Ramda currying: how to apply argument to multiple parameters

Question

I have a situation where I need to do this:

const f = (obj) => assoc('list', createList(obj), obj)

Due to the fact that I need the argument for the second and the third parameter, prohibits me from doing something like:

const f = assoc('list', somehowGetObj())

I also tried this, but that didn't work:

const f = assoc('list', createList(__))
const f = converge(assoc, [createList, identity])

Is there a proper way to do this by currying?

Answer 1

Another option is

chain(createList, assoc('list'))

which you can see in action on the Ramda REPL.

Update

For further explanation of how this works, I'll use the variation which will work with the next release of Ramda:

chain(assoc('list'), createList)

to show how it matches the current signature:

chain :: Chain m => (a -> m b) -> m a -> m b

Ramda treats functions as FantasyLand Monads, and therefore thus also as Chains. So to specialize the above to functions, we have

chain :: (a -> Function x b) -> Function x a -> Function x -> b

but Function x y can be written more simply as x -> y, so the above can written more simply as

chain :: (a -> x -> b) -> (x -> a) -> (x -> b)

Then you can use these (specialized) types:

createList :: OriginalData -> YourList                              (x -> a)
assoc :: String -> YourList -> OriginalData -> EnhancedData
assoc('list') :: YourList -> OriginalData -> EnhancedData           (a -> x -> b)

and hence

chain(assoc('list'), createList) :: OriginalData -> EnhancedData    (x -> b)

Answer 2

const f = converge(assoc('list'), [createList, identity])