最小二乘最小化复数

Question

我一直在使用我的Matlab,但我的愿景是最终切换到用Python完成所有分析,因为它是一种实际的编程语言和其他一些原因.

我一直试图解决的最近问题是对复杂数据进行最小二乘最小化.我是一名工程师,我们经常处理复杂的阻抗,我正在尝试使用曲线拟合将简单的电路模型拟合到测量数据中.

阻抗方程如下:

Z(w)= 1 /(1/R + j*w*C)+ j*w*L.

然后我试图找到R,C和L的值,以便找到最小二乘曲线.

我已经尝试使用优化包,例如optimize.curve_fit或optimize.leastsq,但它们不适用于复数.

然后我尝试使我的残差函数返回复杂数据的大小,但这也不起作用.

Answer 1

参考unutbu答案,没有必要通过取功能残差的幅度平方来减少可用信息,因为leastsq不关心数字是真实的还是复杂的,只是它们被表示为一维数组,保持了完整性功能关系.

这是替换残差函数:

def residuals(params, w, Z):
    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

这是唯一需要做出的改变.种子2013的答案是:[2.96564781,1.99929516,4.00106534].

相对于unutbu答案的错误显着减少了一个数量级.

Answer 2

最终，目标是减少模型与观测值之间的平方差之和的绝对值Z：

abs(((model(w, *params) - Z)**2).sum())

我最初的答案 建议应用于leastsq一个residuals函数，该函数返回一个表示实部和虚部差的平方和的标量：

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残差函数。

以下是使用这些残差函数的结果的比较：

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
    # make Z non-contiguous and of a different complex dtype
    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):
    """
    /sf/answers/1407311811/ (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 = optimize.leastsq(resfunc, p_guess, args=(w, Z))
        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

因此看来使用哪个残差函数可能取决于模型函数。model1鉴于和 的相似性，我无法解释结果的差异model2。