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def CROSS(S1, S2): # 判断向上金叉穿越 CROSS(MA(C,5),MA(C,10)) 判断向下死叉穿越 CROSS(MA(C,10),MA(C,5))
return np.concatenate(([False], np.logical_not((S1>S2)[:-1]) & (S1>S2)[1:])) # 不使用0级函数,移植方便 by jqz1226
改成:
def CROSS(S1, S2): # 判断向上金叉穿越 CROSS(MA(C,5),MA(C,10)) 判断向下死叉穿越 CROSS(MA(C,10),MA(C,5))
return np.concatenate(([False], np.logical_not((S1 > S2)[:-1]) & (S1 > S2)[1:].reset_index(drop=True)))
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