Set certain values to NA with dplyr

Question

I'm trying to figure out a simple way to do something like this with dplyr (data set = dat, variable = x):

day$x[dat$x<0]=NA

Should be simple but this is the best I can do at the moment. Is there an easier way?

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

Answer 1

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

Answer 2

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)

Answer 3

The most natural approach in dplyr is to use the na_if function.

For one variable:

dat %<>% mutate(x = na_if(x, x < 0))

For all variables:

dat %<>% mutate_all(~ na_if(., . < 0))

If interested in replacing a specific value, instead of a range for all variables:

dat %<>% mutate_all(na_if, 0)

Note that I am using the %<>% operator from the magrittr package.

Answer 4

If you are using data.table, the below code is faster

library(data.table)
setDT(dat)[x<0,x:=NA]

Benchmarks

Using data.table_1.9.5 and dplyr_0.3.0.9000

library(microbenchmark)
set.seed(285)
dat <- data.frame(x=sample(-5:5, 1e7, replace=TRUE), y=rnorm(1e7))

dtbl1 <- function() {as.data.table(dat)[x<0,x:=NA]}
dplr1 <- function() {dat %>% mutate(x = replace(x, x<0, NA))}

microbenchmark(dtbl1(), dplr1(), unit='relative', times=20L)
#Unit: relative
#expr     min       lq     mean   median       uq      max neval cld
#dtbl1() 1.00000 1.000000 1.000000 1.000000 1.000000 1.000000    20  a 
#dplr1() 2.06654 2.064405 1.927762 1.795962 1.881821 1.885655    20   b

Updated Benchmarks

Using data.table_1.9.5 and dplyr_0.4.0. I used a slightly bigger dataset and replaced as.data.table with setDT (Included @Sven Hohenstein's faster function as well.)

set.seed(285)
dat <- data.frame(x=sample(-5:5, 1e8, replace=TRUE), y=rnorm(1e8))
dat1 <- copy(dat)
dtbl1 <- function() {setDT(dat)[x<0,x:=NA]}
dplr1 <- function() {dat1 %>% mutate(x = replace(x, x<0, NA))}
dplr2 <- function() {dat1 %>% mutate(x = NA ^ (x < 0) * x)} 

microbenchmark(dtbl1(), dplr1(), dplr2(), unit='relative', times=20L)
#Unit: relative
#  expr      min       lq     mean   median       uq      max neval cld
#dtbl1() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    20  a 
#dplr1() 2.523945 2.542412 2.536255 2.579379 2.518336 2.486757    20   b
#dplr2() 1.139216 1.089992 1.088753 1.058653 1.093906 1.100690    20  a 

Updated Benchmarks2

At the request of @docendo discimus, benchmarking again his "new" version of dplyrusing data.table_1.9.5 and dplyr_0.4.0.

NOTE: Because there is a change in @docendo discimus code, I changed 0 to 0L for the data.table`

set.seed(285)
dat <- data.frame(x=sample(-5:5, 1e8, replace=TRUE), y=rnorm(1e8))
dat1 <- copy(dat)
dtbl1 <- function() {setDT(dat)[x<0L, 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     mean   median       uq      max neval cld
#dtbl1() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    20 a  
#dplr1() 2.186055 2.183432 2.142293 2.222458 2.194450 1.442444    20  b 
#dplr2() 2.919854 2.925795 2.852528 2.942700 2.954657 1.904249    20   c

data

set.seed(24)
dat <- data.frame(x=sample(-5:5, 25, replace=TRUE), y=rnorm(25))

Answer 5

Using replace directly on the x column and not using mutate also works.

dat$x <- replace(dat$x, dat$x<0, NA)