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Comments on How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?

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How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?

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I'm learning monadic composition through Scott Wlaschin's Railway-oriented Programming post. Oncebind, switch, and >=> functions are defined, he introduces map to show how to "turn a one-track function into a two-track function". That is:

f: a -> b     =>    f': T<a,c> -> T<b,c>

The implementation in the article is the following:

let map oneTrackFunction twoTrackInput =
    match twoTrackInput with
    | Success s -> Success (oneTrackFunction s)
    | Failure f -> Failure f

Did an an equivalent implementation using switch and bind as an exercise,

let map' f = bind (switch f)

but when I tried to implement map with >=>, I arrived at this ugly mess:

let map'' f result =
    match result with
        | Ok o -> ((fun _ -> result) >=> (switch f)) o
        | Error e -> Error e

Note to self: o could be any value of type 'a (if result : Result<'a,'c>), because f's input is already saved in the closure used as >=>'s first operand, but this was the only way I could think of to keep it more generic.

Is there a "cleaner" implementation similar to map's?


Notes

I used the following example to test the maps above:

map  ((+) 2) ((Ok 27) : Result<int,string>)

Used implementations of bind, switch, >=>:

let bind
    (     f : 'a -> Result<'b,'c>)
    (result :       Result<'a,'c>)
    =
    match result with 
   |    Ok o -> f o
   | Error e -> Error e

let switch
   (f : 'a -> 'b)
   (x : 'a      )
   =
   f x |> Ok

let (>=>)
    (f : 'a -> Result<'b,'error>)
    (g : 'b -> Result<'c,'error>)
    =
    f >> (bind g)
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Is there a "cleaner" implementation similar to map's?

Yes:

let map f = id >=> switch f

This follows from two of your other equations:

map f = bind (switch f)
g >=> h = g >> bind h

So if you want to get bind (switch f) out of (>=>), you can start by making h = switch f, then get rid of the superfluous composition by letting g be id and you're done.

 g >=> h        =  g >> bind h    <------ (h = switch f)
 g >=> switch f =  g >> bind (switch f)
 g >=> switch f =  g >> map f     <------ (g = id)
id >=> switch f = id >> map f
id >=> switch f = map f
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1 comment thread

Works for me (5 comments)
Works for me
toraritte‭ wrote about 1 month ago

Wow, the simplicity of this blew my mind. Did you come up with this on the spot or is this a well known formula in other (pure) functional programming languages?

I'm also trying to make sense how id can be the first argument to >=>, so here's another question if you are interested: Why does let map f = id >=> switch f work in F#?

r~~‭ wrote about 1 month ago

I don't think this particular formulation is typical in other FPLs—the switch combinator is a bit unusual. But it follows easily from your other definitions; I'll expand my answer to detail how.

toraritte‭ wrote about 1 month ago · edited about 1 month ago

Thank you for the clarification! I made an edit suggestion lining up the terms - it took me unnecessarily long to understand it as I have no experience in theorem proving, so "explain it to me like I'm 5" is how I roll for now.. Is my assumption correct that you picked h = switch f intuitively?

r~~‭ wrote 30 days ago

Intuitively, sure, although it's not a very big intuitive leap if you have bind h and you know you want bind (switch f). It's as mechanical as algebra once you're used to it.

toraritte‭ wrote 29 days ago

Well, it surely wasn't obvious to me right away, but once it clicked, I was surprised how I could've missed such an easy step. Thanks again and have a great day!