Initialise a complete empty data frame (no rows, no columns) [duplicate]

Question

Here's a hack to create an empty data frame with no rows and no columns:

iris[FALSE, FALSE]
#> data frame with 0 columns and 0 rows

Smarter-looking code creates a spurious column:

x <- list(NULL)
class(x) <- c("data.frame")
attr(x, "row.names") <- integer(0)
str(x)
#> 'data.frame':    0 obs. of  1 variable:
#>  $ : NULL

Is there a non-hack alternative?

The reason to create such a thing is to satisfy a function that can handle empty data frames but not NULLs.

This is different from similar questions because it is about having no columns as well as no rows.

structure(list(),class="data.frame") would be a way to go your original method of trying to add a class to a list. — thelatemail
– thelatemail, Commented Jun 7, 2016 at 1:32
OP says they're trying "to satisfy a function that can handle empty data frames but not NULLs"... if OP is the one writing the function then can I suggest they're attacking this from the wrong side? What about testing inherits(x, "data.frame")? which will pass for a data.frame (empty or not) but will fail for NULL. If they're trying to pass data into an existing function, then data.frame() should bypass the test (which could very likely be the above one anyway). — Jonathan Carroll
– Jonathan Carroll, Commented Jun 7, 2016 at 1:39
@JonathanCarroll OP is trying to satisfy unnest in tidyr package ;) — nacnudus
– nacnudus, Commented Jun 7, 2016 at 2:13

akuiper · Accepted Answer · 2016-06-07 01:16:12Z

7

df <- data.frame()
str(df)
'data.frame':   0 obs. of  0 variables

answered Jun 7, 2016 at 1:16

akuiper

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1 Comment

nacnudus Over a year ago

Why didn't I think of that?

nacnudus · Accepted Answer · 2016-06-07 02:23:01Z

2

empty.data.frame <- function() {
  structure(NULL,
            names = character(0),
            row.names = integer(0),
            class = "data.frame")
}
empty.data.frame()
#> data frame with 0 columns and 0 rows

# thelatemail's suggestion in a comment (fastest)
empty.data.frame2 <- function() {
  structure(NULL, class="data.frame")
}

library(microbenchmark)
microbenchmark(data.frame(), empty.data.frame(), empty.data.frame2())
#> Unit: microseconds
#>                 expr    min      lq     mean median     uq    max neval
#>         data.frame() 12.831 13.4485 15.18162 13.879 14.378 65.967   100
#>   empty.data.frame()  8.323  9.0515  9.76106  9.363  9.732 19.427   100
#>  empty.data.frame2()  5.884  6.9650  7.63442  7.240  7.540 17.746   100

edited Jun 7, 2016 at 2:23

answered Jun 7, 2016 at 1:32

nacnudus

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5 Comments

Jonathan Carroll Over a year ago

Is performance really an issue here? The only possible scaling is with repetition.

Pierre L Over a year ago

@JonathanCarroll how would you scale an empty data frame?

Jonathan Carroll Over a year ago

@PierreLafortune that's my point. This isn't going to be a costly procedure no matter which way you do it. You may potentially do it lots of times (under some odd scenario) but even then it's not slow.

Pierre L Over a year ago

His test is most likely for learning to see which function call gets to the point with the least internal moves.

nacnudus Over a year ago

"least internal moves" -- precisely, to get as close as possible to a variable declaration like "int a;". I should have said that in the question. In practice it's far faster to create one and copy it.

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