Python中边缘零斜率约束的曲线拟合

Question

我希望对以下数据进行曲线拟合，这样我就可以使其拟合一个趋势，其边缘的斜率为零。polyfit的输出将拟合该数据，但边缘处的坡度不是零。你知道吗

这是我想要的输出-请原谅我的油漆工作。我需要它像这样适合，这样我就可以适当地消除正弦/余弦偏差的数据，这是不是真正的中心。 数据如下：

[0.23353535 0.25586247 0.26661164 0.26410896 0.24963951 0.22670266
 0.19955422 0.17190263 0.1598439  0.17351905 0.18212444 0.18438673
 0.17952432 0.18314894 0.19265689 0.19432385 0.19605163 0.20326011
 0.20890851 0.20590997 0.21856518 0.23771665 0.24530019 0.23940831
 0.22078396 0.23075128 0.2346082  0.22466281 0.24384843 0.26339594
 0.26414153 0.24664183 0.24278978 0.31023648 0.3614195  0.37773436
 0.3505998  0.28893167 0.23965877 0.24063917 0.27922502 0.32716477
 0.36553767 0.42293146 0.50968856 0.5458872  0.52192533 0.45243764
 0.36313155 0.3683921  0.40942553 0.4420537  0.46145585 0.4648034
 0.4523771  0.4272876  0.39404616 0.3570107  0.35060245 0.3860975
 0.3996996  0.44551122 0.46611032 0.45998383 0.4309985  0.38563925
 0.37105605 0.4074444  0.48815584 0.5704579  0.6448988  0.7018853
 0.73397845 0.73739105 0.7122451  0.6618154  0.591451   0.5076601
 0.48578677 0.47347385 0.4791471  0.48306277 0.47025493 0.43479836
 0.44380915 0.45868078 0.5341566  0.57549906 0.55790776 0.56244135
 0.57668275 0.561856   0.67564166 0.7512851  0.76957643 0.7266262
 0.734133   0.7231936  0.6776926  0.60511285 0.51599765 0.5579323
 0.56723005 0.5440337  0.5775593  0.5950776  0.5722321  0.57858473
 0.5652703  0.54723704 0.59561515 0.7071321  0.8169259  0.91443264
 0.9883759  1.0275097  1.0235045  0.9737119  1.029139   1.1354861
 1.1910824  1.1826864  1.1092159  0.9832138  0.9643041  0.92324203
 0.9093703  0.88915515 1.0007693  1.0542978  1.0857164  1.0211861
 0.88474303 0.8458009  0.76522666 0.7478076  0.90081936 1.0690157
 1.1569089  1.1493248  1.0622779  1.0327609  0.9805119  0.9583969
 0.8973544  0.9543319  0.9777171  0.94951093 0.97323567 1.0244237
 1.0569099  1.0951824  1.0771195  1.3078191  1.7212077  2.09409
 2.320331   2.3279085  2.125451   1.7908521  1.4180487  1.0744424
 1.0218129  1.0916439  1.1255138  1.125803   1.1139745  1.2187989
 1.300092   1.3025533  1.2312403  1.221301   1.2535597  1.2298189
 1.1458241  1.1012102  1.0889369  1.1558667  1.3051153  1.4143198
 1.6345526  1.8093723  1.9037704  1.8961821  1.7866236  1.5958548
 1.3865516  1.5308585  1.6140417  1.627337   1.5733193  1.4981418
 1.5048542  1.4935548  1.4798748  1.4131776  1.3792214  1.3728334
 1.3683671  1.3593615  1.2995907  1.2965002  1.366058   1.4795257
 1.5462885  1.61591    1.5968509  1.5222199  1.6210756  1.7074443
 1.8351102  2.3187535  2.6568012  2.7676315  2.6480794  2.3636303
 2.0673316  1.9607923  1.8074365  1.713272   1.5893831  1.4734347
 1.507817   1.5213271  1.6091452  1.7162323  1.7608733  1.7497622
 1.9187828  2.0197518  2.0487514  2.01107    1.9193696  1.7904462
 1.8558109  2.1955926  2.4700975  2.6562278  2.675197   2.6645825
 2.6295316  2.4182043  2.2114453  2.2506614  2.2086055  2.0497518
 1.9557768  1.901191   2.067513   2.1077373  2.0159333  1.8138607
 1.5413624  1.600069   1.7631899  1.9541935  1.9340311  1.805134
 2.0671906  2.2247658  2.2641945  2.3594956  2.2504601  1.9749025
 1.8905054  2.0679731  2.1193469  2.0307171  2.0717037  2.0340347
 1.925536   1.7820351  1.9467943  2.315468   2.4834185  2.3751369
 2.0240622  1.9363666  2.1732547  2.3113241  2.3264208  2.22015
 2.0187428  1.7619076  1.796859   1.8757095  2.0501778  2.44711
 2.6179967  2.508112   2.1694388  1.7242104  1.7671669  1.862043
 1.8392721  1.7120028  1.6650634  1.6319505  1.482931   1.5240219
 1.5815579  1.5691646  1.4766116  1.3731087  1.4666644  1.4061015
 1.3652745  1.425564   1.4006845  1.5000012  1.581379   1.6329607
 1.6444355  1.6098644  1.5300899  1.6876912  1.8968476  2.048039
 2.1006014  2.0271482  1.8300935  1.6986666  1.9628603  2.0521066
 1.9337255  1.6407858  1.2583638  1.2110122  1.2476432  1.2360718
 1.2886397  1.2862154  1.2343681  1.1458222  1.209224   1.2475786
 1.2353342  1.1797879  1.0963987  1.0928186  1.1553882  1.1569618
 1.1932304  1.3002363  1.3386917  1.2973225  1.1816871  1.0557054
 0.9350373  0.896656   0.8565816  0.90168726 0.9897751  1.02342
 1.0232298  1.1199353  1.1466643  1.1081418  1.0377598  1.0348651
 1.0223045  1.0607077  1.0089502  0.885213   1.023178   1.1131796
 1.1331098  1.0779471  0.9626393  0.81472665 0.85455835 0.87542623
 0.87286425 0.89130884 0.9545931  1.0355722  1.0201533  0.93568784
 0.9180018  0.8202782  0.7450139  0.72550577 0.68578506 0.6431666
 0.66193295 0.6386373  0.7060119  0.7650972  0.80093855 0.803342
 0.76590335 0.7151591  0.6946282  0.7136788  0.7714012  0.8022328
 0.79840165 0.8543819  0.8586749  0.8028453  0.7383879  0.73423904
 0.65107304 0.61139977 0.5940311  0.6151931  0.59349155 0.54995483
 0.5837645  0.5891752  0.56406695 0.5638191  0.5762535  0.58305734
 0.5830114  0.57470953 0.5568098  0.52852243 0.49031836 0.45275375
 0.47168964 0.46634504 0.4600581  0.45332378 0.41508177 0.3834329
 0.4137769  0.41392407 0.3824464  0.36310086 0.434278   0.48041886
 0.49433306 0.475708   0.43060693 0.36886734 0.34740242 0.34108457
 0.36160505 0.40907663 0.43613982 0.4394311  0.42070773 0.38575593
 0.3827834  0.4338096  0.46581286 0.45669746 0.40830874 0.3505502
 0.32584783 0.3381971  0.33949164 0.36409503 0.3759155  0.3610108
 0.37174097 0.39990777 0.38925973 0.34376588 0.32478797 0.32705626
 0.3228174  0.30941254 0.28542265 0.2687348  0.25517422 0.26127565
 0.27331188 0.3028561  0.31277937 0.29953563 0.2660389  0.27051866
 0.2913383  0.30363902 0.30684754 0.3011791  0.28737035 0.26648855
 0.26413882 0.25501928 0.23947525 0.21937743 0.19659272 0.18965112
 0.21511254 0.23329383 0.24157354 0.2391297  0.22697571 0.20739041
 0.1855308  0.18856761 0.19565174 0.20542233 0.21473111 0.22244582
 0.22726117 0.22789808 0.22336568 0.21322969 0.20314343 0.2031754
 0.19738965 0.1959791  0.20284075 0.20859875 0.21363212 0.21804498
 0.22160804 0.22381367]

这很接近，但不完全是因为边缘不是零斜率：How do I fit a sine curve to my data with pylab and numpy?

有什么可以让我这样做，而不必写一个自定义算法来处理这个吗？谢谢。你知道吗

Answer 1

从您自己的基于正弦拟合的示例开始，我添加了一些约束，使得模型的导数在端点处必须为零。我使用symfit完成了这项工作，我编写了一个包来简化这类工作。如果您喜欢使用scipy来实现这一点，您可以根据需要对示例进行调整，symfit只是它们的极小值的包装器，它使用sympy添加符号操作。你知道吗

# Make variables and parameters
x, y = variables('x, y')
a, b, c, d = parameters('a, b, c, d')
# Initial guesses
b.value = 1e-2
c.value = 100

# Create a model object
model = Model({y: a * sin(b * x + c) + d})

# Take the derivative and constrain the end-points to be equal to zero.
dydx = D(model[y], x).doit()
constraints = [Eq(dydx.subs(x, xdata[0]), 0),
               Eq(dydx.subs(x, xdata[-1]), 0)]

# Do the fit!
fit = Fit(model, x=xdata, y=ydata, constraints=constraints)
fit_result = fit.execute()
print(fit_result)

plt.plot(xdata, ydata)
plt.plot(xdata, model(x=xdata, **fit_result.params).y)
plt.show()

这将打印：（来自当前的symfitPR#221，它可以更好地报告结果。）

Parameter Value        Standard Deviation
a         8.790393e-01 1.879788e-02
b         1.229586e-02 3.824249e-04
c         9.896017e+01 1.011472e-01
d         1.001717e+00 2.928506e-02
Status message         Optimization terminated successfully.
Number of iterations   10
Objective              <symfit.core.objectives.LeastSquares object at 0x0000016F670DF080>
Minimizer              <symfit.core.minimizers.SLSQP object at 0x0000016F78057A58>

Goodness of fit qualifiers:
chi_squared            29.72125657199736
objective_value        14.86062828599868
r_squared              0.8695978050586373

Constraints:
          
Question: a*b*cos(c) == 0?
Answer:   1.5904051811454707e-17

Question: a*b*cos(511*b + c) == 0?
Answer:   -6.354261416082215e-17

Answer 2

这是一个洛伦兹式的峰值方程，适合你的数据，对于“x”值，我使用了一个索引，类似于我在你文章的示例输出图上看到的。我还放大了峰值中心，以便更好地显示您提到的正弦曲线形状。您可以从该峰值方程中减去预测值，以调节或预处理您所讨论的原始数据。你知道吗

a =  1.7056067124682076E+02
b =  7.2900803359572393E+01
c =  2.5047064423525464E+02
d =  1.4184767800540945E+01
Offset = -2.4940994412221318E-01

y = a/ (b + pow((x-c)/d, 2.0)) + Offset