Fastest way to register namespaces programatically

Question

I'm building a function that uses getAnywhere() to retrieve the different packages from which a given function can be exported.

Unfortunately, getAnywhere() is limited to functions that exist in a loaded namespace, so I need to register the one I use.

Time is a constraint here, as loading a lot of namespaces can be a bit long (>10s), so using library() or require() is not realistic.

In RStudio, simply printing a prefixed function is enough to register the whole namespace without loading the library:

getAnywhere("align")$where
#> character(0)

flextable::as_flextable
#> function (x, ...) 
#> {
#>     UseMethod("as_flextable")
#> }
#> <bytecode: 0x000000002158e7f0>
#> <environment: namespace:flextable>

getAnywhere("align")$where
#> [1] "namespace:xtable"    "namespace:flextable"

pkgload::unload("flextable")
getAnywhere("align")$where
#> [1] "namespace:xtable"

^{Created on 2023-06-29 with reprex v2.0.2}

As you can see:

• at first, no loaded namespace contain a function named "align"
• after printing any function from flextable, xtable and flextable are stated (xtable is a dependency of flextable)
• for the example, after unloading flextable, only xtable remains

This seems fast enough, but it is hard-coded.

How could I register the flextable namespace programmatically (along with others)?
Is there an even faster way to do so?

Answer 1

Actually, there is a fast little function obviously named base::loadNamespace(). Don't know how I missed it...

getAnywhere("align")$where
#> character(0)
loadNamespace("flextable")
#> <environment: namespace:flextable>
getAnywhere("align")$where
#> [1] "namespace:xtable"    "namespace:flextable"

^{Created on 2023-06-30 with reprex v2.0.2}