replace_na() not pipeable?

[foundinblank]

y <- replace_na(df$x, 0) works

y <- df %>% replace_na(x, 0) does not.

Seems like a no-brainer to have this pipeable...or is it a brainer? ;)

[lionel-]

Pipeable does not mean that a function can refer to data. You can pipe with any sort of function. In the case of replace_na(), it's a normal function (i.e. not a quoting function) that takes a vector and not a data frame. It's meant to be used within mutate() for example, which takes a data frame and evaluates its arguments in the data.

[foundinblank]

Thank you, @lionel- I see now that I can do

y <- df %>% replace_na(list(x = 0))

but that seems kludgy. I feel it's more intuitive to have replace_na(x, 0) - what would that be called? Tidy code?

[lionel-]

You have to distinguish between data frame verbs (like mutate() and vector verbs (like replace_na()). The way to do this is:

df %>% mutate(x = replace_na(x))

We are thinking about a more systematic way of approaching the data frame vs vector issue, but nothing concrete yet.

[foundinblank]

That's a good problem - how to clearly explain data frame v. vector verbs. In the meantime, the examples at the end of each vector verb could include a mutate example, as that'd be the most familiar usage scenario for most people who use tidyverse but don't understand the distinction yet between the two verb types.

I've put in a pull request for this.