我如何擬合這條曲線？

Question

我正在嘗試使用 scipy 的曲線擬合 function 找到 function 到 model 我的曲線數據集，這讓我得到一條線並吐出錯誤消息“OptimizeWarning：無法估計參數的協方差”

這是我的數據：

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit


def func(x, a, b, c):
    return a * np.exp(-b * x) + c

x = [6529,6530,6531,6532,6533,6534,6535,6536,6537,6538,6539,6540,6541,6542,6543,6544,6545,6546,6547,6548,6549,6550,6551,6552,6553,6554,6555,6556,6557,6558,6559,6560,6561,6562,6563,6564,6565,6566,6567,6568,6569,6570,6571,6572,6573,6574,6575,6576,6577,6578,6579,6580,6581,6582,6583,6584,6585,6586,6587,6588,6589,6590,6591,6592,6593,6594,6595,6596,6597,6598,6599,6600,6601,6602,6603,6604,6605,6606,6607,6608,6609,6610,6611,6612,6613,6614,6615,6616,6617,6618,6619,6620,6621,6622,6623,6624,6625,6626,6627,6628,6629]
xdata = np.array(x)
xdata

y =[0.547409184506862,0.548439334507089,0.548707663683029,0.549517457040911,0.549928046272648,0.552133076225856,0.553680404362389,0.554546610865359,0.556183559635298,0.55702393264449,0.558127603440548,0.56005701419833,0.567268696243044,0.567959158455945,0.56888383955428,0.570002946460703,0.571738099634388,0.571781066314887,0.572492904162659,0.573363360104158,0.575401312020501,0.579556244995718,0.581790310757611,0.583406644934125,0.583431253838386,0.584646128487338,0.591634693424932,0.593621181449664,0.596918227952608,0.597537299010122,0.597822020010253,0.598891783912097,0.599584877929425,0.600869677067233,0.605499427101361,0.60658392374002,0.607408603951367,0.608003672935112,0.612816417541406,0.614176393253985,0.615691612727725,0.617134841882831,0.624502603183639,0.627504005062751,0.62811483139368,0.630923224681103,0.631913519350306,0.632861774084856,0.633396927216081,0.634723100574364,0.636823036518848,0.637335872514631,0.641989703420432,0.645848889736627,0.65711344945379,0.657128729116295,0.663572015593525,0.663936607768306,0.664261916284895,0.665497047151865,0.675768594810369,0.676207425367557,0.67770213122942,0.684362066388147,0.686091459831405,0.68923449405901,0.696555953074396,0.699523852803358,0.700114629853266,0.700439968032363,0.70422912133774,0.709054418203755,0.718323675919034,0.720874110082631,0.731420313805414,0.740339970645214,0.743941832408661,0.744098264483215,0.755637715557576,0.770607412727517,0.772170144147719,0.77386595160397,0.782346429315201,0.793536926034814,0.799814727138212,0.811806170429752,0.820389958104833,0.823442687233217,0.836431157138314,0.837230603981482,0.84457830759536,0.849472002016762,0.853064650303987,0.864143132487712,0.875848713252245,0.879457500784825,0.906226501241044,0.938124902423877,0.957251151492073,0.984140865987764,1]
ydata = np.array(y)
ydata

plt.plot(xdata, ydata, 'b.', label='data')

我試過 curve_fit function

popt, pcov = curve_fit(func, xdata, ydata)
popt

plt.plot(xdata, func(xdata, *popt), 'r-',
         label='fit: a=%5.3f, b=%5.3f, c=%5.3f' % tuple(popt))

但它只是繪制了一條水平線。

我最接近的是在 R 中使用線性回歸 model（類似於 model <- lm(log(y)~x, data=df），看起來像這樣：

但我希望它看起來更像這樣：

這樣我就可以在之后對曲線求導。 抱歉，如果缺少我的代碼，我是 python 的新手。

Answer 1

您可以對xdata應用簡單的轉換：

max_xdata = xdata.max()
xdata_t = max_xdata - xdata

popt, pcov = curve_fit(func, xdata_t, ydata)
print(popt)

fix, ax = plt.subplots(1, 1)
ax.plot(xdata_t, ydata, 'b.', label='data')
ax.plot(xdata_t, func(xdata_t, *popt), 'r-',
         label='fit: a=%5.3f, b=%5.3f, c=%5.3f' % tuple(popt))

這讓你：

或者，要擬合原始數據集：

fix, ax = plt.subplots(1, 1)
ax.plot(xdata, ydata, 'b.', label='data')
ax.plot(xdata, func(max_xdata - xdata, *popt), 'r-',
         label='fit: a=%5.3f, b=%5.3f, c=%5.3f' % tuple(popt))

Answer 2

如果在進行擬合之前轉換數據，則可以使用線性回歸。

你要的function長這樣：

y = a*exp(b*x)

取兩邊的自然對數：

ln(y) = ln(a) + b*x

您可以看到這看起來像一個線性回歸，其中 y 截距為ln(a)且斜率為b 。

在進行線性回歸之前，您需要對 y 數組中的每一項取自然對數。

計算完成后，您將通過exp(ln(a))獲得a的值。