Question

Williams Fractal通过发现5巴的高/低，滞后2巴来绘制分形。

我想建立一个指标，以找到3个柱线的低/高。

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

//@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：为什么默认值2？这是指滞后吗？另外，我认为设置input允许用户在TradingView上操作此数字以了解指标的偏好（例如设置MA的回溯期），但是我看不到内置的Williams Fractal的输入。

然后我们获得以下参数：

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]))

我能够确定high和low是当前的高价和低价，并且括号中的值可以访问以前的高/低值。但是，high[n+3]（举个例子）是否意味着高价3支蜡烛，而high[n-1]意味着高1支蜡烛？

无论如何，我都很难遵循代码的逻辑。我得到upFractal是将在屏幕上绘制的变量，但让我尝试将以下行输入单词：

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

这就是我的意思：

“如果前两根柱的高点小于当前柱的当前高点，并且前一根柱的高点小于当前柱的高点，则绘制upFractal后面的一根柱的高点小于当前柱的高点，后面两个柱的柱的高点小于当前柱的高点。”

我说对了吗？我可以简单地遵循这种逻辑来制作3条线吗？

谢谢。

Answer 1

在Pine中使用在时间序列上引用[]运算符的历史记录时，行为与您的理解相反；过去的索引更大，零（或没有运算符的变量名）引用当前柱。

只要您还同时阅读文档，那么从一开始就开始玩代码无疑是学习的一种好方法，而Stack Overflow并不能替代这些文档。开始学习松树的两个好地方是： https://www.tradingview.com/pine-script-docs/en/v4/Quickstart_guide.html 和 http://www.pinecoders.com/learning_pine_roadmap/

如何在Pine Script中编辑Williams Fractal指标以显示3条形分形？

1 个答案: