Optimize.fmin在行为良好的连续函数

Question

我正试图找到以下函数的最小值：

这是电话：

>>> optimize.fmin(residualLambdaMinimize, 0.01, args=(u, returnsMax, Param, residualLambdaExtended),
                               disp=False, full_output=True, xtol=0.00001, ftol = 0.0001)
Out[19]: (array([ 0.0104]), 0.49331109755304359, 10, 23, 0)
>>> residualLambdaMinimize(0.015, u, returnsMax, Param, residualLambdaExtended)
Out[22]: 0.46358005517761958
>>> residualLambdaMinimize(0.016, u, returnsMax, Param, residualLambdaExtended)
Out[23]: 0.42610470795409616

正如您所看到的，直接邻域中存在产生较小值的点。为什么我的解算器不考虑它们？

Answer 1

这是一个可以帮助您调试情况的建议。 如果你添加类似的东西 data.append((x, result))到residualLambdaMinimize，您可以收集optimize.fmin正在评估的所有点residualLambdaMinimize：

data = []
def residualLambdaMinimize(x, u, returnsMax, Param, residualLambdaExtended):
    result = ...
    data.append((x, result))
    return result

然后，如果您发布fmin，我们可能会更好地了解data正在做什么（并且可能会重现问题），而我们无需确切知道residualLambdaMinimize的定义方式。< / p>

此外，您可以想象出＆＃34;路径＆＃34; fmin正在尝试找到最低要求：

import numpy as np
import scipy.optimize as optimize
import matplotlib.pyplot as plt

data = []

def residualLambdaMinimize(x, u, returnsMax, Param, residualLambdaExtended):
    result = (x-0.025)**2
    data.append((x, result))
    return result

u, returnsMax, Param, residualLambdaExtended = range(4)
retval = optimize.fmin(
    residualLambdaMinimize, 0.01, 
    args=(u, returnsMax, Param, residualLambdaExtended),
    disp=False, full_output=True, xtol=0.00001, ftol = 0.0001)

data = np.squeeze(data)
x, y = data.T
plt.plot(x, y)
plt.show()