Question

我有x和y的以下数据：

x   y
1.71    0.0
1.76    5.0
1.81    10.0
1.86    15.0
1.93    20.0
2.01    25.0
2.09    30.0
2.20    35.0
2.32    40.0
2.47    45.0
2.65    50.0
2.87    55.0
3.16    60.0
3.53    65.0
4.02    70.0
4.69    75.0
5.64    80.0
7.07    85.0
9.35    90.0
13.34   95.0
21.43   100.0

对于上述数据，我试图以下列形式拟合数据：

$y = a{x}^{n} + b$

然而，x和y存在某些不确定性，其中x具有50％x的不确定性，y具有固定的不确定性。我试图用uncertainties package来确定拟合参数的不确定性。但是，我遇到了曲线拟合与scipy optimize曲线拟合函数的问题。我收到以下错误：

minpack.error：函数调用的结果不是正确的数组浮动。

如何修复以下错误并确定拟合参数（a，b和n）的不确定性？

MWE

from __future__ import division
import numpy as np
import re
from scipy import optimize, interpolate, spatial
from scipy.interpolate import UnivariateSpline
from uncertainties import unumpy


def linear_fit(x, a, b):
    return a * x + b


uncertainty = 0.5
y_error = 1.2
x = np.array([1.71, 1.76, 1.81, 1.86, 1.93, 2.01, 2.09, 2.20, 2.32, 2.47, 2.65, 2.87, 3.16, 3.53, 4.02, 4.69, 5.64, 7.07, 9.35, 13.34, 21.43])
x_uncertainty = x * uncertainty
x = unumpy.uarray(x, x_uncertainty)
y = np.array([0.0, 5.0, 10.0, 15.0, 20.0, 25.0, 30.0, 35.0, 40.0, 45.0, 50.0, 55.0, 60.0, 65.0, 70.0, 75.0, 80.0, 85.0, 90.0, 95.0, 100.0])
y = unumpy.uarray(y, y_error)


n = np.arange(0, 5, 0.005)
coefficient_determination_on = np.empty(shape = (len(n),))
for j in range(len(n)):
    n_correlation = n[j]
    x_fit = 1 / ((x) ** n_correlation)
    y_fit = y
    fit_a_raw, fit_b_raw = optimize.curve_fit(linear_fit, x_fit, y_fit)[0]
    x_prediction = (fit_a_raw / ((x) ** n_correlation)) + fit_b_raw
    y_residual_squares = np.sum((x_prediction - y) ** 2)
    y_total_squares = np.sum((y - np.mean(y)) ** 2)
    coefficient_determination_on[j] = 1 - (y_residual_squares / y_total_squares)

Answer 1

让我先说一下这个问题是不可能解决的问题＆＃34;很好地＆＃34;鉴于您要解决a，b 和 n。这是因为对于固定的n，您的问题会接受封闭的表单解决方案，而如果您让n空闲，则不会，并且实际上问题可能有多个解决方案。因此，经典的错误分析（例如uncertanities使用的分析）会崩溃，你必须求助于其他方法。

案例`n`已修复

如果修复了n，则问题在于您调用的库不支持uarray，因此您必须制定解决方法。值得庆幸的是，线性拟合（在l2-距离下）只是Linear least squares，它允许一个封闭的形式解决方案，只需要用1填充值，然后求解normal equations。

其中：

你可以这样做：

import numpy as np
from uncertainties import unumpy

uncertainty = 0.5
y_error = 1.2
n = 1.0

# Define x and y
x = np.array([1.71, 1.76, 1.81, 1.86, 1.93, 2.01, 2.09, 2.20, 2.32, 2.47, 2.65,
              2.87, 3.16, 3.53, 4.02, 4.69, 5.64, 7.07, 9.35, 13.34, 21.43])
# Take power of x values according to n
x_pow = x ** n
x_uncertainty = x_pow * uncertainty
x_fit = unumpy.uarray(np.c_[x_pow, np.ones_like(x)],
                      np.c_[x_uncertainty, np.zeros_like(x_uncertainty)])

y = np.array([0.0, 5.0, 10.0, 15.0, 20.0, 25.0, 30.0, 35.0, 40.0, 45.0, 50.0,
              55.0, 60.0, 65.0, 70.0, 75.0, 80.0, 85.0, 90.0, 95.0, 100.0])
y_fit = unumpy.uarray(y, y_error)

# Use normal equations to find coefficients
inv_mat = unumpy.ulinalg.pinv(x_fit.T.dot(x_fit))
fit_a, fit_b = inv_mat.dot(x_fit.T.dot(y_fit))

print('fit_a={}, fit_b={}'.format(fit_a, fit_b))

结果：

fit_a=4.8+/-2.6, fit_b=28+/-10

案例`n`未知

n未知，因为问题是非凸的，所以你真的遇到了麻烦。在这里，线性误差分析（由uncertainties执行）将不起作用。

一种解决方案是使用Bayesian inference之类的包来执行pymc。如果你对此感兴趣，我可以尝试写一篇文章，但它不会像上面一样干净。

Answer 2

稍等一下linear function的情况，我认为可能会做类似的事情。然而，解决拉格朗日似乎是非常乏味的，当然可能。一个看似合理的不同措施，应该给出非常相似的结果。取误差椭圆，我将其重新缩放，使得图形变为切线。我将距离触摸点（X_k,Y_k）作为卡方的度量，这是从(x_k-X_k/sx_k)**2+(y_k-Y_k/sy_k)**2计算得出的。在纯y的情况下这是合理的 - 错误这是标准的最小二乘拟合。对于纯x - 错误，它只是切换。对于相等的x,y - 误差，它将给出垂直规则，即最短距离。使用相应的卡方函数scipy.optimize.leastsq已经提供了近似于Hesse的协方差矩阵。但是，一个人必须scale it。另请注意，参数密切相关。

我的程序如下：

import matplotlib
matplotlib.use('Qt5Agg')

import matplotlib.pyplot as plt
import numpy as np
import myModules.multipoleMoments as mm 
from random import random
from scipy.optimize import minimize,leastsq


###for gaussion distributed errors
def boxmuller(x0,sigma):
    u1=random()
    u2=random()
    ll=np.sqrt(-2*np.log(u1))
    z0=ll*np.cos(2*np.pi*u2)
    z1=ll*np.cos(2*np.pi*u2)
    return sigma*z0+x0, sigma*z1+x0


###function to fit
def f(t,c,d,n):
    return c+d*np.abs(t)**n


###to create some test data
def test_data(c,d,n, xList,const_sx,rel_sx,const_sy,rel_sy):
    yList=[f(t,c,d,n) for t in xList]
    xErrList=[ boxmuller(x,const_sx+x*rel_sx)[0] for x in xList]
    yErrList=[ boxmuller(y,const_sy+y*rel_sy)[0] for y in yList]
    return xErrList,yErrList


###how to rescale the ellipse to make fitfunction a tangent
def elliptic_rescale(x,c,d,n,x0,y0,sa,sb):
    y=f(x,c,d,n)
    r=np.sqrt((x-x0)**2+(y-y0)**2)
    kappa=float(sa)/float(sb)
    tau=np.arctan2(y-y0,x-x0)
    new_a=r*np.sqrt(np.cos(tau)**2+(kappa*np.sin(tau))**2)
    return new_a


###for plotting ellipses
def ell_data(a,b,x0=0,y0=0):
    tList=np.linspace(0,2*np.pi,150)
    k=float(a)/float(b)
    rList=[a/np.sqrt((np.cos(t))**2+(k*np.sin(t))**2) for t in tList]
    xyList=np.array([[x0+r*np.cos(t),y0+r*np.sin(t)] for t,r in zip(tList,rList)])
    return xyList


###residual function to calculate chi-square
def residuals(parameters,dataPoint):#data point is (x,y,sx,sy)
    c,d,n= parameters
    theData=np.array(dataPoint)
    best_t_List=[]
    for i in range(len(dataPoint)):
        x,y,sx,sy=dataPoint[i][0],dataPoint[i][1],dataPoint[i][2],dataPoint[i][3]
        ###getthe point on the graph where it is tangent to an error-ellipse
        ed_fit=minimize(elliptic_rescale,0,args=(c,d,n,x,y,sx,sy) )
        best_t=ed_fit['x'][0]
        best_t_List+=[best_t]
    best_y_List=[f(t,c,d,n) for t in best_t_List]
    ##weighted distance not squared yet, as this is done by scipy.optimize.leastsq
    wighted_dx_List=[(x_b-x_f)/sx for x_b,x_f,sx in zip( best_t_List,theData[:,0], theData[:,2] ) ]
    wighted_dy_List=[(x_b-x_f)/sx for x_b,x_f,sx in zip( best_y_List,theData[:,1], theData[:,3] ) ]
    return wighted_dx_List+wighted_dy_List


###some start values
cc,dd,nn=2.177,.735,1.75
ssaa,ssbb=1,3
xx0,yy0=2,3
csx,rsx=.1,.05
csy,rsy=.4,.00

####some data
xThData=np.linspace(0,3,15)
yThData=[ f(t, cc,dd,nn) for t in xThData]

###some noisy data
xNoiseData,yNoiseData=test_data(cc,dd,nn, xThData, csx,rsx, csy,rsy)
xGuessdError=[csx+rsx*x for x in xNoiseData]
yGuessdError=[csy+rsy*y for y in yNoiseData]

###plot that
fig1 = plt.figure(1)
ax=fig1.add_subplot(1,1,1)
ax.plot(xThData,yThData)
ax.errorbar(xNoiseData,yNoiseData, xerr=xGuessdError, yerr=yGuessdError, fmt='none',ecolor='r')

#### and plot...
for i in range(len(xNoiseData)):
    ###...an elliple on the error values
    el0=ell_data(xGuessdError[i],yGuessdError[i],x0=xNoiseData[i],y0=yNoiseData[i])
    ax.plot(*zip(*el0),linewidth=1, color="#808080",linestyle='-')
    ####...as well as a scaled version that touches the original graph. This gives the error shortest distance to that graph
    ed_fit=minimize(elliptic_rescale,0,args=(cc,dd,nn,xNoiseData[i],yNoiseData[i],xGuessdError[i],yGuessdError[i]) )
    best_t=ed_fit['x'][0]
    best_a=elliptic_rescale(best_t,cc,dd,nn,xNoiseData[i],yNoiseData[i],xGuessdError[i],yGuessdError[i])
    el0=ell_data(best_a,best_a/xGuessdError[i]*yGuessdError[i],x0=xNoiseData[i],y0=yNoiseData[i])
    ax.plot(*zip(*el0),linewidth=1, color="#a000a0",linestyle='-')

###Now fitting
zipData=zip(xNoiseData,yNoiseData, xGuessdError, yGuessdError)    
estimate = [2.0,1,2.5]
bestFitValues, cov,info,mesg, ier = leastsq(residuals, estimate,args=(zipData), full_output=1)
print bestFitValues
####covariance matrix
####note e.g.: https://stackoverflow.com/questions/14854339/in-scipy-how-and-why-does-curve-fit-calculate-the-covariance-of-the-parameter-es
s_sq = (np.array(residuals(bestFitValues, zipData))**2).sum()/(len(zipData)-len(estimate))
pcov = cov * s_sq
print pcov



#### and plot the result...
ax.plot(xThData,[f(x,*bestFitValues) for x in xThData])
for i in range(len(xNoiseData)):
    ####...as well as a scaled ellipses that touches the fitted graph. 
    ed_fit=minimize(elliptic_rescale,0,args=(bestFitValues[0],bestFitValues[1],bestFitValues[2],xNoiseData[i],yNoiseData[i],xGuessdError[i],yGuessdError[i]) )
    best_t=ed_fit['x'][0]
    best_a=elliptic_rescale(best_t,bestFitValues[0],bestFitValues[1],bestFitValues[2],xNoiseData[i],yNoiseData[i],xGuessdError[i],yGuessdError[i])
    el0=ell_data(best_a,best_a/xGuessdError[i]*yGuessdError[i],x0=xNoiseData[i],y0=yNoiseData[i])
    ax.plot(*zip(*el0),linewidth=1, color="#f0a000",linestyle='-')

#~ ax.grid(None)
plt.show()

蓝色图是原始功能。具有红色误差条的数据点由此计算。灰色椭圆显示恒定概率密度的＆＃34;线＃34;。紫色椭圆的原始图形为切线，橙色椭圆的拟合为切线

这里最合适的值是（不是你的数据！）：

[ 2.16146783  0.80204967  1.69951865]

协方差矩阵具有以下形式：

[[ 0.0644794  -0.05418743  0.05454876]
 [-0.05418743  0.07228771 -0.08172885]
 [ 0.05454876 -0.08172885  0.10173394]]

修改思考椭圆距离＆＃34;我相信这正是链接文章中的拉格朗日方法所做的。只有直线，你可以写下一个确切的解决方案，而在这种情况下你不能。

<强>更新我的OP数据存在一些问题。但它适用于重新缩放。由于斜率和指数高度相关，我首先要弄清楚协方差矩阵如何改变重新缩放的数据。详细信息可在J. Phys. Chem. 105 (2001) 3917
中找到
使用上面的函数定义，数据处理如下：

###some start values cc,dd,nn=.2,1,7.5 csx,rsx=2.0,.0 csy,rsy=0,.5 ###some noisy data yNoiseData=np.array([1.71,1.76, 1.81, 1.86, 1.93, 2.01, 2.09, 2.20, 2.32, 2.47, 2.65, 2.87, 3.16, 3.53, 4.02, 4.69, 5.64, 7.07, 9.35,13.34,21.43]) xNoiseData=np.array([0.0,5.0,10.0,15.0,20.0,25.0,30.0,35.0,40.0,45.0,50.0,55.0,60.0,65.0,70.0,75.0,80.0,85.0,90.0,95.0,100.0]) xGuessdError=[csx+rsx*x for x in xNoiseData] yGuessdError=[csy+rsy*y for y in yNoiseData] ###plot that fig1 = plt.figure(1) ax=fig1.add_subplot(1,2,2) bx=fig1.add_subplot(1,2,1) ax.errorbar(xNoiseData,yNoiseData, xerr=xGuessdError, yerr=yGuessdError, fmt='none',ecolor='r') ####rescaling print "\n++++++++++++++++++++++++ scaled ++++++++++++++++++++++++\n" alpha=.05 beta=.01 xNoiseDataS = [ beta*x for x in xNoiseData ] yNoiseDataS = [ alpha*x for x in yNoiseData ] xGuessdErrorS = [ beta*x for x in xGuessdError ] yGuessdErrorS = [ alpha*x for x in yGuessdError ] xtmp=np.linspace(0,1.1,25) bx.errorbar(xNoiseDataS,yNoiseDataS, xerr=xGuessdErrorS, yerr=yGuessdErrorS, fmt='none',ecolor='r') ###Now fitting zipData=zip(xNoiseDataS,yNoiseDataS, xGuessdErrorS, yGuessdErrorS) estimate = [.1,1,7.5] bestFitValues, cov,info,mesg, ier = leastsq(residuals, estimate,args=(zipData), full_output=1) print bestFitValues plt.plot(xtmp,[ f(x,*bestFitValues)for x in xtmp]) ####covariance matrix s_sq = (np.array(residuals(bestFitValues, zipData))**2).sum()/(len(zipData)-len(estimate)) pcov = cov * s_sq print pcov #### scale back print "\n++++++++++++++++++++++++ scaled back ++++++++++++++++++++++++\n" realBestFitValues= [bestFitValues[0]/alpha, bestFitValues[1]/alpha*(beta)**bestFitValues[2],bestFitValues[2] ] print realBestFitValues uMX = np.array( [[1/alpha,0,0],[0,beta**bestFitValues[2]/alpha,bestFitValues[1]/alpha*beta**bestFitValues[2]*np.log(beta)],[0,0,1]] ) uMX_T = uMX.transpose() realCov = np.dot(uMX, np.dot(pcov,uMX_T)) print realCov for i,para in enumerate(["b","a","n"]): print para+" = "+"{:.2e}".format(realBestFitValues[i])+" +/- "+"{:.2e}".format(np.sqrt(realCov[i,i])) ax.plot(xNoiseData,[f(x,*realBestFitValues) for x in xNoiseData]) plt.show()

所以数据合理。不过，我认为也有一个线性术语。

输出提供：

++++++++++++++++++++++++ scaled ++++++++++++++++++++++++ [ 0.09788886 0.69614911 5.2221032 ] [[ 1.25914194e-05 2.86541696e-05 6.03957467e-04] [ 2.86541696e-05 3.88675025e-03 2.00199108e-02] [ 6.03957467e-04 2.00199108e-02 1.75756532e-01]] ++++++++++++++++++++++++ scaled back ++++++++++++++++++++++++ [1.9577772055183396, 5.0064036934715239e-10, 5.2221031993990517] [[ 5.03656777e-03 -2.74367539e-11 1.20791493e-02] [ -2.74367539e-11 8.69854174e-19 -3.90815222e-10] [ 1.20791493e-02 -3.90815222e-10 1.75756532e-01]] b = 1.96e+00 +/- 7.10e-02 a = 5.01e-10 +/- 9.33e-10 n = 5.22e+00 +/- 4.19e-01

可以看出协方差矩阵中斜率和指数的强相关性。另请注意，斜率误差很大。

<强>顺便说一句使用模型b+a*x**n + e*x我得到了

++++++++++++++++++++++++ scaled ++++++++++++++++++++++++ [ 0.08050174 0.78438855 8.11845402 0.09581568] [[ 5.96210962e-06 3.07651631e-08 -3.57876577e-04 -1.75228231e-05] [ 3.07651631e-08 1.39368435e-03 9.85025139e-03 1.83780053e-05] [ -3.57876577e-04 9.85025139e-03 1.85226736e-01 2.26973118e-03] [ -1.75228231e-05 1.83780053e-05 2.26973118e-03 7.92853339e-05]] ++++++++++++++++++++++++ scaled back ++++++++++++++++++++++++ [1.6100348667765145, 9.0918698097511416e-16, 8.1184540175879985, 0.019163135651422442] [[ 2.38484385e-03 2.99690170e-17 -7.15753154e-03 -7.00912926e-05] [ 2.99690170e-17 3.15340690e-30 -7.64119623e-16 -1.89639468e-18] [ -7.15753154e-03 -7.64119623e-16 1.85226736e-01 4.53946235e-04] [ -7.00912926e-05 -1.89639468e-18 4.53946235e-04 3.17141336e-06]] b = 1.61e+00 +/- 4.88e-02 a = 9.09e-16 +/- 1.78e-15 n = 8.12e+00 +/- 4.30e-01 e = 1.92e-02 +/- 1.78e-03

当然，如果添加参数，总是会变得更好，但我认为这看起来有点合理（甚至可能是b + a*x**n+e*x**m，但这导致太过分了。）

如何用Python确定拟合参数的不确定性？

2 个答案:

案例`n`已修复

案例`n`未知

如何用Python确定拟合参数的不确定性？

2 个答案:

案例n已修复

案例n未知

案例`n`已修复

案例`n`未知