Set certain values to NA with dplyr

You can use replace which is a bit faster than ifelse:

dat <-  dat %>% mutate(x = replace(x, x<0, NA))

You can speed it up a bit more by supplying an index to replace using which:

dat <- dat %>% mutate(x = replace(x, which(x<0L), NA))

On my machine, this cut the time to a third, see below.

Here's a little comparison of the different answers, which is only indicative of course:

set.seed(24)
dat <- data.frame(x=rnorm(1e6))
system.time(dat %>% mutate(x = replace(x, x<0, NA)))
       User      System     elapsed
       0.03        0.00        0.03 
system.time(dat %>% mutate(x=ifelse(x<0,NA,x)))
       User      System     elapsed
       0.30        0.00        0.29 
system.time(setDT(dat)[x<0,x:=NA])
       User      System     elapsed
       0.01        0.00        0.02 
system.time(dat$x[dat$x<0] <- NA)
       User      System     elapsed
       0.03        0.00        0.03 
system.time(dat %>% mutate(x = "is.na<-"(x, x < 0)))
       User      System     elapsed
       0.05        0.00        0.05 
system.time(dat %>% mutate(x = NA ^ (x < 0) * x))
       User      System     elapsed
       0.01        0.00        0.02 
system.time(dat %>% mutate(x = replace(x, which(x<0), NA)))
       User      System     elapsed
       0.01        0.00        0.01 

(I'm using dplyr_0.3.0.2 and data.table_1.9.4)

---

Since we're always very interested in benchmarking, especially in the course of data.table-vs-dplyr discussions I provide another benchmark of 3 of the answers using microbenchmark and the data by akrun. Note that I modified dplyr1 to be the updated version of my answer:

set.seed(285)
dat1 <- dat <- data.frame(x=sample(-5:5, 1e8, replace=TRUE), y=rnorm(1e8))
dtbl1 <- function() {setDT(dat)[x<0,x:=NA]}
dplr1 <- function() {dat1 %>% mutate(x = replace(x, which(x<0L), NA))}
dplr2 <- function() {dat1 %>% mutate(x = NA ^ (x < 0) * x)}
microbenchmark(dtbl1(), dplr1(), dplr2(), unit='relative', times=20L)
#Unit: relative
#    expr      min       lq   median       uq      max neval
# dtbl1() 1.091208 4.319863 4.194086 4.162326 4.252482    20
# dplr1() 1.000000 1.000000 1.000000 1.000000 1.000000    20
# dplr2() 6.251354 5.529948 5.344294 5.311595 5.190192    20

You can use the is.na<- function:

dat %>% mutate(x = "is.na<-"(x, x < 0))

Or you can use mathematical operators:

dat %>% mutate(x = NA ^ (x < 0) * x)