很沒有面子吧?
是的!
自詡做時序算法的我,竟然對“九轉(zhuǎn)”思想毫無所知,在太多次的聽聞之后、在網(wǎng)友的提問之下,急忙補(bǔ)課、列流程、做代碼,整整一天的時間,有了一個可入門檻的算法,如若不棄,還請笑納。
不要太驚喜喲~
思路步驟:
算法建立:
KX:=
TOTALBARSCOUNT-CURRBARSCOUNT+1;{KX K線的序號}
C4:=IF(MOD(KX,4)=1,REFX(C,3),IF(MOD(KX,4)=2,REFX(C,2),IF(MOD(KX,4)=3,REFX(C,1),IF(MOD(KX,4)=0,C,DRAWNULL))));{C4 收盤價的模4計算}
MW:=MOD(KX,4)=0;{MW 模位}
KH4:=IF(MW,HHV(H,4),DRAWNULL);{KH4 K線的最高價 模4(下同)}
KO4:=IF(MW,REF(O,3),DRAWNULL);
KL4:=IF(MW,LLV(L,4),DRAWNULL);
KC4:=IF(MW,C,DRAWNULL);
DRAWKLINE(KH4,KO4,KL4,KC4);
3.向上方向的排序
QD:=CONST(LLVBARS(L,40));{QD 起點}
ZXH:= QD-CURRBARSCOUNT;{ZXH 漲序號}
KC4Z0:=IF(CURRBARSCOUNT<=QD AND QD>=9*4,CONST(REF(KC4,9*4)),DRAWNULL);
KC4Z1:=IF(CURRBARSCOUNT<=QD AND QD>=8*4,CONST(REF(KC4,8*4)),DRAWNULL);
KC4Z2:=IF(CURRBARSCOUNT<=QD AND QD>=7*4,CONST(REF(KC4,7*4)),DRAWNULL);
KC4Z3:=IF(CURRBARSCOUNT<=QD AND QD>=6*4,CONST(REF(KC4,6*4)),DRAWNULL);
KC4Z4:=IF(CURRBARSCOUNT<=QD AND QD>=5*4,CONST(REF(KC4,5*4)),DRAWNULL);
KC4Z5:=IF(CURRBARSCOUNT<=QD AND QD>=4*4,CONST(REF(KC4,4*4)),DRAWNULL);
KC4Z6:=IF(CURRBARSCOUNT<=QD AND QD>=3*4,CONST(REF(KC4,3*4)),DRAWNULL);
KC4Z7:=IF(CURRBARSCOUNT<=QD AND QD>=2*4,CONST(REF(KC4,2*4)),DRAWNULL);
KC4Z8:=IF(CURRBARSCOUNT<=QD AND QD>=1*4,CONST(REF(KC4,1*4)),DRAWNULL);
KC4Z9:=IF(CURRBARSCOUNT<=QD AND QD>=0*4,CONST(REF(KC4,0*4)),DRAWNULL);
4.排除法邏輯條件
ZS:=IF(KC4Z9<KC4Z8 OR KC4Z8<KC4Z7 OR KC4Z7<KC4Z6 OR KC4Z6<KC4Z5 OR KC4Z5<KC4Z4 OR KC4Z4<KC4Z3 OR KC4Z3<KC4Z2 OR KC4Z2<KC4Z1 OR KC4Z1<KC4Z0,0,1);
5.若真,開始數(shù)9
ZXH0:=IF(MOD (INTPART(ZXH/4) ,9)=0,9, MOD (INTPART(ZXH/4) ,9));
DRAWNUMBER(ZS=1,KC4, ZXH0);
代碼流程:
代碼流程
效果示意:
效果示意
寫在結(jié)尾:江豐魚的文章在今日頭條首發(fā),文中代碼均已運(yùn)行通過。