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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