如何确定股票价格数据的转折点 [英] how to identify turning points in stock price data

问题描述

这个问题是一个延续的这一个。

我的目标是找到转折点的股票价格数据。

My goal is to find the turning points in stock price data.

到目前为止，我：

试图区分平滑价格集，博士的帮助安德鲁·伯内特汤普森使用该中心的五点法，作为解释这里 。

Tried differentiating the smoothed price set, with the help of Dr. Andrew Burnett-Thompson using the centered five-point method, as explained here.

我用刻度数据的EMA20平滑数据集。

I use the EMA20 of tick data for smoothing the data set.

有关图表我得到的一阶导数（DY / DX）上的每个点。 我创建第二个图表转折点。 每次DY / DX之间[-some_small_value]和[+ some_small_value] - 我补充一点，这个图表。

For each point on the chart I get the 1st derivative (dy/dx). I create a second chart for the turning points. Each time the dy/dx is between [-some_small_value] and [+some_small_value] - I add a point to this chart.

的问题是：   我没有得到真正的转折点，我得到的东西接近。   我得到过多或过少点 - depening上[some_small_value]

The problems are: I don't get the real turning points, I get something close. I get too much or too little points - depening on [some_small_value]

我尝试添加一个点的第二个方法时DY / DX由负转正，这也创造了太多了点，也许是因为我用打勾的数据EMA（而不是1分钟收盘价）

I tried a second method of adding a point when dy/dx turns from negative to positive, which also creates too many points, maybe because I use EMA of tick data (and not of 1 minute closing price)

一个第三种方法是将套入的n个点的切片的数据，并找到最小和最大点。这工作正常（不理想），但它的落后。

A third method is to divide the data set into slices of n points, and to find the minimum and maximum points. This works fine (not ideal), but it's lagging.

任何人有一个更好的方法？

Anyone has a better method?

我附上2张图片的输出（一阶导数和n个最小/最大）

I attached 2 pictures of the output (1st derivative and n points min/max)

推荐答案

您可以采取的二阶导数考虑在内，这意味着你应该额外（您的一阶导数）计算（Y_ {I-1} + Y_ {I + 1} - 2y_i）/（DX）2 。如果这是高于某个阈值你有一个最大值，如果是下面你有一个最小和其他可以丢弃它。这应该扔掉了很多分，您继续使用寻找极值的方法（ Y'= 0 ），因为这种情况也适用于鞍点。

You could take the second derivative into account, meaning you should additionally (to your first derivative) evaluate (y_{i-1} + y_{i+1} - 2y_i) / (dx)². If this is above a certain threshold you have a maximum, if it is below you have a minimum and else you can discard it. This should throw out a lot of points that you keep using your method of finding extrema (y' = 0), because this condition is also valid for saddle points.

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