在Scipy最小化2次迭代之间执行了多少个功能评估

发布于 2025-02-05 08:06:36 字数 551 浏览 4 评论 0原文

我正在使用Scipy最小化功能，主要是BFGS方法。我需要找到在迭代2之间进行2个功能评估。此功能评估通常旨在计算数值衍生物。

如果可以找到在迭代之间计算了多少梯度评估，那就更好了。

代码的示例：

def callback_params(theta):
    global params
    params = np.vstack((params, theta))

def rosen(X):
    return (1.0 - X[0])**2 + 100.0 * (X[1] - X[0]**2)**2 + \
           (1.0 - X[1])**2 + 100.0 * (X[2] - X[1]**2)**2

init = np.random.rand(3)
params = np.empty([0, 3])

res = minimize(rosen,init, method='BFGS',options = {'disp': True}, callback=callback_params)

我如何知道 params 中2行之间的函数评估数？

原文

I am using scipy minimize function, mostly the BFGS method. I need to find how many function evaluation were executed between 2 following iterations. This function evaluation usually aim to calculate numerical derivatives.

If it is possible to find how many gradient evaluation were calculated between iterations, that would be even better.

Example of code:

def callback_params(theta):
    global params
    params = np.vstack((params, theta))

def rosen(X):
    return (1.0 - X[0])**2 + 100.0 * (X[1] - X[0]**2)**2 + \
           (1.0 - X[1])**2 + 100.0 * (X[2] - X[1]**2)**2

init = np.random.rand(3)
params = np.empty([0, 3])

res = minimize(rosen,init, method='BFGS',options = {'disp': True}, callback=callback_params)

How can I know the number of function evaluation between 2 rows in params?

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又爬满兰若 2025-02-12 08:06:37

功能返回 OptimizerEsult 其中一个成员在哪里

nfev，njev，njev：int
目标函数及其雅各布和黑森州的评估数量。

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遮云壑 2025-02-12 08:06:36

您可以利用scipy.optimize.minimize在每次迭代之后调用传递的回调：

def callback(xk, d):
    d['obj_evals'].append(0)
    d['iters'] += 1

def rosen(X, d):
    d['obj_evals'][d['iters']] += 1    
    return (1.0 - X[0])**2 + 100.0 * (X[1] - X[0]**2)**2 + \
           (1.0 - X[1])**2 + 100.0 * (X[2] - X[1]**2)**2

init = np.random.rand(3)

d = {'obj_evals': [0], 'iters': 0}
obj_fun = lambda x: rosen(x, d)
cb = lambda xk: callback(xk, d)

res = minimize(obj_fun, init, method='BFGS',options = {'disp': True}, callback=cb)

然后，d ['obj_evals']包含一个事实。每次迭代的目标函数评估数量。可以轻松地扩展此想法以计算您将梯度传递给最小化的梯度评估的数量。

You can exploit the fact that scipy.optimize.minimize calls the passed callback after each iteration:

def callback(xk, d):
    d['obj_evals'].append(0)
    d['iters'] += 1

def rosen(X, d):
    d['obj_evals'][d['iters']] += 1    
    return (1.0 - X[0])**2 + 100.0 * (X[1] - X[0]**2)**2 + \
           (1.0 - X[1])**2 + 100.0 * (X[2] - X[1]**2)**2

init = np.random.rand(3)

d = {'obj_evals': [0], 'iters': 0}
obj_fun = lambda x: rosen(x, d)
cb = lambda xk: callback(xk, d)

res = minimize(obj_fun, init, method='BFGS',options = {'disp': True}, callback=cb)

Then, d['obj_evals'] contains the number of objective function evaluations for each iteration. This idea can easily be extended to count the number of gradient evaluations given you pass the gradient to minimize as well.

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