我想在我正在使用的算法中添加参数选择的交叉熵方法。问题是我不太懂数学表示法,也找不到用代码写成的交叉熵方法的这个版本。
在这张图像中,可以看到伪代码中的算法:
http://i.imgur.com/PXbFOhq.png (我不能把它粘贴在这里,因为它有很多胶乳)
摘自本文:thesis.pdf (第69页)
你能帮我把它翻译成c#或其他语言,甚至英语吗?
谢谢!
发布于 2014-03-17 19:03:11
在罗伯特·多迪尔( Robert Dodier )澄清后,这在某种程度上帮助了我,但在另一些方面却让我更加困惑。之后,我回到了我所见过的交叉熵法的红宝石代码中,但我认为它与我试图“翻译”的算法不完全一样。随着新发现的知识(从澄清),我看到它确实是相同的算法,并将它转化为c#。
原始红宝石代码:entropy.html
我把它翻译成c#:
class CrossEntropyMethod
{
Random r = new Random();
double objective_function(double[] vector)
{
double sum=0f;
foreach (var f in vector)
{
sum+=(double)Math.Pow(f,2);
}
return -sum;
}
double QuadraticEquation(double[] vector)
{
// 5X^2 + 10X - 2 = 0 -> X=-2.183216 || X=0.183216
double sum = 5 * Math.Pow(vector[0],2) + 10 * vector[0] - 2;
return - Math.Abs(sum);
}
double QuadraticEquation2(double[] vector)
{
// 5X^2 + 10X - 2 = 0 -> X=-2.183216 || X=0.183216
double sum1 = vector[0] * Math.Pow(0.183216, 2) + vector[1] * 0.183216 + vector[2];
double sum2 = vector[0] * Math.Pow(-2.183216, 2) + vector[1] * -2.183216 + vector[2];
return - (Math.Abs(sum1) + Math.Abs(sum2));
}
double random_variable(double min, double max)
{
return min + ((max - min) * r.NextDouble());
}
double random_gaussian(double mean=0.0, double stdev=1.0)
{
double u1, u2, w;
u1 = u2 = w = 0;
do{
u1 = 2 * r.NextDouble() - 1;
u2 = 2 * r.NextDouble() - 1;
w = u1 * u1 + u2 * u2;
} while (w >= 1);
w = Math.Sqrt((-2.0 * Math.Log(w)) / w);
return mean + (u2 * w) * stdev;
}
double[] generate_sample(double[][] search_space, double[] means, double[] stdevs)
{
double[] vector = new double[search_space.Length];
for (int i=0; i<vector.Length; i++)
{
vector[i] = random_gaussian(means[i], stdevs[i]);
vector[i] = Math.Max(vector[i] ,search_space[i][0]);
vector[i] = Math.Min(vector[i], search_space[i][1]);
}
return vector;
}
void update_distribution(double[][] samples, double alpha, ref double[] means, ref double[] stdevs)
{
for (int i=0; i< means.Length; i++)
{
double[] tArray = new double[samples.Length];
for (int z = 0; z < samples.Length; z++)
{
tArray[z] = samples[z][i];
}
means[i] = alpha * means[i] + ((1.0 - alpha) * tArray.Average());
stdevs[i] = alpha * stdevs[i] + ((1.0 - alpha) * MyExtensions.StandardDeviation(tArray));
}
}
double[] search(double[][] bounds, int max_iter, int num_samples, int num_update, double learning_rate)
{
double[] means = new double[bounds.Length];
double[] stdevs = new double[bounds.Length];
for (int i=0; i< means.Count(); i++)
{
means[i]=random_variable(bounds[i][0], bounds[i][1]);
stdevs[i]=bounds[i][1]-bounds[i][0];
}
double[] best=null;
double bestScore=double.MinValue;
for (int t=0; t<max_iter; t++)
{
double[][] samples= new double[num_samples][];
double[] scores=new double[num_samples];
for (int s=0; s<num_samples; s++)
{
samples[s]=generate_sample(bounds, means, stdevs);
scores[s]=QuadraticEquation(samples[s]);
}
Array.Sort(scores,samples);
Array.Reverse(scores);
Array.Reverse(samples);
if (best==null || scores.First() > bestScore)
{
bestScore=scores.First();
best=samples.First();
}
double[][] selected = new double[num_update][];
Array.Copy(samples,selected,num_update);
update_distribution(selected, learning_rate, ref means, ref stdevs);
Console.WriteLine("iteration={0}, fitness={1}", t, bestScore);
}
return best;
}
public void Run()
{
double[][] parameters = new double[][] { new double[] { -500, 500 }}; //QuadraticEquation parameters
//double[][] parameters = new double[][] { new double[] { 4, 6 }, new double[] { 9, 11 }, new double[] { -3, -1} }; //QuadraticEquation2 parameters
//double[][] parameters = new double[][] { new double[] { -5, 5 }, new double[] { -5, 5 }, new double[] { -5, 5 } }; //object_function parameters
int maxIter = 100;
int nSamples = 50;
int nUpdate = 5;
double alpha = 1;
double[] best = search(parameters, maxIter, nSamples, nUpdate, alpha);
string str = string.Join(" | ", best.Select(a => a.ToString("N10")).ToArray());
Console.WriteLine("Best: " + str);
}
}发布于 2014-03-16 20:46:33
嗯,它们的表示法并不是最清晰的,但我将尝试解释那些可能会令人困惑的部分。我想循环和分配对你来说不是什么问题。
希望这足够让你走了。如有任何问题,请随时跟进。
https://stackoverflow.com/questions/22441172
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