我希望能够执行允许我将任意曲线函数拟合到数据的拟合,并允许我在参数上设置任意边界,例如我想拟合函数:
f(x) = a1(x-a2)^a3\cdot\exp(-\a4*x^a5)
并说:
a2
位于以下范围内:(-1, 1)
a3
和a5
为正有很好的scipy curve_fit函数,但它不允许指定参数边界。还有一个很好的http://code.google.com/p/pyminuit/库可以执行泛型最小化,它允许在参数上设置边界,但在我的情况下它没有覆盖。
答案 0 :(得分:10)
注意:SciPy版本0.17中新增
假设您希望将模型拟合到如下所示的数据:
y=a*t**alpha+b
并使用alpha
约束0<alpha<2
而其他参数a和b保持空闲。然后我们应该以下列方式使用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 = curve_fit(func, xdata, ydata,bounds=param_bounds)
来源为here。
答案 1 :(得分:3)
解决方法:使用变量转换,例如a2 = tanh(a2'),a3 = exp(a3')或a5 = a5'^ 2.
答案 2 :(得分:3)
正如Rob Falck已经提到的,您可以使用scipy.minimize中的scipy非线性优化例程来最小化任意误差函数,例如basin-hopping。均方误差。
请注意,您提供的功能不一定具有实际值 - 也许这就是您在pyminuit中的最小化没有收敛的原因。你必须更明确地对待它,见例2。
下面的示例都使用L-BFGS-B
最小化方法,该方法支持有界参数区域。我将这个答案分为两部分:
下面的示例显示了此功能略微修改版本的优化。
import numpy as np
import pylab as pl
from scipy.optimize import minimize
points = 500
xlim = 3.
def f(x,*p):
a1,a2,a3,a4,a5 = p
return a1*np.abs(x-a2)**a3 * np.exp(-a4 * np.abs(x)**a5)
# generate noisy data with known coefficients
p0 = [1.4,-.8,1.1,1.2,2.2]
x = (np.random.rand(points) * 2. - 1.) * xlim
x.sort()
y = f(x,*p0)
y_noise = y + np.random.randn(points) * .05
# mean squared error wrt. noisy data as a function of the parameters
err = lambda p: np.mean((f(x,*p)-y_noise)**2)
# bounded optimization using scipy.minimize
p_init = [1.,-1.,.5,.5,2.]
p_opt = minimize(
err, # minimize wrt to the noisy data
p_init,
bounds=[(None,None),(-1,1),(None,None),(0,None),(None,None)], # set the bounds
method="L-BFGS-B" # this method supports bounds
).x
# plot everything
pl.scatter(x, y_noise, alpha=.2, label="f + noise")
pl.plot(x, y, c='#000000', lw=2., label="f")
pl.plot(x, f(x,*p_opt) ,'--', c='r', lw=2., label="fitted f")
pl.xlabel("x")
pl.ylabel("f(x)")
pl.legend(loc="best")
pl.xlim([-xlim*1.01,xlim*1.01])
pl.show()
将上述最小化扩展到复杂域可以通过显式地转换为复数并调整误差函数来完成:
首先,您将值x显式地转换为复数值,以确保f返回复数值,并且实际上可以计算负数的小数指数。其次,我们在实部和虚部上计算一些误差函数 - 一个简单的候选者是平方复数绝对值的平均值。
import numpy as np
import pylab as pl
from scipy.optimize import minimize
points = 500
xlim = 3.
def f(x,*p):
a1,a2,a3,a4,a5 = p
x = x.astype(complex) # cast x explicitly to complex, to ensure complex valued f
return a1*(x-a2)**a3 * np.exp(-a4 * x**a5)
# generate noisy data with known coefficients
p0 = [1.4,-.8,1.1,1.2,2.2]
x = (np.random.rand(points) * 2. - 1.) * xlim
x.sort()
y = f(x,*p0)
y_noise = y + np.random.randn(points) * .05 + np.random.randn(points) * 1j*.05
# error function chosen as mean of squared absolutes
err = lambda p: np.mean(np.abs(f(x,*p)-y_noise)**2)
# bounded optimization using scipy.minimize
p_init = [1.,-1.,.5,.5,2.]
p_opt = minimize(
err, # minimize wrt to the noisy data
p_init,
bounds=[(None,None),(-1,1),(None,None),(0,None),(None,None)], # set the bounds
method="L-BFGS-B" # this method supports bounds
).x
# plot everything
pl.scatter(x, np.real(y_noise), c='b',alpha=.2, label="re(f) + noise")
pl.scatter(x, np.imag(y_noise), c='r',alpha=.2, label="im(f) + noise")
pl.plot(x, np.real(y), c='b', lw=1., label="re(f)")
pl.plot(x, np.imag(y), c='r', lw=1., label="im(f)")
pl.plot(x, np.real(f(x,*p_opt)) ,'--', c='b', lw=2.5, label="fitted re(f)")
pl.plot(x, np.imag(f(x,*p_opt)) ,'--', c='r', lw=2.5, label="fitted im(f)")
pl.xlabel("x")
pl.ylabel("f(x)")
pl.legend(loc="best")
pl.xlim([-xlim*1.01,xlim*1.01])
pl.show()
似乎最小化器可能对初始值有点敏感 - 因此我把我的第一个猜测(p_init)放在离最佳值不太远的地方。如果你不得不与此作斗争,除了全局优化循环之外,你可以使用相同的最小化程序,例如: brute或{{3}}。
答案 3 :(得分:3)
您可以使用lmfit来解决这些问题。因此,我添加了一个示例(使用了另一个功能,但它可以轻松调整),以便在有人对此主题感兴趣的情况下使用它。
假设您有一个数据集如下:
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)))
具有
的约束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,参数在请求的范围内。
以下是使用一些附加注释重现该图的完整代码:
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:
如果您使用版本0.9.x,则需要相应地调整代码;检查here已从0.8.3更改为0.9.x。
答案 4 :(得分:1)
您是否考虑将其视为优化问题并使用scipy中的一个非线性优化例程来通过改变函数系数来最小化最小二乘误差?优化中的许多例程允许对自变量进行约束约束。