FAQ - Error: Input must be a vector

This error occurs when a function expects a vector and gets a scalar object instead. This commonly happens when some code attempts to assign a scalar object as column in a data frame:

fn <- function() NULL
tibble::tibble(x = fn)
#> Error in `tibble::tibble()`:
#> ! All columns in a tibble must be vectors.
#> x Column `x` is a function.

fit <- lm(1:3 ~ 1)
tibble::tibble(x = fit)
#> Error in `tibble::tibble()`:
#> ! All columns in a tibble must be vectors.
#> x Column `x` is a `lm` object.

Vectorness in base R and in the tidyverse

In base R, almost everything is a vector or behaves like a vector. In the tidyverse we have chosen to be a bit stricter about what is considered a vector. The main question we ask ourselves to decide on the vectorness of a type is whether it makes sense to include that object as a column in a data frame.

The main difference is that S3 lists are considered vectors by base R but in the tidyverse that’s not the case by default:

fit <- lm(1:3 ~ 1)

typeof(fit)
#> [1] "list"
class(fit)
#> [1] "lm"

# S3 lists can be subset like a vector using base R:
fit[c(1, 4)]
#> $coefficients
#> (Intercept) 
#>           2 
#> 
#> $rank
#> [1] 1

# But not in vctrs
vctrs::vec_slice(fit, c(1, 4))
#> Error in `vctrs::vec_slice()`:
#> ! `x` must be a vector, not a <lm> object.

Defused function calls are another (more esoteric) example:

call <- quote(foo(bar = TRUE, baz = FALSE))
call
#> foo(bar = TRUE, baz = FALSE)

# They can be subset like a vector using base R:
call[1:2]
#> foo(bar = TRUE)
lapply(call, function(x) x)
#> [[1]]
#> foo
#> 
#> $bar
#> [1] TRUE
#> 
#> $baz
#> [1] FALSE

# But not with vctrs:
vctrs::vec_slice(call, 1:2)
#> Error in `vctrs::vec_slice()`:
#> ! `x` must be a vector, not a call.

I get a scalar type error but I think this is a bug

It’s possible the author of the class needs to do some work to declare their class a vector. Consider reaching out to the author. We have written a developer FAQ page to help them fix the issue.