How to remove a tail with NA in r?

Question

I have a vector:

a <- c(NA,1:5,NA,NA,1:3, rep(NA,round(runif(1,0,100))))

I need to remove the trailing NAs. Desired result:

c(NA, 1:5, NA, NA, 1:3)

Related: [Remove leading and trailing NA](https://stackoverflow.com/questions/42759027/remove-leading-and-trailing-na) — Henrik, Feb 02 '21 at 00:00

score 4 · Accepted Answer · answered Apr 10 '19 at 20:10

4

You can do

a[1:max(which(!is.na(a)))]
# [1] NA  1  2  3  4  5 NA NA  1  2  3

We subset the vector from position 1 to the last non NA value.

answered Apr 10 '19 at 20:10

markus

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3

This one fails in the (probably very unlikely) case when `a` contains only `NA`. – Julius Vainora Apr 10 '19 at 20:14

score 3 · Answer 2 · answered Apr 10 '19 at 20:09

One option would be

a[rev(cumprod(rev(is.na(a)))) == 0]
# [1] NA  1  2  3  4  5 NA NA  1  2  3

Here are the steps:

(a <- c(NA, 1:5, NA, NA, 1:3, NA, NA))
# [1] NA  1  2  3  4  5 NA NA  1  2  3 NA NA
is.na(a)
# [1]  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE  TRUE  TRUE
rev(is.na(a))
# [1]  TRUE  TRUE FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE
cumprod(rev(is.na(a)))
# [1] 1 1 0 0 0 0 0 0 0 0 0 0 0
rev(cumprod(rev(is.na(a))))
# [1] 0 0 0 0 0 0 0 0 0 0 0 1 1

score 2 · Answer 3 · answered Apr 10 '19 at 20:10

2

You can find the maximum position which is not an NA and subset accordingly

> a[1:max(which(!is.na(a)))]
 [1] NA  1  2  3  4  5 NA NA  1  2  3

answered Apr 10 '19 at 20:10

CT Hall

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tmfmnk · Answer 4 · 2019-04-10T20:39:37.717

Also a possibility:

a[cumsum(!is.na(a)) != max(cumsum(!is.na(a))) * is.na(a)]

 [1] NA  1  2  3  4  5 NA NA  1  2  3

In idividual steps:

is.na(a)

 [1]  TRUE FALSE FALSE FALSE FALSE

cumsum(!is.na(a))

 [1] 0 1 2 3 4

cumsum(!is.na(a)) != max(cumsum(!is.na(a)))

 [1]  TRUE  TRUE  TRUE  TRUE  TRUE

cumsum(!is.na(a)) != max(cumsum(!is.na(a))) * is.na(a)

 [1]  TRUE  TRUE  TRUE  TRUE  TRUE

Just for fun, a little benchmarking:

library(microbenchmark)

a <- rep(a, 1e5)

 microbenchmark(
 markus = a[1:max(which(!is.na(a)))],
 Julius_Vainora = a[rev(cumprod(rev(is.na(a)))) == 0],
 Kim = rm_NA_tail(a),
 tmfmnk = a[cumsum(!is.na(a)) != max(cumsum(!is.na(a))) * is.na(a)],
 nsinghs = a[1:(length(a) - rle(is.na(rev(a)))$lengths[1])],
 times = 5
)

Unit: milliseconds
           expr      min       lq     mean   median       uq       max neval cld
         markus 150.7346 153.0674 156.4194 153.3031 159.4718  165.5201     5 a  
 Julius_Vainora 393.8520 418.8186 616.3269 703.4022 749.6600  815.9018     5  bc
            Kim 370.7680 382.1826 536.0828 632.0031 642.1882  653.2720     5  bc
         tmfmnk 390.2626 415.2378 466.4245 415.8310 423.3828  687.4082     5  b 
        nsinghs 537.0404 781.1403 798.6929 793.1027 842.6777 1039.5033     5   c

score 0 · Answer 5 · answered Apr 10 '19 at 20:09

0

I think this works:

rm_NA_tail <- function(a) {
  if (is.na(a[length(a)])) {
    return(a[is.na(match(data.table::rleid(a), max(data.table::rleid(a))))])
  } else {
    return(a)
  }
}

answered Apr 10 '19 at 20:09

Kim

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score 0 · Answer 6 · answered Apr 10 '19 at 20:24

This can be done using rle()

a[1:(length(a) - rle(is.na(rev(a)))$lengths[1])]
#  [1] NA  1  2  3  4  5 NA NA  1  2  3

rle(is.na(rev(a)))$lengths[1] gets the count of trailing NA in the vector, then subtract it from the total vector length to get the index up to which you want to keep the vector.

How to remove a tail with NA in r?

6 Answers6