如何使用 python/scipy 對曲線族進行曲線擬合

Question

我有一個解決以下問題：

• 我想用一個方程擬合一系列曲線（參見代碼）
• 擬合參數 (C1-C4) 應該是常數
• 所以我可以通過改變“Sigma and T”來擬合（或描述）該族的一條曲線

我希望我能描述我的問題，希望你們能提供幫助，我將非常感激！

問題已編輯（由於誤解 - 2020_04_04）

我現在會嘗試更具體一些，因為我附上了一張圖片，您可以在其中看到“曲線族”的示例，該示例會隨着不同的“Sigma”而變化。 我想用一對常數來描述那些曲線族——C1、C2、C3 和 C4，而不改變它們。 線索是找到一個常數的最佳值，它可以僅以改變 Sigma 和 T 作為變量來描述這個曲線族。 因此，我必須以最小的誤差擬合一堆曲線的參數。 之后，只需更改“Sigma 和 T”，方程就應該涵蓋整個曲線族。

曲線示例

此致！

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


#Equation --> Eps_Cr = (C1*Sigma**C2*x**(C3+1)*e(-C4/T))/(C3+1)

def func(x, C1, C2, C3,C4):
    Sigma = 20
    T = 1
    return (C1*Sigma**C2*x**(C3+1)*np.exp(-C4*1/T))/(C3+1)

#Example Data 1
xdata = [1, 10, 100, 1000, 10000, 100000]
ydata = [0.000382,0.000407,0.000658,0.001169,0.002205,0.004304]

#Example Data 2
xdata1 = [1, 10, 100, 1000, 10000, 100000]
ydata1 = [0.002164,0.002371,0.004441,0.008571,0.016811,0.033261]

#Example Data 3
xdata2 = [1, 10, 100, 1000, 10000, 100000]
ydata2 = [0.001332,0.001457,0.002707,0.005157,0.010007,0.019597]

plt.plot(xdata, ydata, 'b-', label='data')
plt.plot(xdata1, ydata1, 'g-', label='data')
plt.plot(xdata2, ydata2, 'y-', label='data')

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

plt.plot(xdata, func(xdata, *popt), 'r--',
         label='fit: C1=%5.2e, C2=%5.3f, C3=%5.3f,C4=%5.3f' % tuple(popt))


plt.xlabel('X')
plt.ylabel('Y')
plt.legend()
plt.show()

Answer 1

從您在“答案”中提供的額外信息來看，您似乎想要適應分層 model。 至少統計學家經常這么稱呼他們。 一些參數在所有數據點之間共享（參數C1到C4 ，一些參數在數據集組內共享（ T和Sigma ）。所有這些參數都需要從數據中估計。

這通常通過為所有數據構建更大的 model 來解決，並在 model 中構建一個 select 來使用分組參數。 如果一個數據點屬於數據組1 ，我們選擇Sigma1和T1等等......

由於您已經在使用curve_fit ，因此我制作了一個可以完成這項工作的代碼版本。 由於我不是scipy方面的專家，因此代碼風格有點要求，但我認為您至少會理解該方法。

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

def func(x_and_grp, C1, C2, C3, C4, Sigma0, Sigma1, Sigma2, T0, T1, T2):
    # We estimate one sigma and one T per group of data points
    x = x_and_grp[:,0]
    grp_id = x_and_grp[:,1]
    # here we select the appropriate T and Sigma for each data point based on their group id
    T = np.array([[T0, T1, T2][int(gid)] for gid in grp_id])
    Sigma = np.array([[Sigma0, Sigma1, Sigma2][int(gid)] for gid in grp_id])
    return (C1*Sigma**C2*x**(C3+1)*np.exp(-C4*1/T))/(C3+1)

#Example Data in 3 groups
xdata0 = [1, 10, 100, 1000, 10000, 100000]
ydata0 = [0.000382,0.000407,0.000658,0.001169,0.002205,0.004304]
xdata1 = [1, 10, 100, 1000, 10000, 100000]
ydata1 = [0.002164,0.002371,0.004441,0.008571,0.016811,0.033261]
xdata2 = [1, 10, 100, 1000, 10000, 100000]
ydata2 = [0.001332,0.001457,0.002707,0.005157,0.010007,0.019597]

#  merge all the data and add the group id to the x-data vectors
y_all = np.concatenate([ydata0, ydata1, ydata2])
x_and_grp_all = np.zeros(shape=(3 * 6, 2))
x_and_grp_all[:, 0] = np.concatenate([xdata0, xdata1, xdata2])
x_and_grp_all[0:6, 1] = 0
x_and_grp_all[6:12, 1] = 1
x_and_grp_all[12:18, 1] = 2

# fit a model to all the data together
popt, pcov = curve_fit(func, x_and_grp_all, y_all)

xspace = np.logspace(1,5)
plt.plot(xdata0, ydata0, 'b-', label='data')
plt.plot(xdata1, ydata1, 'g-', label='data')
plt.plot(xdata2, ydata2, 'y-', label='data')
for gid,color in zip([0,1,2],['r','k','purple']):
    T = popt[4+gid]
    Sigma = popt[7+gid]
    x_and_grp = np.column_stack([xspace,np.ones_like(xspace)*gid])
    plt.plot(xspace,
             func(x_and_grp, *popt),
             linestyle='dashed', color=color,
             label='fit: T=%5.2e, Sigma=%5.3f' % (T,Sigma))

plt.xlabel('X')
plt.ylabel('Y')
plt.title('fit: C1=%5.2e, C2=%5.3f, C3=%5.3f,C4=%5.3f' % tuple(popt[0:4]))
plt.legend()
plt.show()

output 看起來像這樣：

最后，我想補充一點，如果您有很多不同的組， curve_fit不太適合這項任務。 考慮一些其他可能相關的庫。 Statmodels 是可能的。 一種替代方法是使用scipy.optimize.minimze ，因為它為您提供了更大的靈活性。 您需要手動進行置信區間估計......

我還想補充一點，如果您知道每組數據的T和Sigma ，則上述方法過於復雜。 在這種情況下，我們將Sigma和T的相關值添加到 x 向量，而不是組 id。

Answer 2

根據您的查詢，我可以理解您需要分別為三個不同的數據集擬合一個方程。 因此，我通過保持 sigma 和 T 相同來更新您的代碼。 請看一下，讓我進一步了解。

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


#Equation --> Eps_Cr = (C1*Sigma**C2*x**(C3+1)*e(-C4/T))/(C3+1)

def func(x, C1, C2, C3,C4):
    Sigma = 20
    T = 1
    return (C1*Sigma**C2*x**(C3+1)*np.exp(-C4*1/T))/(C3+1)

#Example Data 1
xdata = [1, 10, 100, 1000, 10000, 100000]
ydata = [0.000382,0.000407,0.000658,0.001169,0.002205,0.004304]

#Example Data 2
xdata1 = [1, 10, 100, 1000, 10000, 100000]
ydata1 = [0.002164,0.002371,0.004441,0.008571,0.016811,0.033261]

#Example Data 3
xdata2 = [1, 10, 100, 1000, 10000, 100000]
ydata2 = [0.001332,0.001457,0.002707,0.005157,0.010007,0.019597]

plt.plot(xdata, ydata, 'b-', label='data 1')
plt.plot(xdata1, ydata1, 'g-', label='data 2')
plt.plot(xdata2, ydata2, 'y-', label='data 3')

popt, pcov = curve_fit(func, xdata, ydata)
popt1, pcov1 = curve_fit(func, xdata1, ydata1)
popt2, pcov2 = curve_fit(func, xdata2, ydata2)

plt.plot(xdata, func(xdata, *popt), 'r.',
         label='fit for Data 1: C1=%5.2e, C2=%5.3f, C3=%5.3f,C4=%5.3f' % tuple(popt))

plt.plot(xdata1, func(xdata1, *popt1), 'r+',
         label='fit for Data 2: C1=%5.2e, C2=%5.3f, C3=%5.3f,C4=%5.3f' % tuple(popt1))

plt.plot(xdata2, func(xdata2, *popt2), 'r--',
         label='fit for Data 3 : C1=%5.2e, C2=%5.3f, C3=%5.3f,C4=%5.3f' % tuple(popt2))


plt.xlabel('X')
plt.ylabel('Y')
plt.legend(loc='upper left',prop={'size': 8})
plt.show()