Scipy的优化曲线拟合限制

Question

有什么方法可以为Scipy的优化曲线拟合提供限制吗？

我的例子:

    def optimized_formula(x, m_1, m_2, y_1, y_2, ratio_2):
        return (log(x[0]) * m_1 + m_2)*((1 - x[1]/max_age)*(1-ratio_2)) + ((log(x[1]) * y_1 + y_2)*(x[1]/max_age)*ratio_2)

    popt, pcov = optimize.curve_fit(optimized_formula, usage_and_age, prices)

x [0]是年龄,max_age是常数.考虑到这一点,当x [0]接近最大值时,x [1]/max_age接近1.

是否有可能提供约束/限制,其中x [1]/max_age> 0.3和x [1]/max_age <0.7以及其他约束,例如m_1 <0,m_2> 0,等等.

Answer 1

正如另一个答案中所建议的那样,你可以使用lmfit来解决这些问题.因此,我添加了一个如何使用它的示例,以防有人对此主题感兴趣.

假设您有一个数据集,如下所示:

xdata = np.array([177.,180.,183.,187.,189.,190.,196.,197.,201.,202.,203.,204.,206.,218.,225.,231.,234.,
          252.,262.,266.,267.,268.,277.,286.,303.])

ydata = np.array([0.81,0.74,0.78,0.75,0.77,0.81,0.73,0.76,0.71,0.74,0.81,0.71,0.74,0.71,
      0.72,0.69,0.75,0.59,0.61,0.63,0.64,0.63,0.35,0.27,0.26])

并且您希望将模型拟合到如下所示的数据:

model = n1 + (n2 * x + n3) * 1./ (1. + np.exp(n4 * (n5 - x)))

有限制

0.2 < n1 < 0.8
-0.3 < n2 < 0

使用lmfit(版本0.8.3),然后获得以下输出:

n1:   0.26564921 +/- 0.024765 (9.32%) (init= 0.2)
n2:  -0.00195398 +/- 0.000311 (15.93%) (init=-0.005)
n3:   0.87261892 +/- 0.068601 (7.86%) (init= 1.0766)
n4:  -1.43507072 +/- 1.223086 (85.23%) (init=-0.36379)
n5:   277.684530 +/- 3.768676 (1.36%) (init= 274)

如您所见,拟合非常好地再现数据,并且参数在请求的范围内.

以下是使用一些附加注释重现绘图的完整代码:

from lmfit import minimize, Parameters, Parameter, report_fit
import numpy as np

xdata = np.array([177.,180.,183.,187.,189.,190.,196.,197.,201.,202.,203.,204.,206.,218.,225.,231.,234.,
      252.,262.,266.,267.,268.,277.,286.,303.])

ydata = np.array([0.81,0.74,0.78,0.75,0.77,0.81,0.73,0.76,0.71,0.74,0.81,0.71,0.74,0.71,
      0.72,0.69,0.75,0.59,0.61,0.63,0.64,0.63,0.35,0.27,0.26])

def fit_fc(params, x, data):

    n1 = params['n1'].value
    n2 = params['n2'].value
    n3 = params['n3'].value
    n4 = params['n4'].value
    n5 = params['n5'].value

    model = n1 + (n2 * x + n3) * 1./ (1. + np.exp(n4 * (n5 - x)))

    return model - data #that's what you want to minimize

# create a set of Parameters
# 'value' is the initial condition
# 'min' and 'max' define your boundaries
params = Parameters()
params.add('n1', value= 0.2, min=0.2, max=0.8)
params.add('n2', value= -0.005, min=-0.3, max=10**(-10))
params.add('n3', value= 1.0766, min=-1000., max=1000.)
params.add('n4', value= -0.36379, min=-1000., max=1000.)
params.add('n5', value= 274.0, min=0., max=1000.)

# do fit, here with leastsq model
result = minimize(fit_fc, params, args=(xdata, ydata))

# write error report
report_fit(params)

xplot = np.linspace(min(xdata), max(xdata), 1000)
yplot = result.values['n1'] + (result.values['n2'] * xplot + result.values['n3']) * \
                              1./ (1. + np.exp(result.values['n4'] * (result.values['n5'] - xplot)))
#plot results
try:
    import pylab
    pylab.plot(xdata, ydata, 'k+')
    pylab.plot(xplot, yplot, 'r')
    pylab.show()
except:
    pass

编辑:

如果使用版本0.9.x,则需要相应地调整代码; 检查这里它已经作了修改,从0.8.3到0.9.x.

Answer 2

注意:SciPy版本0.17中的新功能

假设您想要将模型拟合到如下所示的数据:

y=a*t**alpha+b

以及对alpha的约束

0<alpha<2

而其他参数a和b仍然是免费的.然后我们应该使用optimize.curve_fit的bounds选项:

import numpy as np
from scipy.optimize import curve_fit
def func(t, a,alpha,b):
     return a*t**alpha+b
param_bounds=([-np.inf,0,-np.inf],[np.inf,2,np.inf])
popt, pcov = optimize.curve_fit(func, xdata,ydata,bounds=param_bounds)

来源就在这里