0
我想重現一個模型方程使用非線性最小二乘法得出一個「測試」值的曲線。 y和x實驗數據是兩個相同大小的1D numpy數組,分別是「a」和「angle_plot」。我正在使用的代碼產生一個錯誤:「'float'對象不可調用」。我的代碼有什麼問題?謝謝擬合曲線與模型方程numpy
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import leastsq
a = array([ 0.04022493, 0.04287536, 0.03983657, 0.0393201 , 0.03810298,
0.0363814 , 0.0331144 , 0.03074823, 0.02795767, 0.02413816,
0.02180802, 0.01861309, 0.01632699, 0.01368056, 0.01124232,
0.01005323, 0.00867196, 0.00940864, 0.00961282, 0.00892419,
0.01048963, 0.01199101, 0.01533408, 0.01855704, 0.02163586,
0.02630014, 0.02971127, 0.03511223, 0.03941218, 0.04280329,
0.04689105, 0.04960554, 0.05232003, 0.05487037, 0.05843364,
0.05120701])
angle_plot = array([ 0. , 0.08975979, 0.17951958, 0.26927937, 0.35903916,
0.44879895, 0.53855874, 0.62831853, 0.71807832, 0.80783811,
0.8975979 , 0.98735769, 1.07711748, 1.16687727, 1.25663706,
1.34639685, 1.43615664, 1.52591643, 1.61567622, 1.70543601,
1.7951958 , 1.88495559, 1.97471538, 2.06447517, 2.15423496,
2.24399475, 2.33375454, 2.42351433, 2.51327412, 2.60303391,
2.6927937 , 2.78255349, 2.87231328, 2.96207307, 3.05183286,
3.14159265])
def residual(vars, x, data):
beta = vars[0]
model = 1/(4*np.pi)(1+beta*(3/2*np.cos(x)**2-1/2))
return data-model
vars = [0.2]
out = leastsq(residual, vars, args=(angle_plot, a))
的確如此。我看不到它。現在似乎工作,但我想繪製實驗與模型,看看合適的好處。我可以從哪裏獲得配件?謝謝 – diegus 2014-09-12 10:33:27
這是一個**完全不同的問題,你應該單獨詢問。 – Ffisegydd 2014-09-12 10:34:17