I am trying to fit this function to some data:
But when I use my code
import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
def f(x, start, end):
res = np.empty_like(x)
res[x < start] =-1
res[x > end] = 1
linear = np.all([[start <= x], [x <= end]], axis=0)[0]
res[linear] = np.linspace(-1., 1., num=np.sum(linear))
return res
if __name__ == '__main__':
xdata = np.linspace(0., 1000., 1000)
ydata = -np.ones(1000)
ydata[500:1000] = 1.
ydata = ydata + np.random.normal(0., 0.25, len(ydata))
popt, pcov = curve_fit(f, xdata, ydata, p0=[495., 505.])
print(popt, pcov)
plt.figure()
plt.plot(xdata, f(xdata, *popt), 'r-', label='fit')
plt.plot(xdata, ydata, 'b-', label='data')
plt.show()
I get the error
OptimizeWarning: Covariance of the parameters could not be estimated
Output:
In this example start and end should be closer to 500, but they dont change at all from my initial guess.
The warning (not error) of
OptimizeWarning: Covariance of the parameters could not be estimated
means that the fit could not determine the uncertainties (variance) of the fitting parameters.
The main problem is that your model function f
treats the parameters start
and end
as discrete values -- they are used as integer locations for the change in functional form. scipy's curve_fit
(and all other optimization routines in scipy.optimize
) assume that parameters are continuous variables, not discrete.
The fitting procedure will try to take small steps (typically around machine precision) in the parameters to get a numerical derivative of the residual with respect to the variables (the Jacobian). With values used as discrete variables, these derivatives will be zero and the fitting procedure will not know how to change the values to improve the fit.
Parece que está intentando ajustar una función de paso a algunos datos. Permítame recomendarle que pruebe lmfit
( https://lmfit.github.io/lmfit-py ) que proporciona una interfaz de nivel superior para el ajuste de curvas y tiene muchos modelos incorporados. Por ejemplo, incluye un StepModel
que debería poder modelar sus datos.
Para una ligera modificación de sus datos (para que tenga un paso finito), el siguiente script lmfit
puede ajustarse a dichos datos:
#!/usr/bin/python
import numpy as np
from lmfit.models import StepModel, LinearModel
import matplotlib.pyplot as plt
np.random.seed(0)
xdata = np.linspace(0., 1000., 1000)
ydata = -np.ones(1000)
ydata[500:1000] = 1.
# note that a linear step is added here:
ydata[490:510] = -1 + np.arange(20)/10.0
ydata = ydata + np.random.normal(size=len(xdata), scale=0.1)
# model data as Step + Line
step_mod = StepModel(form='linear', prefix='step_')
line_mod = LinearModel(prefix='line_')
model = step_mod + line_mod
# make named parameters, giving initial values:
pars = model.make_params(line_intercept=ydata.min(),
line_slope=0,
step_center=xdata.mean(),
step_amplitude=ydata.std(),
step_sigma=2.0)
# fit data to this model with these parameters
out = model.fit(ydata, pars, x=xdata)
# print results
print(out.fit_report())
# plot data and best-fit
plt.plot(xdata, ydata, 'b')
plt.plot(xdata, out.best_fit, 'r-')
plt.show()
que imprime un informe de
[[Model]]
(Model(step, prefix='step_', form='linear') + Model(linear, prefix='line_'))
[[Fit Statistics]]
# fitting method = leastsq
# function evals = 49
# data points = 1000
# variables = 5
chi-square = 9.72660131
reduced chi-square = 0.00977548
Akaike info crit = -4622.89074
Bayesian info crit = -4598.35197
[[Variables]]
step_sigma: 20.6227793 +/- 0.77214167 (3.74%) (init = 2)
step_center: 490.167878 +/- 0.44804412 (0.09%) (init = 500)
step_amplitude: 1.98946656 +/- 0.01304854 (0.66%) (init = 0.996283)
line_intercept: -1.00628058 +/- 0.00706005 (0.70%) (init = -1.277259)
line_slope: 1.3947e-05 +/- 2.2340e-05 (160.18%) (init = 0)
[[Correlations]] (unreported correlations are < 0.100)
C(step_amplitude, line_slope) = -0.875
C(step_sigma, step_center) = -0.863
C(line_intercept, line_slope) = -0.774
C(step_amplitude, line_intercept) = 0.461
C(step_sigma, step_amplitude) = 0.170
C(step_sigma, line_slope) = -0.147
C(step_center, step_amplitude) = -0.146
C(step_center, line_slope) = 0.127
Lmfit tiene muchas características adicionales. Por ejemplo, si desea establecer límites en algunos de los valores de los parámetros o corregir algunos para que no varíen, puede hacer lo siguiente:
# make named parameters, giving initial values:
pars = model.make_params(line_intercept=ydata.min(),
line_slope=0,
step_center=xdata.mean(),
step_amplitude=ydata.std(),
step_sigma=2.0)
# now set max and min values for step amplitude"
pars['step_amplitude'].min = 0
pars['step_amplitude'].max = 100
# fix the offset of the line to be -1.0
pars['line_offset'].value = -1.0
pars['line_offset'].vary = False
# then run fit with these parameters
out = model.fit(ydata, pars, x=xdata)
Si sabe que el modelo debe ser Step+Constant
y que la constante debe ser fija, también puede modificar el modelo para que sea
from lmfit.models import ConstantModel
# model data as Step + Constant
step_mod = StepModel(form='linear', prefix='step_')
const_mod = ConstantModel(prefix='const_')
model = step_mod + const_mod
pars = model.make_params(const_c=-1,
step_center=xdata.mean(),
step_amplitude=ydata.std(),
step_sigma=2.0)
pars['const_c'].vary = False
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