r - data.table:j 中的匿名函数

Question

我试图让匿名函数返回 j 中的多列data.table 的参数.下面是一个例子:

## sample data
tmpdt <- data.table(a = c(rep("a", 5), rep("b", 5)),
                    b = c(rep("f", 3), rep("r", 7)),
                    c = 1:10,
                    d = 21:30)
tmpdt[c %in% c(2,4), c := NA]

## this works fine
tmpdt[ , list(testout =
                (function(x) {
                    model <- lm(c ~ d, x)
                    residuals(model)
                })(.SD)),
      by = a]

## but I want to return a data.frame from the
## anonymous function

tmpdt[ , list(testout =
                (function(x) {
                    model <- lm(c ~ d, x)
                    tmpresid <- residuals(model)
                    tmpvalue <- x$b[as.numeric(names(tmpresid))]
                    data.frame(tmpvalue, tmpresid)
                })(.SD)),
      by = a]

第二个版本不起作用，因为该函数返回 data.frame而不仅仅是一个向量。有什么方法可以在不编写 data.table j 之外的函数调用的情况下完成这项工作争论？

Answer 1

你不需要匿名函数 - 你可以在 { } 中包含任何你想要的表达式(匿名机构)在 j .

tmpdt[, {
          model <- lm(c ~ d, .SD)
          tmpresid <- residuals(model)
          tmpvalue <- b[as.numeric(names(tmpresid))]
          list(tmpvalue, tmpresid) # every element of the list becomes a column in result
        }
      , by = a]

关于匿名体使用的一些文档 { }在 j :

评论在 示例 在 ?data.table :

anonymous lambda in j: j accepts any valid expression. TO REMEMBER: every element of the list becomes a column in result.

data.table FAQ 2.8 What are the scoping rules for j expressions?

No anonymous function is passed to j. Instead, an anonymous body [{ }] is passed to j [...] Some programming languages call this a lambda.

Andrew Brooks 关于 { } 使用的博客文章在 j :Suppressing intermediate output with {}