虧三次加倉,盈三次減倉軟件難實現嗎?
作者：金字塔 來源：cxh99.com 發布時間：2017年01月29日
- 咨詢內容：
下面是突破20日高低點策略,想實現連續盈三次后減倉,直到連續虧三次后恢復初始倉位,老師修改二次也不行,老樣回測倉位錯亂,軟件難實現還是寫法有錯?input :k(20,5,30,1) input:avglen(30,10,100,5);input : atrlen(20,5,30,1) ;input : ss(2,1,1000,1) ;
//聲明變量nt := 1 ; //調試信息帶時間戳buyorderthisbar := 0 ; //當前bar有過交易variable : _debug = 1 ; //是否輸出前臺交易指令variable : turtleunits=0 ; //交易單位variable : myentryprice1=0; variable : myentryprice2=0; variable : myexitprice=0;variable : n=0;variable : m=0;variable : yk=0; //準備需要計算的變量zhh:ref(high,k);zll:ref(low,k);kqls:=enterbars+1;khh:=hhv(high,kqls);kll:=llv(low,kqls);
avgtr :=ref(ma(tr,atrlen),1);
//開始執行時 初始化數據if barpos=1 then begin position :=0 ; posnum:=ss; end
//如果當前是沒有持倉的狀態if position=0 and barpos>20 and h>l then begin //建立多頭進場條件 long:=high>=zhh ; //多頭進場 if long and holding=0 then begin myentryprice1 :=if(open>zhh+0.2 ,open+0.6 ,zhh+0.6); buy( _debug,posnum,limitr,myentryprice1); jqsh:=1; position := 1 ; turtleunits := 1 ; // n:= avgtr ; buyorderthisbar := 1; end //建立空頭進場條件 short:=low<=zll ; //空頭進場 if short and position=0 and holding=0 then begin myentryprice2:= if(open<zll-0.2 ,open-0.6 ,zll-0.6) ; buyshort( _debug,posnum,limitr,myentryprice2); jqsh:=1; position := -1 ; turtleunits := 1 ; //n := avgtr ; buyorderthisbar := 1;
end end
//如果當前持有多頭倉位的狀態
if position=1 and barpos>20 and h>l then begin
//多頭加倉條件 {while (close>myentryprice+0.5*n) and turtleunits<4 and jqsh=1 do begin myentryprice := if(open>myentryprice+0.5*n ,open+0.6 ,myentryprice+0.5*n+0.6 ) ; myentryprice:= ceiling(myentryprice/mindiff)*mindiff ; buy( _debug, posnum, limitr,myentryprice ); turtleunits := turtleunits+1 ; buyorderthisbar := 1; end } //建立多頭離場條件 longx1 := low<zll-mindiff; myexitprice:=zll-mindiff; if longx1 and buyorderthisbar=0 and holding>0 then begin myexitprice := if(open<myexitprice-0.2 ,open-0.6 ,myexitprice-0.6) ; sell( _debug ,0,limitr,myexitprice); position := 0 ; turtleunits := 0 ; yk:=myexitprice- myentryprice2; if yk>0 then begin n:=0; m:=m+1; end if yk<0 then begin n:=n+1; m:=0; end // if n>=3 then posnum:=ss; end // if m>=3 then posnum:=0.5*ss; //建立多頭止損條件 longx2 := low<(khh-2*n); if longx2 and position=1 and buyorderthisbar=0 and holding>0 then begin myexitprice := if(open<khh-2*n ,open-0.6 ,khh-2*n-0.6) ; myexitprice := floor(myexitprice/mindiff)*mindiff ; sell( _debug ,0,limitr,myexitprice); position := 0 ; turtleunits := 0 ; yk:=myexitprice- myentryprice2; if yk>0 then begin n:=0; m:=m+1; end if yk<0 then begin n:=n+1; m:=0; end end end
//如果當前持有空頭倉位的狀態
if position = -1 and barpos>20 and h>l then begin
//空頭加倉條件 {while (close<myentryprice-0.5*n) and turtleunits<4 and jqsh=1 do begin myentryprice := if(open<myentryprice-0.5*n ,open-0.6 ,myentryprice-0.5*n-0.6 ) ; kkj := floor(myentryprice/mindiff)*mindiff ; buyshort( _debug,posnum, limitr,myentryprice ); turtleunits := turtleunits+1 ; buyorderthisbar := 1; end }
//建立空頭離場條件 shortx1 := high>zhh+mindiff; myexitprice:=zhh+mindiff; if shortx1 and buyorderthisbar=0 and holding<0 then begin myexitprice := if(open>myexitprice,open+0.6, myexitprice+0.6); sellshort( _debug,0,limitr,myexitprice); position := 0 ; turtleunits := 0 ; yk:=myentryprice2-myexitprice; if yk>0 then begin n:=0; m:=m+1; end if yk<0 then begin n:=n+1; m:=0; end //if n>=3 then posnum:=ss; //if m>=3 then posnum:=0.5*ss; end
//建立空頭止損條件 shortx2 := high>kll+2*n ; if shortx2 and position = -1 and buyorderthisbar=0 and holding<0 then begin myexitprice:= if(open>kll+2*n ,open+0.6 ,kll+2*n+0.6) ; myexitprice := ceiling(myexitprice/mindiff)*mindiff ; sellshort( _debug,0,limitr,myexitprice); position := 0 ; turtleunits := 0 ; yk:=myentryprice2-myexitprice; if yk>0 then begin n:=0; m:=m+1; end if yk<0 then begin n:=n+1; m:=0; end // if n>=3 then posnum:=ss; //if m>=3 then posnum:=0.5*ss; end end if n>=3 and ref(n,1)=2 then posnum:=ss;if m>=3 and ref(m,1)=2 then posnum:=0.5*ss;
- 金字塔客服：
不要重復發帖，在之前那個帖子里面進行處理
打印本文
關閉窗口