I've created an example data.table
library(data.table)
set.seed(1)
siz <- 10
my <- data.table(
  AA=c(rep(NA,siz-1),"11/11/2001"),
  BB=sample(c("wrong", "11/11/2001"),siz, prob=c(1000000,1), replace=T),
  CC=sample(siz),
  DD=rep("11/11/2001",siz),
  EE=rep("HELLO", siz)
)
my[2,AA:=1]
         NA wrong  3 11/11/2001 HELLO
          1 wrong  2 11/11/2001 HELLO
         NA wrong  6 11/11/2001 HELLO
         NA wrong 10 11/11/2001 HELLO
         NA wrong  5 11/11/2001 HELLO
         NA wrong  7 11/11/2001 HELLO
         NA wrong  8 11/11/2001 HELLO
         NA wrong  4 11/11/2001 HELLO
         NA wrong  1 11/11/2001 HELLO
 11/11/2001 wrong  9 11/11/2001 HELLO
If I run this code
patt <-  "^\\d\\d?/\\d\\d?/\\d{4}$" 
sapply(my, function(x)   (grepl(patt,x )))
I get a table with TRUE whenever there is a date.
         AA    BB    CC   DD    EE
 [1,] FALSE FALSE FALSE TRUE FALSE
 [2,] FALSE FALSE FALSE TRUE FALSE
 [3,] FALSE FALSE FALSE TRUE FALSE
 [4,] FALSE FALSE FALSE TRUE FALSE
 [5,] FALSE FALSE FALSE TRUE FALSE
 [6,] FALSE FALSE FALSE TRUE FALSE
 [7,] FALSE FALSE FALSE TRUE FALSE
 [8,] FALSE FALSE FALSE TRUE FALSE
 [9,] FALSE FALSE FALSE TRUE FALSE
[10,]  TRUE FALSE FALSE TRUE FALSE
But if I do it like this:
my[,lapply(.SD, grepl, patt)]
I just get this result:
   AA    BB    CC    DD    EE
1: NA FALSE FALSE FALSE FALSE
Why? How can I get the same result writing wverything inside the brackets?
We need to specify the pattern argument if we are not using anonymous function call
my[,lapply(.SD, grepl, pattern = patt)]
Or otherwise with an anonymous function call
my[,lapply(.SD, function(x) grepl(patt, x))]
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