scipy 最小化约束 [英] scipy minimize with constraints

问题描述

我知道这个问题应该在scipy.optimize的手册中处理，但我对它的理解不够好.也许你能帮上忙

I know that this question should be handled in the manual of scipy.optimize, but I don't understand it well enough. Maybe you can help

我有一个函数(这只是一个例子，不是真正的函数，但我需要在这个层面上理解它):

I have a function (this is just an example, not the real function, but I need to understand it at this level):

编辑(更好的例子):

假设我有一个矩阵

arr = array([[0.8, 0.2],[-0.1, 0.14]])

带有目标函数

def matr_t(t):
    return array([[t[0], 0],[t[2]+complex(0,1)*t[3], t[1]]]

def target(t):
    arr2 = matr_t(t)
    ret = 0
    for i, v1 in enumerate(arr):
          for j, v2 in enumerate(v1):
               ret += abs(arr[i][j]-arr2[i][j])**2
    return ret

现在我想在 t[i] 是实数的假设下最小化这个目标函数，比如 t[0]+t[1]=1

now I want to minimize this target function under the assumption that the t[i] are real numbers, and something like t[0]+t[1]=1

推荐答案

此约束

t[0] + t[1] = 1

将是一个等式 (type='eq') 约束，您可以在其中创建一个必须等​​于零的函数:

would be an equality (type='eq') constraint, where you make a function that must equal zero:

def con(t):
    return t[0] + t[1] - 1

然后你创建一个 dict 你的约束(如果多个 dicts 列表):

Then you make a dict of your constraint (list of dicts if more than one):

cons = {'type':'eq', 'fun': con}

我从未尝试过，但我相信要保持 t 真实，您可以使用:

I've never tried it, but I believe that to keep t real, you could use:

con_real(t):
    return np.sum(np.iscomplex(t))

并使您的 cons 包含两个约束:

And make your cons include both constraints:

cons = [{'type':'eq', 'fun': con},
        {'type':'eq', 'fun': con_real}]

然后将 cons 输入 minimize 为:

Then you feed cons into minimize as:

scipy.optimize.minimize(func, x0, constraints=cons)

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