最终目标是减少模型和观测值之间平方差的绝对值之和,其中包括Z
。
abs(((model(w, *params) - Z)**2).sum())
我的原始回答建议将leastsq
应用于一个残差
函数,该函数返回一个标量,表示实数和虚数差的平方和:
def residuals(params, w, Z):
R, C, L = params
diff = model(w, R, C, L) - Z
return diff.real**2 + diff.imag**2
Mike Sulzer建议使用返回浮点数向量的剩余函数。
以下是使用这些残差函数的结果比较:
from __future__ import print_function
import random
import numpy as np
import scipy.optimize as optimize
j = 1j
def model1(w, R, C, L):
Z = 1.0/(1.0/R + j*w*C) + j*w*L
return Z
def model2(w, R, C, L):
Z = 1.0/(1.0/R + j*w*C) + j*w*L
Z = np.repeat(Z, 2)
Z = Z[::2]
Z = Z.astype(np.complex64)
return Z
def make_data(R, C, L):
N = 10000
w = np.linspace(0.1, 2, N)
Z = model(w, R, C, L) + 0.1*(np.random.random(N) + j*np.random.random(N))
return w, Z
def residuals(params, w, Z):
R, C, L = params
diff = model(w, R, C, L) - Z
return diff.real**2 + diff.imag**2
def MS_residuals(params, w, Z):
"""
https://dev59.com/3nLYa4cB1Zd3GeqPY4cO#20104454 (Mike Sulzer)
"""
R, C, L = params
diff = model(w, R, C, L) - Z
z1d = np.zeros(Z.size*2, dtype=np.float64)
z1d[0:z1d.size:2] = diff.real
z1d[1:z1d.size:2] = diff.imag
return z1d
def alt_residuals(params, w, Z):
R, C, L = params
diff = model(w, R, C, L) - Z
return diff.astype(np.complex128).view(np.float64)
def compare(*funcs):
fmt = '{:15} | {:37} | {:17} | {:6}'
header = fmt.format('name', 'params', 'sum(residuals**2)', 'ncalls')
print('{}\n{}'.format(header, '-'*len(header)))
fmt = '{:15} | {:37} | {:17.2f} | {:6}'
for resfunc in funcs:
params, cov, infodict, mesg, ier = optimize.leastsq(
resfunc, p_guess, args=(w, Z),
full_output=True)
ssr = abs(((model(w, *params) - Z)**2).sum())
print(fmt.format(resfunc.__name__, params, ssr, infodict['nfev']))
print(end='\n')
R, C, L = 3, 2, 4
p_guess = 1, 1, 1
seed = 2013
model = model1
np.random.seed(seed)
w, Z = make_data(R, C, L)
assert np.allclose(model1(w, R, C, L), model2(w, R, C, L))
print('Using model1')
compare(residuals, MS_residuals, alt_residuals)
model = model2
print('Using model2')
compare(residuals, MS_residuals, alt_residuals)
产量
Using model1
name | params | sum(residuals**2) | ncalls
residuals | [ 2.86950167 1.94245378 4.04362841] | 9.41 | 89
MS_residuals | [ 2.85311972 1.94525477 4.04363883] | 9.26 | 29
alt_residuals | [ 2.85311972 1.94525477 4.04363883] | 9.26 | 29
Using model2
name | params | sum(residuals**2) | ncalls
residuals | [ 2.86590332 1.9326829 4.0450271 ] | 7.81 | 483
MS_residuals | [ 2.85422448 1.94853383 4.04333851] | 9.78 | 754
alt_residuals | [ 2.85422448 1.94853383 4.04333851] | 9.78 | 754
看起来使用哪个剩余函数可能取决于模型函数。考虑到model1
和model2
的相似性,我无法解释结果的差异。