R - How to test for character(0) in IF statement

Question

I imagine this is incredibly simple but I cant seem to find the answer.

I am writing an IF statement but the test is if the object returns a character(0) value. I am not sure how to deal with character(0) in the statement.

Assuming Test <- character(0) or a value:

if (identical(Pols4,character(0))) {
  print('Empty')
} else {
  print('Not Empty')
}

It still does not seem to work.....

Answer 1

Use the identical function to check this.

a <- character(0)
identical(a, character(0))  # returns TRUE

identical(a, "")           # returns FALSE
identical(a, numeric(0))   # returns also FALSE

Answer 2

Adding the obligatory tidyverse answer. The rlang package has the function is_empty(), which does exactly what you want.

Test <- character(0)
rlang::is_empty(Test)
#[1] TRUE

This also works for empty vectors that aren't characters. For example, it works in the case that Patrick Roocks describes in comments.

Test <- as.Date(character(0))
rlang::is_empty(Test)
#[1] TRUE

Loading the 'tidyverse' package also loads is_empty().

Answer 3

Use the length() method:

> check <- function(value) {
+ if (length(value)==0) {
+ print('Empty')
+ } else {
+ print('Not Empty')
+ }
+ }
> check("Hello World")
[1] "Not Empty"
> check("")
[1] "Not Empty"
> check(character(0))
[1] "Empty"

Answer 4

Many people forget that functions like str_locate_all() require %>% unlist() or %>% .[[1]].

Then you can easily detect character(0) with the length() function if > 0 one can safely use the output of str_locate_all() for example.

Example:

value <- str_extract_all("kvk :", ascii_digit(8,8)) %>% unlist()
if (length(value) > 0) {
  # Do something
}

Answer 5

instead of length(), worked for me nrows()

Answer 6

My method to solve this problem is dealing with that column only by cast it into a character again and find "character(0)" instead.

For example:

df$interest_column <- as.character(df$interest_column) # Cast it into a character again
df[df$interest_column == "character(0)","interest_column"] <- NA  # Assign new value

I hope this is a solution you have asked for.

Answer 7

I was doing some micro-optimization today for a related problem: checking if a numeric vector is empty (e.g. equivalent to numeric(0)) when it can either be empty or have a value (it is never NA or NULL). In my problem the check occurs hundreds of millions of time, so it seemed reasonable to benchmark the right approach. Length benchmarks quite a bit better than other options:

vec = numeric(0)

bench::mark(
  x = { !length(vec) },
  y = { rlang::is_empty(vec) },
  z = { identical(vec, numeric(0)) },
  check = FALSE,
  min_time = 5,
  min_iterations = 100000,
  max_iterations = 100000
)

# A tibble: 3 x 6
  expression      min   median `itr/sec` `gc/sec`  n_gc
  <bch:expr> <bch:tm> <bch:tm>     <dbl>    <dbl> <dbl>
1 x             200ns    300ns  3621037.     0        0
2 y             5.2us    5.8us   166594.     8.33     5
3 z             1.3us    1.5us   618090.    12.4      2

Length checking beating identical checking by 6x and by is_empty by 4x over that. The results for the case where the vector is non-empty are similar, so irrespective of the distribution of your data, just use length.

I am cognizant that there are probably edge cases where the behaviours of these three functions aren't identical, but if like me it's just a matter of a value being either c(some, number) or numeric(0) and you want to quickly check which, use length.