lmfit 不探索参数空间

Question

我正在尝试使用lmfit 为使用Model 和Parameters 类的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。

代码如下：

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)

知道为什么它执行的函数评估很少返回不合适的结果吗？对此的任何帮助将不胜感激！

Answer 1

这是与fitting step function with variation in the step location with scipy optimize curve_fit 类似的问题。见https://stackoverflow.com/a/59504874/5179748。

基本上，scipy.optimize/lmfit 中的求解器假定参数是连续的——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 wid 参数的微小变化不会对结果产生影响，因为 argmin() 将始终返回一个整数值。

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()