幂律曲线拟合scipy，numpy不工作
我想出了一个在我的数据上拟合幂律曲线的问题。 我有两个数据集：bins1和bins2
（使用np.exp(coefs[0])*x**coefs[1]来获得幂律方程），然后使用np.exp(coefs[0])*x**coefs[1]
另一方面，bin2performance奇怪，并显示一个不好的R平方
这两个数据有不同的方程比excel显示我（和更糟糕的R平方）。
这里是代码（和数据）：
import numpy as np import matplotlib.pyplot as plt bins1 = np.array([[6.769318871738219667e-03, 1.306418618130891773e-02, 1.912138120913448383e-02, 2.545189874466026111e-02, 3.214689891729670401e-02, 4.101898933375244805e-02, 5.129862592803200588e-02, 6.636505322669797313e-02, 8.409809827572585494e-02, 1.058164348650862258e-01, 1.375849753230810046e-01, 1.830664031837437311e-01, 2.682454535427478137e-01, 3.912508246490400410e-01, 5.893271848997768680e-01, 8.480213305038615257e-01, 2.408136266017391058e+00, 3.629192766488219313e+00, 4.639246557509275171e+00, 9.901792214343277720e+00], [8.501658465758301112e-04, 1.562697718429977012e-03, 1.902062808421856087e-04, 4.411817741488644959e-03, 3.409236963162485048e-03, 1.686099657013027898e-03, 3.643231240239608402e-03, 2.544120616413291154e-04, 2.549036204611017029e-02, 3.527340723977697573e-02, 5.038482027310990652e-02, 5.617932487522721979e-02, 1.620407270423956103e-01, 1.906538999080910068e-01, 3.180688368126549093e-01, 2.364903188268162038e-01, 3.267322385964683273e-01, 9.384571074801122403e-01, 4.419747716107813029e-01, 9.254710022316929852e+00]]).T bins2 = np.array([[6.522512685133712192e-03, 1.300415548684437199e-02, 1.888928895701269539e-02, 2.509905819337970856e-02, 3.239654633369139919e-02, 4.130706234846069635e-02, 5.123820846515786398e-02, 6.444380072984744190e-02, 8.235238352205621892e-02, 1.070907072127811749e-01, 1.403438221033725120e-01, 1.863115065963684147e-01, 2.670209758710758163e-01, 4.003337413814173074e-01, 6.549054078382223754e-01, 1.116611087124244062e+00, 2.438604844718367914e+00, 3.480674117919704269e+00, 4.410201659398489404e+00, 6.401903059926267403e+00], [1.793454543936148608e-03, 2.441092334386309615e-03, 2.754373929745804715e-03, 1.182752729942167062e-03, 1.357797177773524414e-03, 6.711673916715021199e-03, 1.392761674092503343e-02, 1.127957613093066511e-02, 7.928803089359596004e-03, 2.524609593305639915e-02, 5.698702885370290905e-02, 8.607729156137132465e-02, 2.453761830112021203e-01, 9.734443815196883176e-02, 1.487480479168299119e-01, 9.918002699934079791e-01, 1.121298151253063535e+00, 1.389239135742518227e+00, 4.254082922056571237e-01, 2.643453492951096440e+00]]).T bins = bins1 #change to bins2 to see results for bins2 def fit(x,a,m): # power-law fit (based on previous studies) return a*(x**m) coefs= np.linalg.lstsq(np.vstack([np.ones(len(bins[:,0])), np.log(bins[:,0]), bins[:,0]]).T, np.log(bins[:,1]))[0] # calculating fitting coefficients (a,m) y_predict = fit(bins[:,0],np.exp(coefs[0]),coefs[1]) # prediction based of fitted model model_plot = plt.loglog(bins[:,0],bins[:,1],'o',label="error") fit_line = plt.plot(bins[:,0],y_predict,'r', label="fit") plt.ylabel('Y (bins[:,1])') plt.xlabel('X (bins[:,0])') plt.title('model') plt.legend(loc='best') plt.show(model_plot,fit_line) def R_sqr (y,y_predict): # calculating R squared value to measure fitting accuracy rsdl = y - y_predict ss_res = np.sum(rsdl**2) ss_tot = np.sum((y-np.mean(y))**2) R2 = 1-(ss_res/ss_tot) R2 = np.around(R2,decimals=4) return R2 R2= R_sqr(bins[:,1],y_predict) print ('(R^2 = %s)' % (R2))
bins1 [[x]，[y]]： python: y = 0.337*(x)^1.223 (R^2 = 0.7773), excel: y = 0.289*(x)^1.174 (R^2 = 0.8548)
bins2 [[x]，[y]]： python: y = 0.509*(x)^1.332 (R^2 = -1.753), excel: y = 0.311*(x)^1.174 (R^2 = 0.9116)
这些是30个样本中的两个数据集，我在我的数据中随机地看到了这个拟合问题，还有一些在“-150”左右的R-squared！
试过scipy“curve_fit”，但是我没有得到更好的结果，实际上更糟！
任何人都知道如何在python中获得excel-fit？
您正在尝试使用尚未转换为对数空间的Y计算R平方。 以下变化给出了合理的R平方值：
R2 = R_sqr(np.log(bins[:,1]), np.log(y_predict))