我正在尝试使用Nelder-Mead算法最小化函数mymodel
以适合我的数据。这是在myfit
函数中使用scipy的optimize.fmin
完成的。我觉得我很亲密,但我一定错过了一些东西,因为我一直收到错误:
'操作数无法与形状(80,)(5,)'一起广播。
import numpy as np
import matplotlib.pyplot as plt
from scipy import optimize
from scipy import special
def mymodel(c,t,y):
"""
This function specifies the form to be minimized by fmins in myfit.
c is a 1 x 5 array containing the fit parameters.
"""
m = (np.sin(np.exp(-c[1]*t)*c[0]/2.0))**2
# compute complete elliptic integral of the first kind with ellipk
w = np.pi*c[2]/2.0/special.ellipk(m)
dt = t[1] - t[0]
phase = np.cumsum(w)*dt
z = np.sum((y - c[0] * np.exp(-c[1]*t) * np.cos(phase+c[3])-c[4])**2)
return z
def myfit(c, pos):
"""
Fitting procedure for the amplitude decay of the undriven pendulum
initial fit parameters:
c[0]=theta_m, c[1]=alpha, c[2]=omega_0, c[3]=phi, c[4]=const.
pos = the position data
"""
# convert data to seconds
t = 0.001*np.arange(0,len(pos))
dt = t[1] - t[0]
# Minimise the function mymodel using Nelder-Mead algorithm
c = optimize.fmin(mymodel, c, args=(t,y), maxiter=5000, full_output=True)
m = (np.sin(np.exp(-c[1]*t)*c[0]/2.0))**2
# change of frequency with amplitude
w = np.pi*c[2]/2.0/special.ellipk(m)
phase = np.cumsum(w)*dt
# use values from fmin
fit = c[0]*np.exp(-c[1]*t)*np.cos(phase+c[3])+c[4]
return t, c, fit
t = np.array([ 0., 15., 30., 45., 60., 75., 90., 105.,
120., 135., 150., 165., 180., 195., 210., 225.,
240., 255., 270., 285., 300., 315., 330., 345.,
360., 375., 390., 405., 420., 435., 450., 465.,
480., 495., 510., 525., 540., 555., 570., 585.,
600., 615., 630., 645., 660., 675., 690., 705.,
720., 735., 750., 765., 780., 795., 810., 825.,
840., 855., 870., 885., 900., 915., 930., 945.,
960., 975., 1005., 1020., 1035., 1050., 1065., 1080.,
1095., 1110., 1125., 1140., 1155., 1170., 1185., 1200.,
])
pos = np.array([ 28.95, 28.6 , 28.1 , 27.5 , 26.75, 25.92, 24.78, 23.68,
22.5 , 21.35, 20.25, 19.05, 17.97, 16.95, 15.95, 15.1 ,
14.45, 13.77, 13.3 , 13. , 12.85, 12.82, 12.94, 13.2 ,
13.6 , 14.05, 14.65, 15.45, 16.1 , 16.9 , 17.75, 18.7 ,
19.45, 20.3 , 21.1 , 21.9 , 22.6 , 23.25, 23.75, 24.2 ,
24.5 , 24.75, 24.88, 24.9 , 24.8 , 24.65, 24.35, 23.9 ,
23.55, 22.95, 22.5 , 21.98, 21.3 , 20.65, 20.05, 19.4 ,
18.85, 18.3 , 17.8 , 17.35, 16.95, 16.6 , 16.35, 16.2 ,
16.1 , 16.1 , 16.35, 16.5 , 16.75, 17.02, 17.4 , 17.75,
18.3 , 18.65, 19.1 , 19.55, 20. , 20.45, 20.85, 21.25,
])
# fitting with myfit function
c = np.array([1,1,1,1,1]) # initial guess
t, c, fit = myfit(c, pos)
plt.plot(t,fit)
plt.show()
答案 0 :(得分:2)
问题是fmin
,使用full_output=true
参数调用,不仅返回优化参数,还返回包含参数和其他内容的元组。
参数是元组中的第一个值。您可以使用[0]
:
c = optimize.fmin(mymodel, c, args=(t,pos), maxiter=5000, full_output=True)[0]
或者只需删除full_output=true
参数:
c = optimize.fmin(mymodel, c, args=(t,pos), maxiter=5000)
http://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.optimize.fmin.html