Apply a function to every specified column in a data.table and update by reference

by

This seems to work:

dt[ , (cols) := lapply(.SD, "*", -1), .SDcols = cols]

The result is

    a  b d
1: -1 -1 1
2: -2 -2 2
3: -3 -3 3

There are a few tricks here:

  • Because there are parentheses in (cols) :=, the result is assigned to the columns specified in cols, instead of to some new variable named “cols”.
  • .SDcols tells the call that we’re only looking at those columns, and allows us to use .SD, the Subset of the Data associated with those columns.
  • lapply(.SD, ...) operates on .SD, which is a list of columns (like all data.frames and data.tables). lapply returns a list, so in the end j looks like cols := list(...).

EDIT: Here’s another way that is probably faster, as @Arun mentioned:

for (j in cols) set(dt, j = j, value = -dt[[j]])

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