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python - lmfit 不探索参数空间

问题描述

我正在尝试使用和类lmfit为一些随机数据找到函数的最佳拟合参数。但是，它似乎并没有过多地探索参数空间。它会进行大约 10 次函数评估，然后返回一个非常合适的结果。ModelParameters

这是代码：

import numpy as np
from lmfit.model import Model
from lmfit.parameter import Parameters
import matplotlib.pyplot as plt

def dip(x, loc, wid, dep):
    """Make a line with a dip in it"""
    # Array of ones
    y = np.ones_like(x)

    # Define start and end points of dip
    start = np.abs(x - (loc - (wid/2.))).argmin()
    end = np.abs(x - (loc + (wid/2.))).argmin()

    # Set depth of the dip
    y[start:end] *= dep

    return y

def fitter(x, loc, wid, dep, scatter=0.001, sigma=3):
    """Find the parameters of the dip function in random data"""
    # Make the lmfit model
    model = Model(dip)

    # Make random data and print input values
    rand_loc = abs(np.random.normal(loc, scale=0.02))
    rand_wid = abs(np.random.normal(wid, scale=0.03))
    rand_dep = abs(np.random.normal(dep, scale=0.005))
    print('rand_loc: {}\nrand_wid: {}\nrand_dep: {}\n'.format(rand_loc, rand_wid, rand_dep))
    data = dip(x, rand_loc, rand_wid, rand_dep) + np.random.normal(0, scatter, x.size)

    # Make parameter ranges
    params = Parameters()
    params.add('loc', value=loc, min=x.min(), max=x.max())
    params.add('wid', value=wid, min=0, max=x.max()-x.min())
    params.add('dep', value=dep, min=scatter*10, max=0.8)

    # Fit the data
    result = model.fit(data, x=x, params)
    print(result.fit_report())

    # Plot it
    plt.plot(x, data, 'bo')
    plt.plot(x, result.init_fit, 'k--', label='initial fit')
    plt.plot(x, result.best_fit, 'r-', label='best fit')
    plt.legend(loc='best')
    plt.show()

然后我运行：

fitter(np.linspace(55707.97, 55708.1, 100), loc=55708.02, wid=0.04, dep=0.98)

返回（例如，因为它是随机数据）：

rand_loc: 55707.99659784677
rand_wid: 0.02015076619874132
rand_dep: 0.9849809461153651

[[Model]]
    Model(dip)
[[Fit Statistics]]
    # fitting method   = leastsq
    # function evals   = 9
    # data points      = 100
    # variables        = 3
    chi-square         = 0.00336780
    reduced chi-square = 3.4720e-05
    Akaike info crit   = -1023.86668
    Bayesian info crit = -1016.05117
##  Warning: uncertainties could not be estimated:
    loc:  at initial value
    wid:  at initial value
[[Variables]]
    loc:  55708.0200 (init = 55708.02)
    wid:  0.04000000 (init = 0.04)
    dep:  0.99754082 (init = 0.98)

不合适

知道为什么它执行这么少的函数评估返回不合适吗？对此的任何帮助将不胜感激！

标签： pythoncurve-fittingmodel-fittinglmfit

解决方案

这是一个与使用 scipy optimize curve_fit 拟合阶跃位置变化的阶跃函数类似的问题。请参阅https://stackoverflow.com/a/59504874/5179748。

基本上，求解器scipy.optimize/lmfit假设参数是连续的——而不是离散的——变量。他们对参数进行小幅更改，以查看结果发生了什么变化。loc和参数的微小变化wid不会对结果产生影响，因为argmin()总是会返回一个整数值。

您可能会发现使用有限宽度的矩形模型（请参阅https://lmfit.github.io/lmfit-py/builtin_models.html#rectanglemodel）会有所帮助。我稍微改变了你的例子，但这应该足以让你开始：

import numpy as np
import matplotlib.pyplot as plt
from lmfit.models import RectangleModel, ConstantModel

def dip(x, loc, wid, dep):
    """Make a line with a dip in it"""
    # Array of ones
    y = np.ones_like(x)

    # Define start and end points of dip
    start = np.abs(x - (loc - (wid/2.))).argmin()
    end = np.abs(x - (loc + (wid/2.))).argmin()

    # Set depth of the dip
    y[start:end] *= dep
    return y

x = np.linspace(0, 1, 201)
data = dip(x, 0.3, 0.09, 0.98) + np.random.normal(0, 0.001, x.size)

model = RectangleModel() + ConstantModel()
params = model.make_params(c=1.0, amplitude=-0.01, center1=.100, center2=0.7, sigma1=0.15)

params['sigma2'].expr = 'sigma1' # force left and right widths to be the same size
params['c'].vary = False         # force offset = 1.0 : value away from "dip"


result = model.fit(data, params, x=x)
print(result.fit_report())

plt.plot(x, data, 'bo')
plt.plot(x, result.init_fit, 'k--', label='initial fit')
plt.plot(x, result.best_fit, 'r-', label='best fit')
plt.legend(loc='best')
plt.show()

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