Is there a more efficient way to fill extra column than a 'for' loop?

Question

I have a data.table with about 100k rows. I am going to simplify this to only 3 columns because that is all that is relevant here.

dt <- data.table(indicator = c("x", "y"), 
                 date1 = c("20190111", "20190212", "20190512", "20190723"), 
                 date2 = c("20190105", "20190215", "20190616", "20190623"))

What I want to do is assign either date1 or date2 to a new column, 'final_date' depending on the indicator column. If indicator is "x" assign final_date as date1. If indicator "y" assign final_date as date2.

I am able to do this with a "for" loop and if/else statements, but it takes a few minutes to complete with 100k rows.

for (row in 1:nrow(dt)) {
  if(dt$indicator[row] == "x") {
    dt$final_date[row] <- dt$date1[row]
  } else {
    dt$final_date[row] <- dt$date2[row]
  }
  }

Is there any more efficient way to do this with data.table functionality or anything else?

Answer 1

With data.table , I would do something like this:

dt[, final_date := ifelse(indicator == "x", date1, date2)]

Really quick and simple! I suspect with a large set of data it will be faster than dplyr as well as the solution you have, as data.table mutates in place rather than creating a copy of the data.

Answer 2

With the dplyr pipeline

> dt%>%mutate(final_data=if_else(indicator=="x",date1,date2))
  indicator    date1    date2 final_data
1         x 20190111 20190105   20190111
2         y 20190212 20190215   20190215
3         x 20190512 20190616   20190512
4         y 20190723 20190623   20190623

Answer 3

Try this:

# necessary package
library(dplyr)
library(data.table)
# reproduce your data
dt <- data.table(
  indicator = c("x", "y"),
  date1 = c("20190111", "20190212", "20190512", "20190723"),
  date2 = c("20190105", "20190215", "20190616", "20190623")
)
# create your variable final_date
dt[, final_date := case_when(indicator == "x" ~ date1,
                             TRUE ~ date2)]

Hope it helps