using lapply with emmeans list

Question

I'm using the emmeans package and cannot understand why this works

library(emmeans)
mod <- lm(mpg ~ disp + hp, data = mtcars)
l1 <- emmeans(mod, list("disp","hp"))

for (x in l1) {print(data.frame(x))}

      disp   emmean        SE df lower.CL upper.CL
1 230.7219 20.09062 0.5527102 29 18.96021 21.22104
        hp   emmean        SE df lower.CL upper.CL
1 146.6875 20.09062 0.5527102 29 18.96021 21.22104

but this doesn't?

lapply(l1, function(x) data.frame(x))

 Error in as.data.frame.default(x[[i]], optional = TRUE) : 
cannot coerce class ‘"function"’ to a data.frame

With a normal list I get the expected output (data frames in list):

l2 <- list(a=matrix(c(1,2,3,4)), b=matrix(c(5,6,7,8)))
lapply(l2, function(x) data.frame(x))
$a
  x
1 1
2 2
3 1
4 3

$b
  x
1 4
2 5
3 4
4 6

that seems like odd behavior. in `?emmeans` when `specs` is a list, you get an `?emm_list` object which has an `as.data.frame` method so you could do `as.data.frame(l1)` but it gives warnings — rawr, Mar 18 '23 at 02:28
It works if you use `Map(data.frame, l1)`. Seems like it would be worth submitting a bug report. — Ritchie Sacramento, Mar 18 '23 at 03:00
`lapply(l1, \(x) data.frame(as.emmGrid(x)))` is also an option. — Ritchie Sacramento, Mar 18 '23 at 06:55

score 2 · Accepted Answer · answered Mar 18 '23 at 04:33

It does modify the class from emmGrid to list with lapply

> lapply(l1, class)
$`emmeans of disp`
[1] "list"

$`emmeans of hp`
[1] "list"

versus, looping over the sequence and extraction doesn't

> lapply(seq_along(l1), \(i) class(l1[[i]]))
[[1]]
[1] "emmGrid"
attr(,"package")
[1] "emmeans"

[[2]]
[1] "emmGrid"
attr(,"package")
[1] "emmeans"

Thus, as.data.frame.list will be applied on all the elements of the list (which may have different structure, lengths etc). An option is to use as_tibble with map

library(tibble)
library(purrr)
> map(l1, as_tibble)
$`emmeans of disp`
# A tibble: 1 × 6
   disp emmean    SE    df lower.CL upper.CL
  <dbl>  <dbl> <dbl> <dbl>    <dbl>    <dbl>
1  231.   20.1 0.553    29     19.0     21.2

$`emmeans of hp`
# A tibble: 1 × 6
     hp emmean    SE    df lower.CL upper.CL
  <dbl>  <dbl> <dbl> <dbl>    <dbl>    <dbl>
1  147.   20.1 0.553    29     19.0     21.2

using lapply with emmeans list

1 Answers1