如何在Pine脚本中编辑Williams fractal indicator以显示3-bar分形？

Question

Williams Fractal通过找到5个条形的低/高，滞后2个条形来绘制分形。

我想设置一个指标，找出3根柱子的最低点/最高点。

我已经检查了TradingView的内置Williams Fractal，它显示了以下代码：

//@version=4
study("Williams Fractal", shorttitle="Fractal", format=format.price, precision=0, overlay=true)
// Define "n" as the number of periods and keep a minimum value of 2 for error handling.
n = input(title="Periods", defval=2, minval=2, type=input.integer)

我的第一个问题是关于defval的:为什么默认值是2？这是指滞后吗？此外，我认为设置input允许用户在TradingView上操作这个数字来实现指标偏好(比如设置MA的回看周期)，但我在内置的Williams Fractal上看不到这个输入。

然后我们得到这些参数：

upFractal = (                                                                                                          (high[n+2]  < high[n]) and (high[n+1]  < high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n]))
         or (                                                                               (high[n+3]  < high[n]) and (high[n+2]  < high[n]) and (high[n+1] == high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n]))
         or (                                                    (high[n+4]  < high[n]) and (high[n+3]  < high[n]) and (high[n+2] == high[n]) and (high[n+1] <= high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n]))
         or (                          (high[n+5] < high[n]) and (high[n+4]  < high[n]) and (high[n+3] == high[n]) and (high[n+2] == high[n]) and (high[n+1] <= high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n]))
         or ((high[n+6] < high[n]) and (high[n+5] < high[n]) and (high[n+4] == high[n]) and (high[n+3] <= high[n]) and (high[n+2] == high[n]) and (high[n+1] <= high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n]))

dnFractal = (                                                                                                  (low[n+2]  > low[n]) and (low[n+1]  > low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n]))
         or (                                                                         (low[n+3]  > low[n]) and (low[n+2]  > low[n]) and (low[n+1] == low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n]))
         or (                                                (low[n+4]  > low[n]) and (low[n+3]  > low[n]) and (low[n+2] == low[n]) and (low[n+1] >= low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n]))
         or (                        (low[n+5] > low[n]) and (low[n+4]  > low[n]) and (low[n+3] == low[n]) and (low[n+2] == low[n]) and (low[n+1] >= low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n]))
         or ((low[n+6] > low[n]) and (low[n+5] > low[n]) and (low[n+4] == low[n]) and (low[n+3] >= low[n]) and (low[n+2] == low[n]) and (low[n+1] >= low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n]))

我能够计算出high和low是当前的最高和最低价格，括号中的值提供了以前的最高/最低值。但是，high[n+3] (举个例子)是指高价的3支蜡烛，而high[n-1]是指高1支的回购吗？

无论如何，我都很难理解代码的逻辑。我得到的upFractal是将在屏幕上绘制的变量，但让我试着将以下行放入文字中：

upFractal = (                                                                                                          (high[n+2]  < high[n]) and (high[n+1]  < high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n]))

我想这就是它的意思：

“如果前面两个条形的高度小于当前条形的高度，并且前面一个条形的高度小于当前条形的高度，并且后面一个条形的高度小于当前条形的高度，并且后面两个条形的高度小于当前条形的高度，则绘制upFractal。”

我没搞错吧？我可以简单地遵循这个逻辑来设计一个3栏模式吗？

谢谢。

Answer 1

在Pine中的时间序列上使用历史引用[]运算符时，行为与您的理解相反；更大的指数在过去更远，零(或不带运算符的变量名)引用当前条形图。