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我有一组数据(位移与时间),我使用 optimize.leastsq 方法拟合了几个方程。我现在正在寻找拟合参数的错误值。查看文档,输出的矩阵是雅可比矩阵,我必须将它乘以残差矩阵才能得到我的值。不幸的是,我不是统计学家,所以我有点沉迷于术语。

据我了解,我所需要的只是与我的拟合参数一起使用的协方差矩阵,因此我可以对对角线元素进行平方根以获得拟合参数的标准误差。我有一个模糊的记忆,协方差矩阵是 optimize.leastsq 方法的输出。这个对吗?如果不是,您将如何让残差矩阵将输出的雅可比矩阵乘以得到我的协方差矩阵?

任何帮助将不胜感激。我对 python 很陌生,因此如果这个问题被证明是一个基本问题,我深表歉意。

拟合代码如下:

fitfunc = lambda p, t: p[0]+p[1]*np.log(t-p[2])+ p[3]*t # Target function'

errfunc = lambda p, t, y: (fitfunc(p, t) - y)# Distance to the target function

p0 = [ 1,1,1,1] # Initial guess for the parameters


  out = optimize.leastsq(errfunc, p0[:], args=(t, disp,), full_output=1)

args t 和 disp 是时间和位移值的数组(基本上只有 2 列数据)。我已经在代码顶部导入了所有需要的东西。输出提供的拟合值和矩阵如下:

[  7.53847074e-07   1.84931494e-08   3.25102795e+01  -3.28882437e-11]

[[  3.29326356e-01  -7.43957919e-02   8.02246944e+07   2.64522183e-04]
 [ -7.43957919e-02   1.70872763e-02  -1.76477289e+07  -6.35825520e-05]
 [  8.02246944e+07  -1.76477289e+07   2.51023348e+16   5.87705672e+04]
 [  2.64522183e-04  -6.35825520e-05   5.87705672e+04   2.70249488e-07]]

无论如何,我怀疑这种合身性现在有点可疑。当我可以解决错误时,这将得到确认。

4

3 回答 3

93

更新于 2016 年 4 月 6 日

在大多数情况下,在拟合参数中获得正确的错误可能是微妙的。

让我们考虑拟合一个y=f(x)您有一组数据点的函数(x_i, y_i, yerr_i),其中i是一个在每个数据点上运行的索引。

在大多数物理测量中,误差yerr_i是测量设备或程序的系统不确定性,因此可以将其视为不依赖于 的常数i

使用哪个拟合函数,如何获取参数误差?

optimize.leastsq方法将返回分数协方差矩阵。将此矩阵的所有元素乘以残差方差(即减少的卡方)并取对角元素的平方根将为您提供拟合参数的标准偏差的估计值。我已将执行此操作的代码包含在以下函数之一中。

另一方面,如果您使用optimize.curvefit,则上述过程的第一部分(乘以减少的卡方)在幕后为您完成。然后,您需要对协方差矩阵的对角元素求平方根,以估计拟合参数的标准差。

此外,optimize.curvefit提供可选参数来处理更一般的情况,其中yerr_i每个数据点的值不同。从文档中:

sigma : None or M-length sequence, optional
    If not None, the uncertainties in the ydata array. These are used as
    weights in the least-squares problem
    i.e. minimising ``np.sum( ((f(xdata, *popt) - ydata) / sigma)**2 )``
    If None, the uncertainties are assumed to be 1.

absolute_sigma : bool, optional
    If False, `sigma` denotes relative weights of the data points.
    The returned covariance matrix `pcov` is based on *estimated*
    errors in the data, and is not affected by the overall
    magnitude of the values in `sigma`. Only the relative
    magnitudes of the `sigma` values matter.

我如何确定我的错误是正确的?

确定拟合参数中标准误差的正确估计是一个复杂的统计问题。协方差矩阵的结果,由有关误差的概率分布和参数之间的相互作用的假设实现optimize.curvefit并实际依赖于假设;optimize.leastsq可能存在的交互,具体取决于您的特定拟合函数f(x)

在我看来,处理复杂问题的最佳方法f(x)是使用引导方法,该方法在此链接中进行了概述。

让我们看一些例子

首先,一些样板代码。让我们定义一个波浪线函数并生成一些带有随机误差的数据。我们将生成一个随机误差很小的数据集。

import numpy as np
from scipy import optimize
import random

def f( x, p0, p1, p2):
    return p0*x + 0.4*np.sin(p1*x) + p2

def ff(x, p):
    return f(x, *p)

# These are the true parameters
p0 = 1.0
p1 = 40
p2 = 2.0

# These are initial guesses for fits:
pstart = [
    p0 + random.random(),
    p1 + 5.*random.random(), 
    p2 + random.random()
]

%matplotlib inline
import matplotlib.pyplot as plt
xvals = np.linspace(0., 1, 120)
yvals = f(xvals, p0, p1, p2)

# Generate data with a bit of randomness
# (the noise-less function that underlies the data is shown as a blue line)

xdata = np.array(xvals)
np.random.seed(42)
err_stdev = 0.2
yvals_err =  np.random.normal(0., err_stdev, len(xdata))
ydata = f(xdata, p0, p1, p2) + yvals_err

plt.plot(xvals, yvals)
plt.plot(xdata, ydata, 'o', mfc='None')

图01

现在,让我们使用各种可用的方法来拟合函数:

`optimize.leastsq`

def fit_leastsq(p0, datax, datay, function):

    errfunc = lambda p, x, y: function(x,p) - y

    pfit, pcov, infodict, errmsg, success = \
        optimize.leastsq(errfunc, p0, args=(datax, datay), \
                          full_output=1, epsfcn=0.0001)

    if (len(datay) > len(p0)) and pcov is not None:
        s_sq = (errfunc(pfit, datax, datay)**2).sum()/(len(datay)-len(p0))
        pcov = pcov * s_sq
    else:
        pcov = np.inf

    error = [] 
    for i in range(len(pfit)):
        try:
          error.append(np.absolute(pcov[i][i])**0.5)
        except:
          error.append( 0.00 )
    pfit_leastsq = pfit
    perr_leastsq = np.array(error) 
    return pfit_leastsq, perr_leastsq 

pfit, perr = fit_leastsq(pstart, xdata, ydata, ff)

print("\n# Fit parameters and parameter errors from lestsq method :")
print("pfit = ", pfit)
print("perr = ", perr)


# Fit parameters and parameter errors from lestsq method :
pfit =  [  1.04951642  39.98832634   1.95947613]
perr =  [ 0.0584024   0.10597135  0.03376631]


`optimize.curve_fit`

def fit_curvefit(p0, datax, datay, function, yerr=err_stdev, **kwargs):
    """
    Note: As per the current documentation (Scipy V1.1.0), sigma (yerr) must be:
        None or M-length sequence or MxM array, optional
    Therefore, replace:
        err_stdev = 0.2
    With:
        err_stdev = [0.2 for item in xdata]
    Or similar, to create an M-length sequence for this example.
    """
    pfit, pcov = \
         optimize.curve_fit(f,datax,datay,p0=p0,\
                            sigma=yerr, epsfcn=0.0001, **kwargs)
    error = [] 
    for i in range(len(pfit)):
        try:
          error.append(np.absolute(pcov[i][i])**0.5)
        except:
          error.append( 0.00 )
    pfit_curvefit = pfit
    perr_curvefit = np.array(error)
    return pfit_curvefit, perr_curvefit 

pfit, perr = fit_curvefit(pstart, xdata, ydata, ff)

print("\n# Fit parameters and parameter errors from curve_fit method :")
print("pfit = ", pfit)
print("perr = ", perr)


# Fit parameters and parameter errors from curve_fit method :
pfit =  [  1.04951642  39.98832634   1.95947613]
perr =  [ 0.0584024   0.10597135  0.03376631]


`引导`

def fit_bootstrap(p0, datax, datay, function, yerr_systematic=0.0):

    errfunc = lambda p, x, y: function(x,p) - y

    # Fit first time
    pfit, perr = optimize.leastsq(errfunc, p0, args=(datax, datay), full_output=0)


    # Get the stdev of the residuals
    residuals = errfunc(pfit, datax, datay)
    sigma_res = np.std(residuals)

    sigma_err_total = np.sqrt(sigma_res**2 + yerr_systematic**2)

    # 100 random data sets are generated and fitted
    ps = []
    for i in range(100):

        randomDelta = np.random.normal(0., sigma_err_total, len(datay))
        randomdataY = datay + randomDelta

        randomfit, randomcov = \
            optimize.leastsq(errfunc, p0, args=(datax, randomdataY),\
                             full_output=0)

        ps.append(randomfit) 

    ps = np.array(ps)
    mean_pfit = np.mean(ps,0)

    # You can choose the confidence interval that you want for your
    # parameter estimates: 
    Nsigma = 1. # 1sigma gets approximately the same as methods above
                # 1sigma corresponds to 68.3% confidence interval
                # 2sigma corresponds to 95.44% confidence interval
    err_pfit = Nsigma * np.std(ps,0) 

    pfit_bootstrap = mean_pfit
    perr_bootstrap = err_pfit
    return pfit_bootstrap, perr_bootstrap 

pfit, perr = fit_bootstrap(pstart, xdata, ydata, ff)

print("\n# Fit parameters and parameter errors from bootstrap method :")
print("pfit = ", pfit)
print("perr = ", perr)


# Fit parameters and parameter errors from bootstrap method :
pfit =  [  1.05058465  39.96530055   1.96074046]
perr =  [ 0.06462981  0.1118803   0.03544364]


观察

我们已经开始看到一些有趣的东西,所有三种方法的参数和误差估计几乎一致。那很好!

现在,假设我们想告诉拟合函数我们的数据中还有一些其他的不确定性,可能是系统不确定性会导致 20 倍的额外误差err_stdev。这是一个很大的错误,事实上,如果我们模拟一些具有这种错误的数据,它看起来像这样:

在此处输入图像描述

在这种噪音水平下,我们当然没有希望恢复拟合参数。

首先,让我们意识到leastsq甚至不允许我们输入这个新的系统错误信息。让我们看看curve_fit当我们告诉它错误时会发生什么:

pfit, perr = fit_curvefit(pstart, xdata, ydata, ff, yerr=20*err_stdev)

print("\nFit parameters and parameter errors from curve_fit method (20x error) :")
print("pfit = ", pfit)
print("perr = ", perr)


Fit parameters and parameter errors from curve_fit method (20x error) :
pfit =  [  1.04951642  39.98832633   1.95947613]
perr =  [ 0.0584024   0.10597135  0.03376631]

什么??这肯定是错的!

这曾经是故事的结尾,但最近curve_fit添加了absolute_sigma可选参数:

pfit, perr = fit_curvefit(pstart, xdata, ydata, ff, yerr=20*err_stdev, absolute_sigma=True)

print("\n# Fit parameters and parameter errors from curve_fit method (20x error, absolute_sigma) :")
print("pfit = ", pfit)
print("perr = ", perr)


# Fit parameters and parameter errors from curve_fit method (20x error, absolute_sigma) :
pfit =  [  1.04951642  39.98832633   1.95947613]
perr =  [ 1.25570187  2.27847504  0.72600466]

这有点好,但仍然有点可疑。 curve_fit认为我们可以从噪声信号中得到拟合,p1参数误差为 10%。让我们看看bootstrap有什么要说的:

pfit, perr = fit_bootstrap(pstart, xdata, ydata, ff, yerr_systematic=20.0)

print("\nFit parameters and parameter errors from bootstrap method (20x error):")
print("pfit = ", pfit)
print("perr = ", perr)


Fit parameters and parameter errors from bootstrap method (20x error):
pfit =  [  2.54029171e-02   3.84313695e+01   2.55729825e+00]
perr =  [  6.41602813  13.22283345   3.6629705 ]

啊,这也许是对我们拟合参数误差的更好估计。bootstrap认为它知道p1大约 34% 的不确定性。

概括

optimize.leastsqoptimize.curvefit为我们提供了一种估计拟合参数误差的方法,但我们不能只使用这些方法而不质疑它们。这bootstrap是一种使用蛮力的统计方法,在我看来,它倾向于在可能难以解释的情况下更好地工作。

我强烈建议您查看一个特定问题,然后尝试curvefit使用bootstrap. 如果它们相似,那么curvefit计算起来会便宜得多,因此可能值得使用。如果它们有很大差异,那么我的钱将在bootstrap.

于 2014-02-18T04:58:51.170 回答
12

在尝试回答我自己的类似问题时发现了您的问题。简短的回答。最小平方的cov_x输出应该乘以残差方差。IE

s_sq = (func(popt, args)**2).sum()/(len(ydata)-len(p0))
pcov = pcov * s_sq

curve_fit.py. 这是因为 minimumsq 输出分数协方差矩阵。我的大问题是剩余方差在谷歌搜索时显示为其他内容。

残差只是从您的拟合中减少卡方。

于 2013-02-13T15:55:50.660 回答
2

在线性回归的情况下可以精确计算误差。实际上,leastsq 函数给出了不同的值:

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

A = 1.353
B = 2.145

yerr = 0.25

xs = np.linspace( 2, 8, 1448 )
ys = A * xs + B + np.random.normal( 0, yerr, len( xs ) )

def linearRegression( _xs, _ys ):
    if _xs.shape[0] != _ys.shape[0]:
        raise ValueError( 'xs and ys must be of the same length' )
    xSum = ySum = xxSum = yySum = xySum = 0.0
    numPts = 0

    for i in range( len( _xs ) ):
        xSum += _xs[i]
        ySum += _ys[i]
        xxSum += _xs[i] * _xs[i]
        yySum += _ys[i] * _ys[i]
        xySum += _xs[i] * _ys[i]
        numPts += 1

    k = ( xySum - xSum * ySum / numPts ) / ( xxSum - xSum * xSum / numPts )
    b = ( ySum - k * xSum ) / numPts
    sk = np.sqrt( ( ( yySum - ySum * ySum / numPts ) / ( xxSum - xSum * xSum / numPts ) - k**2 ) / numPts )
    sb = np.sqrt( ( yySum - ySum * ySum / numPts ) - k**2 * ( xxSum - xSum * xSum / numPts ) ) / numPts

    return [ k, b, sk, sb ]

def linearRegressionSP( _xs, _ys ):
    defPars = [ 0, 0 ]
    pars, pcov, infodict, errmsg, success = \
    leastsq( lambda _pars, x, y: y - ( _pars[0] * x + _pars[1] ), defPars, args = ( _xs, _ys ), full_output=1 )

    errs = []
    if pcov is not None:
        if( len( _xs ) > len(defPars) ) and pcov is not None:
            s_sq = ( ( ys - ( pars[0] * _xs + pars[1] ) )**2 ).sum() / ( len( _xs ) - len( defPars ) )
            pcov *= s_sq

        for j in range( len( pcov ) ):
            errs.append( pcov[j][j]**0.5 )
    else:
        errs = [ np.inf, np.inf ]
    return np.append( pars, np.array( errs ) )

regr1 = linearRegression( xs, ys )
regr2 = linearRegressionSP( xs, ys )

print( regr1 )
print( 'Calculated sigma = %f' %  ( regr1[3] * np.sqrt( xs.shape[0] ) ) )
print( regr2 )
#print( 'B = %f must be in ( %f,\t%f )' % ( B, regr1[1] - regr1[3], regr1[1] + regr1[3] ) )


plt.plot( xs, ys, 'bo' )
plt.plot( xs, regr1[0] * xs + regr1[1] )
plt.show()

输出:

[1.3481681543925064, 2.1729338701374137, 0.0036028493647274687, 0.0062446292528624348]
Calculated sigma = 0.237624 # quite close to yerr
[ 1.34816815  2.17293387  0.00360534  0.01907908]

有趣的是,curvefit 和 bootstrap 会给出哪些结果......

于 2017-05-24T14:43:40.037 回答