public int FindStateIndex(int att, int phase)
{
for (int i = 0; i < StateList.Count(); i++)
{
if (StateList[i].Att == att && StateList[i].Phase == phase)
{
return(i);
}
}
return(-1);
}
public void FillSimulatedData()
{
Random ran = new Random();
for (int i = 0; i < StateList.Count(); i++)
{
for (int j = 0; j < FreqList.Count(); j++)
{
Data[i, j] = Math.Cos(j * Math.PI * 2 / 1000 + i) + ran.NextDouble();
//System.Threading.Thread.Sleep(2);
}
}
}
/// <summary>预测状态概率</summary>
public void PredictProb()
{
PredictValue = new double[Count];
PredictValue1 = new Dictionary <int, double>(Count);
//这里很关键，权重和滞时的关系要颠倒，循环计算的时候要注意
//另外，要根据最近几期的出现数，确定概率的状态，必须取出最后几期的数据
//1.先取最后K期数据
var last = StateList.GetRange(StateList.Count - LagPeriod, LagPeriod);
//2.注意last数据是升序,最后一位对于的滞时期 是k =1
for (int i = 0; i < Count; i++)
{
for (int j = 0; j < LagPeriod; j++)
{
//滞时期j的数据状态
var state = last[last.Count - 1 - j] - 1;
PredictValue[i] += Wk[j] * ProbMatrix[j][state, i];
if (double.IsNaN(PredictValue[i]))
{
try
{
var numberCount = StateList.Count(p => p == i + 1);
PredictValue[i] = (double)numberCount / StateList.Count;
}
catch (Exception e)
{
var random = new Random();
PredictValue[i] = (double)random.Next(0, 100) / 100;
}
}
}
PredictValue1.Add(i + 1, PredictValue[i]);
}
}
public void ResetTestResult()
{
Data = new double[StateList.Count(), FreqList.Count()];
}
