在R中的data.table环境中创建公式 [英] create a formula in a data.table environment in R
问题描述
我想在 data.table
中运行回归。 公式
需要动态构造。我尝试了以下方法:
x = data.table(a = 1:20,b = 20:1,id = 1:5)
> x [,as.list(coef(lm(as.formula(a〜b)))),by = id]
eval(expr,envir,
如何将环境指定为实际数据表的环境? / p>
编辑:我意识到我可以做lm(a〜b)。我需要公式是动态的,所以它被建立为一个字符串。通过动态我的意思是公式可以 paste0(var_1,〜,var_2)
其中 var_1 = a
code> var_2 = b
这里有一个解决方案,我认为我们可以做得更好:
txt = parse(text =as.list(coef(lm(a〜b))))
& x [,eval(txt),by = id]
id(拦截)b
1:1 21 -1
2:2 21 -1
3: 1
4:4 21 -1
5:5 21 -1
lm
可以接受字符串作为公式,结合 .SD
像这样:
> x [,as.list(coef(lm(a〜b,.SD))),by = id]
id(拦截)b
1:1 21 -1
2:2 21 -1
3:3 21 -1
4:4 21 -1
5:5 21 -1
I would like to run a regression within a data.table
. The formula
needs to be constructed dynamically. I have tried the following method:
x = data.table(a=1:20, b=20:1, id=1:5)
> x[,as.list(coef(lm(as.formula("a ~ b")))),by=id]
Error in eval(expr, envir, enclos) : object 'a' not found
How does one specify the environment to be that of the actual data.table where the evaluation occurs?
EDIT: I realize I can do lm(a ~ b). I need the formula to be dynamic so it's built up as a character string. By dynamically I mean the formula can be paste0(var_1, "~", var_2)
where var_1 = a
and var_2 = b
Here is one solution thought I think we can do better:
txt = parse(text="as.list(coef(lm(a ~ b)))")
> x[,eval(txt),by=id]
id (Intercept) b
1: 1 21 -1
2: 2 21 -1
3: 3 21 -1
4: 4 21 -1
5: 5 21 -1
lm
can accept a character string as the formula so combine that with .SD
like this:
> x[, as.list(coef(lm("a ~ b", .SD))), by = id]
id (Intercept) b
1: 1 21 -1
2: 2 21 -1
3: 3 21 -1
4: 4 21 -1
5: 5 21 -1
这篇关于在R中的data.table环境中创建公式的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!