Question

对于需要很长时间（几十分钟，几小时，几天或几个月）的优化，在程序出现故障或电源故障的情况下，必须能够使用优化历史记录再次启动优化，但是fmin算法（和其他几个）不提供历史作为输出或输入。如何保存和使用fmin优化历史记录的结果，以确保您的所有计算投资都不会丢失？

我昨天早上有这个问题，无法在网上找到答案，所以我把自己的解决方案放在一起。见下文。

Answer 1

基本上，实时监控，历史记录和恢复fmin的答案导致使用可调用函数将函数输入和输出存储到查找表中。以下是它的完成方式：

import numpy as np
import scipy as sp
import scipy.optimize

要存储历史记录，请为输入创建全局历史记录向量，并为给定输入的目标函数值创建全局历史记录向量。我也在这里初始化了初始输入向量：

x0 = np.array([1.05,0.95])
x_history = np.array([[1e8,1e8]])
fx_history = np.array([[1e8]])

我在这里优化Rosenbrock功能，因为它是典型的优化算法：

def rosen(x):
    """The Rosenbrock function"""
    return sum(100.0*(x[1:]-x[:-1]**2.0)**2.0 + (1-x[:-1])**2.0)

为要优化的函数创建包装器，并在优化函数请求新输入并计算目标函数值时写入全局变量。对于每次迭代，搜索历史记录以查找先前是否已计算所请求矢量的目标函数值，并使用先前计算的值（如果已经过）。要模拟＆＃34;电源故障时，我创建了一个名为powerFailure的变量，在完成收敛之前结束优化。然后我关闭powerFailure以查看优化完成情况。

def f(x):
    global firstPoint, iteration, x_history, fx_history

    iteration = iteration + 1
    powerFailure = True
    failedIteration = 10
    previousPoint = False
    eps = 1e-12

    if powerFailure == True:
        if iteration == failedIteration:
            raise Exception("Optimization Ended Early Due to Power Failure")

    for i in range(len(x_history)):
        if abs(x_history[i,0]-x[0])<eps and abs(x_history[i,1]-x[1])<eps:
            previousPoint = True
            firstPoint=False
            fx = fx_history[i,0]
            print "%d: f(%f,%f)=%f (using history)" % (iteration,x[0],x[1],fx)

    if previousPoint == False:
        fx = rosen(x)
        print "%d: f(%f,%f)=%f" % (iteration,x[0],x[1],fx)

    if firstPoint == True:
        x_history = np.atleast_2d([x])
        fx_history = np.atleast_2d([fx])
        firstPoint = False
    else:
        x_history = np.concatenate((x_history,np.atleast_2d(x)),axis=0)  
        fx_history = np.concatenate((fx_history,np.atleast_2d(fx)),axis=0)

    return fx

最后，我们运行优化。

firstPoint = True
iteration = 0

xopt, fopt, iter, funcalls, warnflag, allvecs = sp.optimize.fmin(f,x0,full_output=True,xtol=0.9,retall=True)

电源故障时，优化在迭代9结束。＆＃34;重新打开电源＆＃34;功能打印

1: f(1.050000,0.950000)=2.328125 (using history)
2: f(1.102500,0.950000)=7.059863 (using history)
3: f(1.050000,0.997500)=1.105000 (using history)
4: f(0.997500,0.997500)=0.000628 (using history)
5: f(0.945000,1.021250)=1.647190 (using history)
6: f(0.997500,1.045000)=0.249944 (using history)
7: f(0.945000,1.045000)=2.312665 (using history)
8: f(1.023750,1.009375)=0.150248 (using history)
9: f(1.023750,0.961875)=0.743420 (using history)
10: f(1.004063,1.024219)=0.025864
11: f(0.977813,1.012344)=0.316634
12: f(1.012266,1.010117)=0.021363
13: f(1.005703,0.983398)=0.078659
14: f(1.004473,1.014014)=0.002569
15: f(0.989707,1.001396)=0.047964
16: f(1.006626,1.007937)=0.002916
17: f(0.995347,1.003577)=0.016564
18: f(1.003806,1.006847)=0.000075
19: f(0.996833,0.990333)=0.001128
20: f(0.998743,0.996253)=0.000154
21: f(1.005049,1.005600)=0.002072
22: f(0.999387,0.999525)=0.000057
Optimization terminated successfully.
         Current function value: 0.000057
         Iterations: 11
         Function evaluations: 22

如何停止scipy.optimize.fmin函数并使用优化历史记录再次启动它？

1 个答案: