faq-error-scalar-type | R Documentation |
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.
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.
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.
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.