Function composition of form f(x)(g(x)) using Ramda

Question

I can't figure out a transform for f(x)(g(x)) using the functions available in Ramda. I'm hoping for something like R.something(f, g)(x) ideally - just so long as x only appears once and is the final argument.

f is a function taking x that returns a function

g is a function taking x that returns a value

I've tried pipe,compose,chain, but tbh I'm guessing pretty hard and no one of them work. I recently asked a similar question and I'd love to know a resource with loads of useful identities if one exists so I don't need to keep asking on SO :)

Btw, if you consider `constant = x => y => x` this is an applicative functor. — , Oct 05 '17 at 06:11

Bergi · Accepted Answer · 2017-10-04T23:58:37.923

2

chain was close: it does chain(f, g)(x) == f(g(x))(x) on functions. So you just need to flip f before passing it into chain:

R.chain(R.flip(f), g)(x)

or

const something = R.compose(R.chain, R.flip)
something(f, g)(x)

edited Oct 04 '17 at 23:58

answered Oct 04 '17 at 23:53

Bergi

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I just realised this myself! Thanks – RichK Oct 04 '17 at 23:56
Let's call it `ap`. – Oct 05 '17 at 06:14
@ftor Oh. I realised that `R.chain` is Haskell's `=<<`, but given that everything is flipped I didn't think there's an equivalent to `(=<<) . flip`. You definitely should post [that](http://ramdajs.com/docs/#ap) as an answer. – Bergi Oct 05 '17 at 13:46
@ftor please add that as full answer :) – RichK Oct 08 '17 at 20:39

Function composition of form f(x)(g(x)) using Ramda

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