在分配（通过引用）的情况下，'with = FALSE'可以由括号“（）”中的LHS替换。简单地在没有赋值的情况下对列进行子集化时，这个好的功能不起作当然有.SD / .SDcols或get（）/ mget（）的工作量，但是以相同的方式对列进行子集会很好，无论是否有赋值。
dt <- data.table(A = 1:3, B = 4:6 )
col <- "A"
cols <- c("A","B")
# assign the old way
dt[, col := 9 , with=FALSE]
dt[, cols := .(9,8), with=FALSE]
# assign the new way
dt[, (col) := 8 ]
dt[, (cols) := .(8,7)]
# But the above syntax does not work for subsetting
dt[, (col)]
dt[, (cols)]
# I know how I can subset col and cols, but that is not the question here,
# e.g.:
dt[, col, with=FALSE]
dt[, cols, with=FALSE]
dt[, .SD, .SDcols=col]
dt[, .SD, .SDcols=cols]
# Below, further (there are even more) types of subsetting but they are not
# the same for col and cols, which is importent for looping where I dont
# know how many cols I call in advance.
dt[, get(col)]
dt[, mget(cols)]
dt[[col]] # Returns a vector, nor running: dt[[cols]]
换句话说：如果dt [，（col）：= 8]运行，作为一个天真的用户，我期望df [，（col）]也能运行。可能在[data.table中存在冲突，因此无法实现？