Data.table: how to get the blazingly fast subsets it promises and apply to a second data.table

Question

I\'m trying to enrich one dataset (adherence) based on subsets from another (lsr). For each individual row in adherence, I want to calculate (as a third column) the medicati

Answer 1

This can be solved by updating in a non-equi join.

This avoids the memory issues caused by a cartesian join or by calling apply() which coerces a data.frame or data.table to a matrix which involves copying the data.

In addition, the OP has mentioned that lsr has a few hundred mio. rows and adherence has 1.5 mio rows (500 timeperiods times 3000 ID's). Therefore, efficient storage of data items will not only reduce the memory footprint but may also reduce the share of processing time which is required for loading data.

library(data.table)
# coerce to data.table by reference, i.e., without copying
setDT(adherence)
setDT(lsr)
# coerce to IDate to save memory
adherence[, year := as.IDate(year)]
cols <- c("eksd", "ENDDATE")
lsr[, (cols) := lapply(.SD, as.IDate), .SDcols = cols]
# update in a non-equi join
adherence[lsr, on = .(ID, year >= eksd, year < ENDDATE), 
                      AH := as.integer(ENDDATE - x.year)][]

   ID       year AH
1:  1 2013-01-01 NA
2:  2 2013-01-01 NA
3:  3 2013-01-01 NA
4:  1 2013-02-01 64
5:  2 2013-02-01 NA
6:  3 2013-02-01 63

Note that NA indicates that no match was found. If required, the AH column can be initialised before the non-equi join by adherence[, AH := 0L].

Data

The code to create the sample datasets can be streamlined:

adherence <- data.frame(
  ID = c("1", "2", "3", "1", "2", "3"), 
  year = as.Date(c("2013-01-01", "2013-01-01", "2013-01-01", "2013-02-01", "2013-02-01", "2013-02-01")),
  stringsAsFactors = FALSE)

lsr <- data.frame(
  ID = c("1", "1", "1", "2", "2", "2", "3", "3"),
  eksd = as.Date(c("2012-03-01", "2012-08-02", "2013-01-06","2012-08-25", "2013-03-22", "2013-09-15", "2011-01-01", "2013-01-05")),
  DDD = as.integer(c("60", "90", "90", "60", "120", "60", "30", "90")),
  stringsAsFactors = FALSE)
lsr$ENDDATE <- lsr$eksd + lsr$DDD

Note that DDD is of type integer which usually requires 4 bytes instead of 8 bytes for type numeric/double.

Also note that the last statement may cause the whole data object lsr to be copied. This can be avoided by using data.table syntax which updates by reference.

library(data.table)
setDT(lsr)[, ENDDATE := eksd + DDD][]