3 
numseq = ['0012000', '0112000', '0212000', '0312000', '1012000', '1112000',                                                                                   '1212000', '1312000', '2012000', '2112000', '2212000', '2312000', '3012000', '3112000',          '3212000', '3312000', '0002000', '0022000', '0032000', '1002000', '1022000', '1032000',     '2002000', '2022000', '2032000', '3002000', '3022000', '3032000', '0010000', '0011000', '0013000', '1010000', '1011000', '1013000', '2010000', '2011000', '2013000', '3010000', '3011000', '3013000', '0012100', '0012200', '0012300', '1012100', '1012200', '1012300', '2012100', '2012200', '2012300', '3012100']
prob = [-0.66474525640568083, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.78361598908750163, -0.66474525640568083, -0.66474525640568083, -0.66474525640568083, -0.66474525640568083, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.66474525640568083, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.66474525640568083, -0.66474525640568083, -0.66474525640568083, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.66474525640568083, -0.66474525640568083, -0.66474525640568083, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.66474525640568083, -0.66474525640568083, -0.66474525640568083, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212, -0.49518440694747212]

numseq并且prob是每个长度为 50 的列表。它们是收集的实验数据。numseq对应于 X 轴值，prob对应于 Y 轴值。

我要最小化的功能是：

def residue(allparams, xdata, ydata):
    chi2 = 0.0
    for i in range(0,len(xdata)):
        x = xdata[i]
        y = 0
        for j in range(len(x)):
            y = y-allparams[int(x[j])][j]
            chi2 = chi2 + (ydata[i]-y)**2
return chi2

所以：

  • allparams是一个 4×7 的矩阵，包含所有要优化的参数。
  • xdata是 X 轴值，即numseq
  • ydata只是一个数字列表，即prob

chi2是实验值和模型值之间的平方差。这是必须最小化的。

参数的初始猜测由下式给出：

x0 = [[-0.6, -0.6, -0.6, -0.6, -0.6, -0.6, -0.6], [-0.6, -0.6, -0.6, -0.6, -0.6, -0.6, -0.6], [-0.6, -0.6, -0.6, -0.6, -0.6, -0.6, -0.6], [-0.6, -0.6, -0.6, -0.6, -0.6, -0.6, -0.6]]

现在我如何调用fmin这个函数？我试过了

fmin(residue, x0, args=(numseq, prob))

但我不断收到错误消息：

Traceback (most recent call last):
  File "<pyshell#362>", line 1, in <module>
    fmin(residue, x0, args=(numseq, prob))
  File "C:\Python31\lib\site-packages\scipy\optimize\optimize.py", line 258, in fmin
    fsim[0] = func(x0)
  File "C:\Python31\lib\site-packages\scipy\optimize\optimize.py", line 177, in function_wrapper
    return function(x, *args)
  File "<pyshell#361>", line 7, in residue
    y = y-allparams[int(x[j])][j]
IndexError: invalid index to scalar variable.

为什么会这样？是因为fmin不能接受二维数组作为初始猜测吗？那么我是否必须更改我的整个代码才能处理一维参数数组？

即使你不能解释这个问题，你至少能告诉我这个fmin模块是如何工作的吗？即如何实现fminN维数组优化的语法？你能解释一下args()是什么吗？我是优化新手，我不知道如何实现它:(

4

1 回答 1

2

“fmin”例程可以接受二维数组作为初始猜测。但它做的第一件事是展平这个数组 [(4,7) --> (28)]。因此，您的残差函数将 (4,7) 数组作为输入，“fmin”例程为其提供了一个长度为 28 的扁平“x0”。这就是您看到错误的原因：
y = y-allparams[int(x[j])][j]
IndexError: invalid index to scalar variable.

请参阅此处的源代码。

因此，您似乎必须更改残差函数以接受向量而不是数组。然而，这似乎并不算太​​糟糕。我尝试了以下似乎可行的方法（注意：请仔细检查！）

def residue_alternative(allparams, inshape, xdata, ydata):
    m, n = inshape
    chi2 = 0.0
    for i in range(0,len(xdata)):
        x = xdata[i]
        y = 0
        for j in range(len(x)):
            idx = int(x[j]) * n +  j #Double check this to 
            y = y-allparams[idx]     #make sure it does what you want
            chi2 = chi2 + (ydata[i]-y)**2
    return chi2

我使用以下方法调用它：

x0 = -0.6 * np.ones((4,7), dtype=np.double)
[xopt, fopt, iter, funcalls, warnflag] = \
    fmin(residue_alternative, x0, args=(x0.shape, numseq, prob),
       maxiter = 100000,
       maxfun  = 100000,
       full_output=True,
       disp=True)

并收到以下结果：

Optimization terminated successfully.
         Current function value: 7.750523
         Iterations: 21570
         Function evaluations: 26076

>>>xopt
array([ 0.57669042, -0.21965861,  0.2635061 , -0.08284016, -0.0779489 ,
   -0.10358114,  0.14041582,  0.72469391, -0.43190214,  0.31269757,
   -0.0338726 , -0.14919739, -2.58314651,  2.74251214,  0.57695759,
   -0.49574628,  0.1490926 ,  0.04912353,  0.02420988,  1.17924051,
   -7.2147027 ,  0.57860843, -0.28386938,  0.2431877 , -0.22674694,
   -0.58308225, -6.05706775, -2.06350063])

您可以将其重塑为 4x7 阵列。试一试，让我知道它是否有效/有帮助。

于 2013-08-28T14:46:20.983 回答