我无法理解为什么以下不起作用：
library(data.table)
dt <- as.data.table(mtcars)
colnum <- grep("disp", tolower(colnames(dt)))
# 3
dt[, colnum, with = F] <- dt[, colnum, with = F]*2
以上引发了错误：
Error in `[<-.data.table`(`*tmp*`, , colnum, with = F, value = list(disp = c(320, :
unused argument (with = F)
删除第二个术语with = F也会输出错误：
dt[, colnum, with = F] <- dt[, colnum] * 2
Error in `[.data.table`(q, , colnum) :
j (the 2nd argument inside [...]) is a single symbol but column name 'colnum' is not found.
Perhaps you intended DT[,..colnum] or DT[,colnum,with=FALSE].
This difference to data.frame is deliberate and explained in FAQ 1.1.
为什么这不起作用？以上只是一个简化示例，但我想在data.table列上执行更复杂的操作。
答案 0 :(得分：3)
data.table - 执行此操作的方式为updating by reference，如下所示：
dt[, (colnum) := lapply(.SD, '*', 2), .SDcols = colnum][]
给出：
mpg cyl disp hp drat wt qsec vs am gear carb 1: 21.0 6 320.0 110 3.90 2.620 16.46 0 1 4 4 2: 21.0 6 320.0 110 3.90 2.875 17.02 0 1 4 4 3: 22.8 4 216.0 93 3.85 2.320 18.61 1 1 4 1 4: 21.4 6 516.0 110 3.08 3.215 19.44 1 0 3 1 5: 18.7 8 720.0 175 3.15 3.440 17.02 0 0 3 2 6: 18.1 6 450.0 105 2.76 3.460 20.22 1 0 3 1 .....
您可以在Getting started页面上找到更多信息。