python多项式曲线拟合 - 系数不对

3 回答

• 1

  在我看来，你使用 poyval 的方式对我来说并不合适 . 尝试使用 numpy.linspace 生成X轴，然后在其上应用 polyval . 类似下面的代码 .

  import numpy as np
  import matplotlib.pyplot as plt
  
  
  data = np.loadtxt('fig3_1_tiltingRL.dat')
  
  x = data[:,0]
  y= data[:,1]
  #plt.xscale('log')#plt.set_xscale('log')
  coefs = np.polyfit(x, y, 10)
  ffit = np.polyval(coefs, x)
  
  new_x = np.linspace(0,26)
  new_ffit = np.polyval(coefs, new_x)
  
  plt.plot(x, y, 'o', label="Raw")
  plt.plot(x, ffit,'x',label="Fit to Raw")
  plt.plot(new_x, new_ffit,label="Fit to LinSpace")
  
  # This is ugly. I'd use list comprehension here!
  arr = np.linspace(0,26,20)
  new_y = []
  for xi in arr:
      total = 0
      for i,v in enumerate(coefs[::-1]):
          total += v*xi**i
      new_y.append(total)
  
  
  plt.plot(arr, new_y, '*', label="Polynomial")
  
  plt.legend(loc=2)
  plt.show()
  

  

  正如你所看到的，有一个驼峰没有出现在你的情节......

  回复于 2024-04-30T04:14:00+08:00
• 1

  您的算法似乎工作正常 . 你应该而不是：

  coefs = poly.polyfit(x, y, 10)
  ffit = poly.polyval(x, coefs)
  

  做这个：

  coefs = poly.polyfit(x, y, 10) # fit data to the polynomial
  new_x = np.linspace(0, 30, 50) # new x values to evaluate
  ffit = poly.polyval(new_x, coefs) # fitted polynomial evaluated with new data
  

  因此，函数 poly.polyval 将评估 new_x 的所有点，而不是您已经知道的 x 坐标 .

  回复于 2024-04-30T04:14:00+08:00
• 0

  非常感谢您回答我的问题 .

  由silgon和RicLeal提供的解决方案都有效 .

  最后，由于我有几条曲线，我已经应用了RicLeal给出的解决方案 .

  我的数据记录在x轴上 . 我刚修改了RicLeal给出的代码，我对结果感到满意 .

  enter image description here

  x = data[:,0]
  y= data[:,1]
  plt.xscale('log')#plt.set_xscale('log')
  logx=np.log10(x)
  coefs = np.polyfit(logx, y, 10)
  ffit = np.polyval(coefs, logx)
  print (coefs)
  
  
  logxmin=math.log10(0.5883596178)
  logxmax=math.log10(26.)
  
  new_x = np.logspace(logxmin, logxmax,50)
  lognew_x=np.log10(new_x)
  new_ffit = np.polyval(coefs, lognew_x)
  plt.semilogx(x, y, 'o', label="Raw")
  plt.semilogx(x, ffit,'x',label="Fit to Raw")
  plt.semilogx(new_x, new_ffit,label="Fit to LogSpace")
  print(lognew_x, new_ffit)
  # This is ugly. I'd use list comprehension here!
  arr = np.logspace(logxmin, logxmax,50)
  arrlog= np.log10(arr)
  new_y = []
  for xi in arrlog:
      total = 0
      for i,v in enumerate(coefs[::-1]):
         #print (v)
         total += v*xi**i
      new_y.append(total) 
  
  
      plt.semilogx(arr, new_y, '*', label="Polynomial")
  
   coeffs=  [6.85869364,     -92.86678553,  343.39375022,           -555.52532934,  434.18179364,
       -152.82724751,     9.71300951,   21.68653301,  -35.62838377,   28.3985976,
        27.04762122]
   new_testy = []
   for xi in arrlog:
   total = 0
      for i,v in enumerate(coeffs[::-1]):
         #print (v)
           total += v*xi**i
      new_testy.append(total)          
   plt.semilogx(arr, new_testy, 'o', label="Polynomial")         
   plt.legend(loc=2)
   plt.show()
  

  回复于 2024-04-30T04:14:00+08:00