使用Pandas .diff（）从时间序列计算二阶导数 [英] Calculating second order derivative from timeseries using Pandas .diff()

问题描述

下面是我的Pandas数据框的示例，以日期作为索引：

An example of my Pandas dataframe is below, with dates as index:

            A
2019-01-09  1.007042
2019-01-10  1.007465
2019-01-11  1.007054
2019-01-12  1.007154
2019-01-13  1.007854
2019-01-14  1.008560

我想每两列A列确定二阶导数。是否适合两次使用Pandas中的 .diff（）函数进行此确定？

I'd like to determine the second order derivative every 2 rows of column A. Would it be appropriate to use the .diff() function in Pandas twice for this determination?

df['A2'] = df['A'].diff(2).diff(2)

推荐答案

这是一个长达9个月的问题，但仍显示在google搜索此材料的顶部，因此值得接受。如果不需要此免责声明，我希望有人为我删除它。

This is a nine month old question, but still appears at the top of google searches for this material and so warrants an acceptable answer. If this disclaimer is unnecessary, I hope someone will remove it for me.

二阶导数可以作为中心，前向或后向导数进行计算，但要基于您的例如，我认为您正在寻找向后导数。可以根据需要在每一行上进行计算，但是，使用 diff（）确实很难。函数 shift（）可以很好地工作，其方法如下：

The second derivative can be calculated either as a central, forward or backward derivative, but based off your example, I think you're looking for the backward derivative. It can be calculated on every row if you want, however, it could be really hard to do with diff(). The function shift() works well though and the method is as follows:

df['A2'] = df['A'] - 2*df['A'].shift(1) + df['A'].shift(2)

该技术依赖于有限差异

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