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我正在尝试使用 优化函数的 SSE（误差平方和）scipy.optimize。为了测试，我创建了一个简单的问题，如下代码。但是 scipy 输出的优化参数永远不会使 SSE=0。有人可以帮我理解，我哪里出错了。

我试图用我的代码计算的 SSE 与在 excel 中计算的 SSE 进行交叉检查。它匹配。然后我使用最小化函数来最小化 SSE 函数，由 Scipy 计算的函数与手动计算的函数不匹配。我以前的函数是形式（y=ax+b）。下面是代码

import numpy as np
from scipy.optimize import minimize


e=np.array([0,2])
sig1=np.array([0,200])
k = [10,10]
#n = 0.2
coe=np.array([k[0],k[1]])

def sig2(e):
    v=(k[0]*e)+ k[1]
    SEzip = zip(sig1, v)
    sse = 0
    for y in SEzip:
        sse += np.power((y[0] - y[1]),2)
    return sse

print (sig2(e))
def f(coe):
   print(coe)
   return f
result = minimize(sig2,coe,method='Nelder-Mead',callback=(f),options={'xtol': 1e-6,'ftol':1e-06,'maxiter':50000,'disp': True,'adaptive' : True})

print(result)
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1 回答 1

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您在这里打印您的x0aka coe，我编辑了您的代码并将您的目标函数缩短sig2()为一行，然后编辑了您的回调以显示测试的变量及其等效的目标函数值。现在你可以清楚地看到sse=0已经达到了。

import numpy as np
from scipy.optimize import minimize

# for prettier numpy prints
np.set_printoptions(precision = 6)

# init 
e    = np.array([0,2])
sig1 = np.array([0,200])
k    = [10, 10]
coe  = np.array([k[0], k[1]])

# define objective function
def sig2(e):
    return sum([np.power((y[0] - y[1]), 2) for y in zip(sig1, (k[0]*e)+ k[1])])

# define callback
def f(e):
   print("e: %25s | sig2(e): %5s" % (e,round(sig2(e), 6)))

# optimize
result = minimize(sig2,
                  coe,
                  method   = 'Nelder-Mead',
                  callback = f,
                  options  = {'xtol': 1e-6,'ftol':1e-06,
                              'maxiter':50000,'disp': True,'adaptive' : True})

print(result)

输出：

...
e:     [-1.000053 18.999751] | sig2(e): 6e-06
e:     [-1.000062 19.000109] | sig2(e): 2e-06
e:     [-1.000062 19.000109] | sig2(e): 2e-06
e:     [-1.000062 19.000109] | sig2(e): 2e-06
e:     [-0.999934 18.999981] | sig2(e):   0.0
e:     [-1.000049 18.999979] | sig2(e):   0.0
e:     [-1.000027 19.000044] | sig2(e):   0.0
e:     [-0.999986 18.999996] | sig2(e):   0.0
e:     [-0.999986 18.999996] | sig2(e):   0.0
e:     [-0.999986 18.999996] | sig2(e):   0.0
e:     [-1.000009 18.999993] | sig2(e):   0.0
e:     [-1.000009 18.999993] | sig2(e):   0.0
e:     [-0.999995 19.      ] | sig2(e):   0.0
e:     [-0.999995 19.      ] | sig2(e):   0.0
e:     [-1.000003 18.999998] | sig2(e):   0.0
e:     [-1.       19.000002] | sig2(e):   0.0
e:     [-0.999998 19.      ] | sig2(e):   0.0
e:     [-1.000001 18.999999] | sig2(e):   0.0
e:     [-1.       19.000001] | sig2(e):   0.0
e:     [-0.999999 19.      ] | sig2(e):   0.0
e:                 [-1. 19.] | sig2(e):   0.0
e:                 [-1. 19.] | sig2(e):   0.0
e:                 [-1. 19.] | sig2(e):   0.0
Optimization terminated successfully.
         Current function value: 0.000000
         Iterations: 56
         Function evaluations: 110
 final_simplex: (array([[-1., 19.],
       [-1., 19.],
       [-1., 19.]]), array([6.221143e-12, 1.914559e-11, 1.946860e-11]))
           fun: 6.2211434216849394e-12
       message: 'Optimization terminated successfully.'
          nfev: 110
           nit: 56
        status: 0
       success: True
             x: array([-1., 19.])
于 2019-04-28T10:35:13.133 回答