松脚本-如果条件为true，则在前6个蜡烛中绘制2次三角形

Question

如果在前面的6个烛台中，至少有2次发生交叉移动，我正在尝试绘制一个三角形。我有以下脚本，但我不知道如何编写if语句。非常感谢您的任何建议！

//@version=4
study(title="Crossover", overlay=true,resolution="")

first = ema(close, 5)
seconds = ema(close, 13)
third = sma(close,21)
fourth = sma(close,34)
fifth = sma(close, 55)
sixth = sma(close,89)


plot(first, title="EMA 5", color=color.red, linewidth=1, transp=0)
plot(seconds, title="Ema 13", color=color.aqua, linewidth=1, transp=0)
plot(third, title="SMA 21", color=color.orange, linewidth=2, transp=0)
plot(fourth, title="SMA 34", color=color.blue, linewidth=2, transp=0)
plot(fifth, title="SMA 55", color=color.black, linewidth=2, transp=0)
plot(sixth, title="SMA 89", color=color.purple, linewidth=2, transp=0)


long1 = (first > seconds) and crossover(first,third)
long2 = (first > third) and crossover(first, seconds)
long3 = (first > fourth) and (first > seconds) and (first > third) and cross(first,fourth)


// if long1 or long2 or long3 is true 2 times in the previous 6 six candles then plot it ("long")
xxxxxx


plotshape(series=long, title="L", style=shape.triangleup, location=location.belowbar, color=color.green, text="L", size=size.small)

Answer 1

有关示例，请参见How can I count the occurrences of a condition in the last x bars?。

Answer 2

使用barssince函数：

LENGTH = 6

count1 = barssince(long1 or long2 or long3)
count2 = count1[count1 + 1]

plotshape(count1 + count2 < LENGTH - 1, title="L", style=shape.triangleup, location=location.belowbar, color=color.green, text="L", size=size.small)

更新

实际上，我的解决方案有点过分设计了，我建议使用sum，如Bjorn Mistiaen所建议的：

LENGTH = 6    

count = sum(long1 or long2 or long3 ? 1 : 0, LENGTH)

plotshape(count >= 2, title="L", style=shape.triangleup, location=location.belowbar, color=color.green, text="L", size=size.small)