Detecting columns containing any value quickly with grep - r

Question

I have a large dataset, 5000 variables and 3 million rows. I want to check what columns contain dates. I'm working with data.table and reading the data with fread. In order to know what columns contain dates I run this:
my[, lapply(.SD,function(xx)
length(grep("^\\d\\d?/\\d\\d?/\\d{4}$",xx))>0 ) ]
or the same with any(grepl())
But it's very slow.
Is there any way to do it faster? Maybe forcing grep to stop the first time it encounters a date? I think (command line) grep has an option to do it:
grep -m 1
But I think it's not available in R.
Any idea? Solutions with base R or other packages are also welcome.
I could also work only with a few rows of the data.table but some columns could have very little values different than NA and there are chances of missing them.
Very simple example with some NA:
library(data.table)
set.seed(1)
siz <- 10000000
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)
)
I've seen there is an option perl = FALSE but I don't know wheter it will allow me to add extra parameters.
Or similarly I want to know among the files supposed to be dates whether there are strange symbols. I could run grep on every column but it would be great to be able to stop as soon as my test is right, without continuing till the end of the column.
Maybe with some extra package such as stringi?

Answer 1

One option would be to check only the first row (assuming that if there is a 'Date' class it would pick it up unless the first one is a missing value)
my[1][, grepl("\\d{2}/\\d{2}/\\d{4}", unlist(.SD))]
In addition to the above, as #Frank mentioned we can check only a subset of character class columns instead of the whole columns by specifying the .SDcols
j1 <- sapply(my, is.character)
my[, lapply(.SD, function(x)
length(grep("\\d{2}/\\d{2}/\\d{4}", x))>1),
.SDcols = j1]
Benchmarks
set.seed(24)
dat <- data.table(col1 = rnorm(1e6), col2 = "05/05/1942",
col3 = rnorm(1e6))
system.time(res <- dat[, lapply(.SD, function(x)
length(grep("\\d{2}/\\d{2}/\\d{4}", x))>1)])
# user system elapsed
# 6.33 0.01 6.35
system.time(res2 <- dat[1][, grepl("\\d{2}/\\d{2}/\\d{4}", unlist(.SD))])
# user system elapsed
# 0 0 0
system.time({
j1 <- sapply(dat, is.character)
res3 <- dat[, lapply(.SD, function(x)
length(grep("\\d{2}/\\d{2}/\\d{4}", x))>1), .SDcols = j1]
res3 <- names(dat) %in% names(res3)
})
# user system elapsed
# 0.43 0.00 0.44
all.equal(unlist(res), res2, check.attributes = FALSE)
#[1] TRUE
all.equal(unlist(res), res3, check.attributes=FALSE)
#[1] TRUE
If there are lots of NAs, then we can check on the first row where it has all non-NA elements
set.seed(24)
dat <- data.table(col1 = sample(c(NA, 1:10), 1e6, replace=TRUE),
col2 = c(NA, "05/05/1942"),
col3 = sample(c(NA, 1:5), 1e6, replace=TRUE))
dt1 <- head(dat, 20)
#Or just a sample of 20 rows from the dataset
#dt1 <- dat[sample(1:.N, 20, replace=TRUE)]
dt1[dt1[, which(!Reduce(`|`, lapply(.SD, is.na)))[1]]
][, grepl("\\d{2}/\\d{2}/\\d{4}", unlist(.SD))]