如何在拟合期间向曲线而不是数据点添加权重？我使用什么导入/函数？

如何解决如何在拟合期间向曲线而不是数据点添加权重？我使用什么导入/函数？

我试过 scipy.optimize import curve_fit 但它似乎只改变了数据点。我想在从我的数据点残差（带权重的最小平方）拟合 Ycurve 期间添加 1/Y^2 权重。我不确定如何定位 yfit 而不是 ydata 或者我是否应该使用其他东西？任何帮助将不胜感激。

xdata = np.array([0.10,0.10,1.12,2.89,6.19,23.30,108.98,255.33,1188.62,1188.62],dtype=float)

ydata = np.array([0.264352,0.412386,0.231238,0.483558,0.613206,0.728528,-1.15391,-1.46504,-0.942926,-2.12808,-2.90962,-1.51093,-3.09798,-5.08591,-4.75703,-4.91317,-5.1966,-4.04019,-13.8455,-16.9911,-11.0881,-10.6453,-15.1288,-52.4669,-68.2344,-74.7673,-70.2025,-65.8181,-55.7344,-271.286,-329.521,-436.097,-654.034,-396.45,-826.195,-1084.43,-984.344,-1124.8,-1076.27,-1072.03,-3968.22,-3114.46,-3771.61,-2805.4,-4078.05],dtype=float)

def fourPL(x,A,B,C,D):
    return ((A-D)/(1.0+((x/C)**B))) + D

params,params_covariance = spo.curve_fit(fourPL,xdata,ydata)
params_list = params
roundy = [round(num,4) for num in params_list]
print(roundy)

popt2,pcov2 = spo.curve_fit(fourPL,ydata,sigma=1/ydata**2,absolute_sigma=True)
yfit2 = fourPL(xdata,*popt2)
params_list2 = popt2
roundy2 = [round(num,4) for num in params_list2]
print(roundy2)

x_min,x_max = np.amin(xdata),np.amax(xdata)
xs = np.linspace(x_min,x_max,1000)
plt.scatter(xdata,ydata)
plt.plot(xs,fourPL(xs,*params),'m--',label='No Weight')
plt.plot(xs,*popt2),'b--',label='Weights')
plt.legend(loc='upper center',bbox_to_anchor=(0.5,1.05,0.1,0.1),ncol=3,fancybox=True,shadow=True)
plt.xlabel('µg/mL)')
plt.ylabel('kHz/s')
#plt.xscale('log')
plt.show()
```

解决方法

这将是我使用手动 least_squares 拟合的版本。我将它与使用简单的 curve_fit 获得的解决方案进行了比较。实际上差异不是很大，curve_fit 的结果对我来说看起来不错。

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

np.set_printoptions( linewidth=250,precision=2 )  ## avoids OPs "roundy"

xdata = np.array(
    [
        0.10,0.10,1.12,2.89,6.19,23.30,108.98,255.33,1188.62,1188.62
    ],dtype=float
)

ydata = np.array(
    [
        0.264352,0.412386,0.231238,0.483558,0.613206,0.728528,-1.15391,-1.46504,-0.942926,-2.12808,-2.90962,-1.51093,-3.09798,-5.08591,-4.75703,-4.91317,-5.1966,-4.04019,-13.8455,-16.9911,-11.0881,-10.6453,-15.1288,-52.4669,-68.2344,-74.7673,-70.2025,-65.8181,-55.7344,-271.286,-329.521,-436.097,-654.034,-396.45,-826.195,-1084.43,-984.344,-1124.8,-1076.27,-1072.03,-3968.22,-3114.46,-3771.61,-2805.4,-4078.05
    ],dtype=float
)

def fourPL( x,a,b,c,d ):
    out = ( a - d ) / ( 1 + ( x / c )**b ) + d
    return out

def residuals( params,xlist,ylist ):
    # ~a,d = params
    yth = np.fromiter( (fourPL( x,*params ) for x in xlist ),np.float )
    diff = np.subtract( yth,ylist )
    weights = 1 / np.abs( yth )**1
    ## not sure if this makes sense,but we weigth with function value
    ## here I put it inverse linear as it gets squared in the chi-square
    ## but other weighting may be required
    return diff * weights

### for initial guess
xl = np.linspace( .1,1200,150 )
yl0 = np.fromiter( (fourPL( x,1,1000,-6000) for x in xl ),np.float )

### standard curve_fit
cfsol,_ = curve_fit( fourPL,xdata,ydata,p0=( 1,-6000 ) )
print( cfsol )
yl1 = np.fromiter( (fourPL( x,*cfsol ) for x in xl ),np.float )

### least square with manual residual function including "unusual" weighting
lssol = least_squares( residuals,x0=( 1,-6000 ),args=( xdata,ydata ) )
print( lssol.x )
yl2 = np.fromiter( (fourPL( x,*( lssol.x ) ) for x in xl ),np.float )

### plotting
fig = plt.figure()
ax = fig.add_subplot( 1,1 )
ax.scatter( xdata,ydata )
ax.plot( xl,yl1 )
ax.plot( xl,yl2 )

plt.show()

查看协方差矩阵，不仅显示出相当大的误差，而且参数的相关性也相当高。这当然是由模型的性质决定的，但应谨慎处理，尤其是在数据解释方面。

当我们只考虑小 x 的数据时，问题变得更加明显。无论如何，大 x 的数据分散很多。对于小x或大c，泰勒展开式是(a - d ) (1 - b x / c ) + d，即a - (a - d ) b / c x，基本上是a + e x。所以 b、c 和 d 基本上都是一样的。