如何使用 CopyRates() 在多个时间范围内搜索和过滤牛市吞没形态

Question

我正在尝试使用 CopyRates() 在几个时间范围内（所有时间范围 H2 到 M10 在 H4 之后的看涨蜡烛内搜索看涨吞没烛台模式（看跌蜡烛，然后是更大的看涨蜡烛）它关闭）。我阅读了CopyRates() 的定义，但我发现实施起来有点挑战性。这里的想法来自我想要过滤具有最大看跌/看涨蜡烛对比率的模式的模式。看看我到目前为止做了什么：

在OnTick()：

for (int i=ArraySize(timeframes); i>=1; i--)   {
    
    if(CopyRates(Symbol(), timeframes[i - 1], 1, MyPeriod, rates)!=MyPeriod) {
        Print("Error CopyRates errcode = ",GetLastError()); 
        return;
    }

    // Using bullish engulfing pattern:
    if ((rates[numCandle].open < rates[numCandle].close) &&
        (rates[numCandle + 1].open > rates[numCandle + 1].close) &&
        (rates[numCandle + 1].open < rates[numCandle].close) &&
        (rates[numCandle + 1].close > rates[numCandle].open)) {

        // Not too certain what should be done here
    }
}

这是其他相关代码：

input int numCandle=0; 
MqlRates rates[];
ENUM_TIMEFRAMES timeframes[7] = {PERIOD_H2, PERIOD_H1, PERIOD_M30, PERIOD_M20, PERIOD_M15, PERIOD_M12, PERIOD_M10};

void OnInit() {
   ArraySetAsSeries(rates, true);
}

---

更新

以下是看涨吞没形态的定义：

如上图所示的看涨吞没形态是看跌蜡烛和看涨蜡烛。看跌蜡烛的开盘价小于看涨蜡烛的收盘价，看跌蜡烛的收盘价大于看涨蜡烛的开盘价。请注意，在某些情况下，看跌蜡烛的收盘价仅比看涨蜡烛的开盘价高一小部分。每根蜡烛的主体尺寸都大于其上下灯芯的总和。

Answer 1

     ENUM_TIMEFRAMES timeframes[7] = {PERIOD_H2, PERIOD_H1, PERIOD_M30, PERIOD_M20, PERIOD_M15, PERIOD_M12, PERIOD_M10};
     //ENUM_TIMEFRAMES timeframes[4] = {PERIOD_H1, PERIOD_M30, PERIOD_M15, PERIOD_M5};
     //---
     const int LONG=1, SHORT=-1, NO_DIR=0;
     const ENUM_TIMEFRAMES timeframeHighest = PERIOD_H4;
     string bestRatioObjectName="bestBullish2BearishPattern!";
     
     datetime lastCandleTime=0;
     
     void OnTick()
     {
        if(!isNewBar(PERIOD_H4))
           return;
        //most likely you will call this block after new bar check?
        MqlRates rates[];
        ArraySetAsSeries(rates,true);
        if(CopyRates(_Symbol,timeframeHighest,0,2,rates)==-1)
        {
           printf("%i %s: failed to load/copy rates on %d. error=%d",__LINE__,__FILE__,PeriodSeconds(timeframeHighest)/60,_LastError);
           return;
        }
        if(getCandleDir(rates[1])!=LONG)
           return;
        const datetime timeStart=rates[1].time, timeEnd=rates[0].time;   //within a bullish H4 candle - DONE
        
        double bestRatio = -1;//once a bearish2bullish ratio is higher, we'll move to new place
        for(int i=ArraySize(timeframes)-1;i>=0;i--)
        {
           if(CopyRates(_Symbol,timeframes[i],timeStart,timeEnd,rates)<0)
           {
              printf("%i %s: failed to copy rates on %d. error=%d",__LINE__,__FILE__,PeriodSeconds(timeframeHighest)/60,_LastError);
              return;
           }
           processRates(rates,bestRatio,bestRatioObjectName);
        }
        printf("%i %s: best=%.5f, objName =%s: %.5f-%.5f",__LINE__,__FILE__,bestRatio,bestRatioObjectName,
         ObjectGetDouble(0,bestRatioObjectName,OBJPROP_PRICE1),ObjectGetDouble(0,bestRatioObjectName,OBJPROP_PRICE2));
        //ExpertRemove();//for scripting, a one time call
     }
     bool isNewBar(const ENUM_TIMEFRAMES tf)
       {
        const datetime time=iTime(_Symbol,tf,0);
        if(time>lastCandleTime)
          {
           lastCandleTime=time;
           return true;
          }
        return false;
       }
     int getCandleDir(const MqlRates& rate) // candle direction: +1 for BULL, -1 for BEAR
       {
        if(rate.close-rate.open>_Point/2.)
           return 1;
        if(rate.open-rate.close>_Point/2.)
           return-1;
        return 0;
       }
     void processRates(const MqlRates& rates[],double &best,const string bestObjName)
     {
        for(int i=ArraySize(rates)-2; i>0; /* no sense to catch last candle - we cant compare it with anybody */ i--)
        {
           if(getCandleDir(rates[i])!=LONG)
              continue;//current - bullish
           if(getCandleDir(rates[i+1])!=SHORT)
              continue;//prev - bearish
           if(rates[i].close-rates[i+1].open>_Point/2.){}
           else continue;
           if(rates[i+1].close-rates[i].open>_Point/2.){}
           else continue;
           const double body=rates[i].close-rates[i].open, twoWicks = rates[i].high-rates[i].low- body;
           if(body<twoWicks)
              continue;   //Each of the candles has a body size bigger than it’s upper and lower wicks combined.
     //---
           const double prevBody = rates[i+1].open - rates[i+1].close;
           const double newRatio = body / prevBody;
           if(newRatio>best) // eventually we'll find best bull2bear ratio
           {
              moveRectangle(rates[i+1],rates[i].time,bestObjName);
              best = newRatio;
           }
        }
     }
     void moveRectangle(const MqlRates& rate,const datetime rectEnd,const string objectName)
     {
        if(ObjectFind(0,objectName)<0)
        {
           if(!ObjectCreate(0,objectName,OBJ_RECTANGLE,0,0,0,0,0))
           {
              printf("%i %s: failed to draw %s. error=%d",__LINE__,__FILE__,objectName,_LastError);
              return;
           }
           //add GUI things like how to display the rectangle
        }
        //moving the rectangle to a new place, even for the first time
        ObjectSetDouble(0,objectName,OBJPROP_PRICE,0,rate.open);
        ObjectSetDouble(0,objectName,OBJPROP_PRICE,1,rate.close);
        ObjectSetInteger(0,objectName,OBJPROP_TIME,0,rate.time);
        ObjectSetInteger(0,objectName,OBJPROP_TIME,1,rectEnd);
     }

Answer 2

假设 MyPeriod 被初始化为 2，其余代码似乎是正确的。您应该创建一个变量来保留具有最大比率的时间范围。在您的if 中，您必须计算candle+1 和candle 的烛台主体大小并计算比率，然后如果计算的比率大于先前计算的比率，则更改值并更新找到它的时间范围.

在for 循环结束时，您可以决定要在哪个时间段下订单。