在data.table中使用“！”运算符选择行时出错

Question

I try to select rows using the ! operator

d = data.table(a = 1:3, b = c(TRUE, FALSE,FALSE))

d [b==FALSE]
#    a     b
# 1: 2 FALSE
# 2: 3 FALSE


d [!b]
# Error in eval(expr, envir, enclos) : object 'b' not found

Shouldn't it be evaluated if it is an expression?

Answer 1

In data.table , when i is a symbol, it is evaluated in the calling scope (see ?data.table i argument explanation) and not within the frame of the data.table. This is because we allow subsetting (or joining) data.tables using another data.table, ie, i can be another data.table.

require(data.table)
dt1 = data.table(x=1:3, y=4:6)
dt2 = data.table(x=2:3, z=7:8)

dt1[dt2, on="x"] # dt2 needs to be looked up first in the calling scope

Because of this feature, symbol in i needs to be wrapped with () so that it's seen as an expression (as opposed to a symbol), which is sufficient to understand that it needs to be evaluated within the frame of a data.table. That is:

dt1[, id := c(TRUE, FALSE, TRUE)]
dt1[(id)] # rows 1 and 3

When you use !<symbol> , the "!" is caught and removed, and the rest of the expression is first evaluated and then "!" is introduced back.. (but done efficiently without materialising intermediate data), ie,

dt1[!dt2, on="x"] 

computes the matching row indices by first computing the matching row indices for dt1[dt2, on="x"] and then obtaining the indices corresponding to "!" by taking the difference.

Therefore, we'll need a () when used with "!" as well so that it's seen as an expression as opposed to a symbol.

dt1[!(id)] # works
dt1[(!id)] # also works

In general, this usage of subsetting a logical vector which is a column in the data.table is extremely rare compared to the usage and advantages of subsetting using another data.table.

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This'll become a lot better when better scoping rules are implemented for i argument. See #633 .

Answer 2

We need to put it inside brackets

d [!(b)]
#   a     b
#1: 2 FALSE
#2: 3 FALSE