使用data.table创建一列回归系数 [英] Using data.table to create a column of regression coefficients

问题描述

我正在努力解决它应该是我问过的上一个问题的简单扩展。

I'm struggling with what seems like it should be a simple extension of a previous question I'd asked here.

我正在尝试在（a）个日期范围内进行汇总（b）因素变量。示例数据可能是：

I'm trying to aggregate over (a) a range of dates and (b) a factor variable. Sample data might be:

Brand    Day     Rev     RVP              
  A      1        2535.00  195.00 
  B      1        1785.45  43.55 
  C      1        1730.87  32.66 
  A      2        920.00   230.00
  B      2        248.22   48.99 
  C      3        16466.00 189.00      
  A      1        2535.00  195.00 
  B      3        1785.45  43.55 
  C      3        1730.87  32.66 
  A      4        920.00   230.00
  B      5        248.22   48.99 
  C      4        16466.00 189.00

借助有用的建议，我已经弄清楚了如何使用data.table查找几天内品牌的平均收入：

Thanks to helpful advice, I've figured out how to find the mean revenue for brands over a period of days using data.table:

new_df<-df[,(mean(Rev)), by=list(Brand,Day)]

现在，我想创建一个新表，其中有一个列，列出了每个品牌的按每日Rev进行的OLS回归得出的系数估算值。我试图这样做，如下所示：

Now, I'd like to create a new table where there is a column listing the coefficient estimate from an OLS regression of Rev by Day for each brand. I tried to do this as follows:

new_df2<-df[,(lm(Rev~Day)), by=list(Brand)]

这似乎不太正确。有什么想法吗？我确信这是我想念的东西。

That doesn't seem quite right. Thoughts? I'm sure it's something obvious I've missed.

推荐答案

我认为这就是您想要的：

I think this is what you want:

new_df2<-df[,(lm(Rev~Day)$coefficients[["Day"]]), by=list(Brand)]

lm 返回完整的模型对象，您需要向下钻取它，以便从每个组中获取单个值，然后将其转换为一列。

lm returns a full model object, you need to drill down into it to get a single value from each group that can be turned into a column.

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