What would be a good example of an endofunctor that is not the identity functor?

问题

In Professor Frisby Introduces Composable Functional JavaScript the identity functor was introduced:

const Box = x => 
   ({ 
       map:  f => Box(f(x)),
       fold: f => f(x)           // for testing
   })

I spent the better part of the day understanding functors and why the above JavaScript code is actually the identity functor. So I thought I would alter it to get a "real" functor that is not the identity functor. I came up with this:

const Endo = x =>
   ({ 
       map:  f => Endo(f(x).split('')),
       fold: f => f(x).split('') // for testing
   })

My reasoning is that with Box, Id_Box: Box -> Box and Id_Box f = f. Endo would also map to itself but Endo(f): Endo(x) -> Endo(y) (if f: x -> y).

Am I on the right track?

EDIT: Replaced string with the more generic x as it was in the original examples.

回答1:

As pointed out in this answer, for our purposes as programmers we can treat all functors as endofunctors so don't get too caught up on the differences.

As for what a functor is, in brief it is

a data structure (Box in your example)
that can support a mapping operation (think Array.prototype.map)
and that mapping operation respects identity: xs === xs.map(x => x)
...and composition: xs.map(f).map(g) === xs.map(f . g) where . is function composition.

That's it. No more, no less. Looking at your Box, it's a data structure that has a map function (check 1 & 2) and that map function looks like it should respect identity and composition (check 3 & 4). So it's a functor. But it doesn't do anything, which is why it's the identity functor. The fold function isn't strictly necessary, it just provides a way to 'unwrap' the box.

For a useful functor, let's look at JavaScript arrays. Arrays actually do something: namely they contain multiple values rather than just a single one. If an array could only have one element, it'd be your Box. For our purposes we'll pretend that they can only hold values of the same type to simply things. So an array is a data structure, that has a map function, that respects identity and composition.

let plus = x => y => x + y;
let mult = x => y => x * y;
let plus2 = plus(2);
let times3 = mult(3);
let id = x => x;
let compose = (...fs) => arg => fs.reverse().reduce((x, f) => { return f(x) }, arg);  

// Here we need to stringify the arrays as JS will compare on 
// ref rather than value. I'm omitting it after the first for
// brevity, but know that it's necessary.
[1,2,3].map(plus2).toString() === [3,4,5].toString(); // true
[1,2,3].map(id) === [1,2,3]; // true
[1,2,3].map(plus2).map(times3) === [1,2,3].map(compose(times3, plus2)); // true

So when we map a function over a functor (array) we get back another instance of the same functor (a new Array) with the function applied to whatever the functor (array) was holding.

So now lets look at another ubiquitous JavaScript data structure, the object. There's no built in map function for objects. Can we make them a functor? Assume again that the object is homogenous (only has keys to one type of value, in this example Number):

let mapOverObj = obj => f => {
  return Object.entries(obj).reduce((newObj, [key, value]) => {
    newObj[key] = f(value);
    return newObj;
  }, {});
};

let foo = { 'bar': 2 };
let fooPrime = mapOverObj(foo)(plus2); // { 'bar': 4 }

And you can continue on to test that the function accurately (as far as is possible in JavaScript) supports identity and composition to satisfy the functor laws.

来源：https://stackoverflow.com/questions/45135978/what-would-be-a-good-example-of-an-endofunctor-that-is-not-the-identity-functor

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