I need to remove from a data.table any row in which column a
contains any NA
nested in a vector:
library(data.table)
a = list(as.numeric(c(NA,NA)), 2,as.numeric(c(3, NA)), c(4,5) )
b <- 11:14
dt <- data.table(a,b)
Thus, rows 1 and 3 should be removed.
I tried three solutions without success:
dt1 <- dt[!is.na(a)]
dt2 <- dt[!is.na(unlist(a))]
dt3 <- dt[dt[,!Reduce(`&`, lapply(a, is.na))]]
Any ideas? Thank you.
CodePudding user response:
You can do the following:
dt[sapply(dt$a, \(l) !any(is.na(l)))]
This alternative also works, but you will get warnings
dt[sapply(dt$a, all)]
Output:
a b
1: 2 12
2: 4,5 14
A third option that you might prefer: You could move the functionality to a separate helper function that ingests a list of lists (nl
), and returns a boolean vector of length equal to length(nl)
, and then apply that function as below. In this example, I explicitly call unlist()
on the result of lapply()
rather than letting sapply()
do that for me, but I could also have used sapply()
f <- \(nl) unlist(lapply(nl,\(l) !any(is.na(l))))
dt[f(a)]
CodePudding user response:
An alternative to *apply()
dt[, .SD[!anyNA(a, TRUE)], by = .I][, !"I"]
# a b
# <list> <int>
# 1: 2 12
# 2: 4,5 14