R数据.表: efficiently access and update a variable column name in j expression with grouping

Question

这个问题已经有答案了:

Apply a function to every specified column in a data.table and update by reference (7 个回答)

已关闭 3 年前。

我想对数据表中的列列表应用转换(其类型，宽松地说，是“向量” -> “向量”)，并且此转换将涉及分组操作。

这是设置以及我想要实现的目标:

library(data.table)

set.seed(123)
n <- 1000
DT <- data.table(
        date = seq.Date(as.Date('2000/1/1'), by='day', length.out = n),
        A = runif(n),
        B = rnorm(n),
        C = rexp(n))

DT[, A.prime := (A - mean(A))/sd(A), by=year(date)]
DT[, B.prime := (B - mean(B))/sd(B), by=year(date)]
DT[, C.prime := (C - mean(C))/sd(C), by=year(date)]

目标是避免键入列名称。在我的实际应用程序中，我有一个想要应用此转换的列列表。

library(data.table)
set.seed(123)
n <- 1000
DT <- data.table(
        date = seq.Date(as.Date('2000/1/1'), by='day', length.out = n),
        A = runif(n),
        B = rnorm(n),
        C = rexp(n))

columns <- c("A", "B", "C")

for (x in columns) {
    # This doesn't work.
    # target <- DT[, (x - mean(x, na.rm=TRUE))/sd(x, na.rm = TRUE), by=year(date)]

    # This doesn't work.
    #target <- DT[, (..x - mean(..x, na.rm=TRUE))/sd(..x, na.rm = TRUE), by=year(date)]

    # THIS WORKS! But it is tedious writing "get(x)" every time. 
    target <- DT[, (get(x) - mean(get(x), na.rm=TRUE))/sd(get(x), na.rm = TRUE), by=year(date)][, V1]

    set(DT, j = paste0(x, ".prime"), value = target)
}

问题:实现上述结果的惯用方法是什么？有两件事可能需要改进:

1. 如何避免输入 get(x)每次我使用 x访问列？
2. 正在访问 [, V1]最有效的方法是什么？可以更新DT直接通过引用，而不创建中间数据表？

Answer 1

您可以使用.SDcols来指定要操作的列:

library(data.table)

columns <- c("A", "B", "C")
newcolumns <- paste0(columns, ".prime")

DT[, (newcolumns) := lapply(.SD, function(x) (x- mean(x))/sd(x)),
                      year(date), .SDcols = columns]

这可以避免每次都使用 get(x) 并通过引用更新 data.table。