,
 0.7583873696081196,
 0.3209681662720337,
 1.158380317785751,
 1.6285714285714286,
 0.44209925841414716,
 0.7396205664008799,
 0.27983539094650206,
 0.575319622012229,
 0.3287671232876712,
 1.382921589688507,
 0.8247978436657682,
 1.315934065934066,
 0.23450134770889489,
 0.5697069296083265,
 1.0015731515469324,
 1.2841602547094721,
 0.645600653772814,
 0.4599483204134367,
 0.8340260459961208,
 0.8992900341835393,
 0.961340206185567,
 0.5845225027442371,
 0.9623773173391493,
 1.3451708366962605,
 0.8418230563002681,
 0.7456025203465477,
 1.9345156889495225]
 yp [0.05202312138728324,
 0.47058823529411764,
 0.04833333333333333,
 0.11392405063291139,
 0.36363636363636365,
 0.020588235294117647,
 0.008823529411764706,
 0.25641025641025644,
 0.12,
 0.47826086956521735,
 0.1826923076923077,
 0.3333333333333333,
 0.01282051282051282,
 0.029411764705882353,
 0.03225806451612903,
 0.26666666666666666,
 0.05,
 0.011428571428571429,
 0.12080536912751677,
 0.11764705882352941,
 0.2926829268292683,
 0.049999999999999996,
 0.06578947368421052,
 0.08024691358024691,
 0.15517241379310343,
 0.024390243902439025,
 0.017543859649122806,
 0.05479452054794521,
 0.03571428571428571,
 0.007142857142857143]

Answer 1

这就是我为什么愿意得到更好的曲线更多的指数衰减

yp = df.final_sulfur/df.start_sulfur
xp = df.mag/df.hm_weight
test_data=xp.to_frame(name = 'xp').join(yp.to_frame(name='yp'))
sns.set(font_scale=1.5, font='DejaVu Sans')
fig, ax = plt.subplots(1,1, figsize=(9,15))
yp = df.final_sulfur/df.start_sulfur
xp = df.mag/df.hm_weight
p = ax.plot(xp, yp, marker='o', markersize=1, alpha=0.2, linestyle='none')
#p = sns.regplot(x=xp, y=yp, data=test_data,logx=True)
model = lambda x, A, x0, sigma, offset:  offset+A*np.exp(-((x-x0)/sigma)**1)
popt, pcov = curve_fit(model, xp.values, yp.values, p0=[0,0.5,1,1])
x = np.linspace(xp.values.min(),xp.values.max())
p =ax.plot(x,model(x,*popt), label="fit",color='red',linestyle='--')
model2 = lambda x, sigma:  model(x,0.5,0,sigma**1,0)
x2 = np.linspace(xp.values.min(),xp.values.max())
popt2, pcov2 = curve_fit(model2, xp.values, 
                              yp.values, p0=[1])
p= ax.plot(x2,model2(x2,*popt2), label="fit2",color='green')
model3 = lambda x, A, x0, sigma, offset:  offset+A*np.exp(-((x+x0)/sigma)**1)
popt, pcov = curve_fit(model3, xp.values, yp.values, p0=[0,0.5,1,1])
x3 = np.linspace(xp.values.min(),xp.values.max())
p = ax.plot(x3,model(x3,*popt), label="fit3",color='blue')