purrr::when() doesn't evaluate first condition

Question

Apart from the more common purrr::case_when that elegantly performs a vectorised if-else statement, there is less known variant purrr::when which is non-vectorised yet still useful in many scenarios. Like purrr::case_when, the purrr::when syntax also contains condition-action pairs.

However, purrr::when doesn't seem to evaluate the first condition. See here several cases with the example data from purrr::case_when.

x <- 35
purrr::when(
  x %% 35 == 0 ~ "fizz buzz",
  x %% 5 == 0 ~ "fizz",
  x %% 7 == 0 ~ "buzz",
  TRUE ~ "no buzz"
)
# should give "fizz buzz" but gives
"fizz"

purrr::when(
  x %% 5 == 0 ~ "fizz",
  x %% 7 == 0 ~ "buzz",
  TRUE ~ "no buzz"
)
# should give "fizz" but gives
"buzz"

purrr::when(
  x %% 7 == 0 ~ "buzz",
  TRUE ~ "no buzz"
)
# should give "buzz" but gives
"no buzz"

Side notes:

1. All conditions evaluate to TRUE.
2. The same behavior happens when TRUE is omitted in the last condition.
3. The same behavior happens when x is a vector (e.g. x <- 1:100).

Is this a bug??

It could be a bug because having only one condition throws this error message:

purrr::when(
  x %% 7 == 0 ~ "buzz"
)
Error: At least one matching condition is needed.

# However, case_when behaves as expected
dplyr::case_when(
  x %% 7 == 0 ~ "buzz"
)
"buzz"

Any ideas?

Answer 1

Ok this is the answer - thanks to jennybc on github

purrr::when requires the value as first argument!

x <- 35
purrr::when(
  x,
  x %% 7 == 0 ~ "buzz",
  TRUE ~ "no buzz"
)

I'm leaving the question and this answer here in case others might also oversee this, assuming the same arguments as for dplyr::case_when.