Python中的曲线拟合，边缘处的斜率为零

Question

我希望对以下数据进行曲线拟合，以使其与边缘斜率为零条件下的趋势相拟合。 polyfit 的输出适合该数据，但边缘的斜率不为零。

这是我要输出的内容 - 请原谅我的绘画工作。我需要它像这样适合，这样我就可以正确地消除对中心不真实的数据的正弦/余弦偏差。 这是数据：

[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?

是否有任何可用的东西可以让我做到这一点而无需编写自定义算法来处理这个问题？谢谢。

Answer 1

从您自己基于正弦拟合的示例开始，我添加了约束，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，这是我编写的一个包，目的是让这种事情变得更容易。如果您更喜欢使用 scipy 执行此操作，您可以根据需要调整示例以适应该语法，symfit 只是其最小化器的包装器，它使用 sympy 添加符号操作。

# 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()

这打印：（来自当前的 symfit PR#221，它有更好的结果报告。）

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

这是适合您的数据的洛伦兹类型的峰值方程，对于“x”值，我使用的索引类似于我在您帖子的示例输出图中看到的索引。我还放大了峰值中心，以更好地显示您提到的正弦形状。您也许可以从这个峰值方程中减去预测值，以便在讨论时对原始数据进行调节或预处理。

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