[Compose] 21. Apply Natural Transformations in everyday work

We see three varied examples of where natural transformations come in handy.

Let's mark the law here for Natural Transformations:

nt(F).map(f) === nf(F.map(f))

Then let's see why we might need to apply Natural Transformations:

1. Natural Transformations can improve the code proformance

For example, we have an array of numbers, we want to get larger number, apply a doulbe function, then only get first element out of the array.

const first = xs => fromNullable(xs[]);
const largeNumbers = xs => xs.filter(x => x > );
const doulbe = x => x * ;

// We first run though the whole array to doulbe the number
// Then apply first to either transform -> Proformance cost
// Since nt(F).map(f) === nt(F.map(f))
const transform1 = xs => first(largeNumbers(xs).map(doulbe));
const app = xs => transform1(xs);
console.log(app([,,,])); // Either { value: 800 }

Imporved one should be get only first of larger number, then apply transform:

const first = xs => fromNullable(xs[]);
const largeNumbers = xs => xs.filter(x => x > );
const doulbe = x => x * ;

// Now we get first of array and transform to either, then apply our doulbe function
const transform2 = xs => first(largeNumbers(xs)).map(doulbe);
const app = xs => transform2(xs);
console.log(app([,,,])); // Either { value: 800 }

2. Solve different nested Functors:

const fake = id =>
    ({id, name: 'user1', best_friend_id: id + });
// Db :: find :: id -> Task(Either)
const Db = ({
    find: id => new Task((rej, res) => {
        res(id >  ? Right(fake(id)): Left('not found'))
    })
});

For example, the code above, we have a Db.find function, which return Task(Either(User)). It would be much easier for us to work with the same Functor type, instead of multi functors. we can think about change

// Either -> Task

const eitherToTask = e =>
    e.fold(Task.rejected, Task.of);

With this tool, let's see how it solve us problem:

Db.find() // Task(Right(User))

Now we chain 'eitherToTask':

Db.find() // Task(Right(User))
    .chain(eitherToTask) // // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)

So after run though 'eitherToTask' we change 'Right to Task' we got 'Task(Task(User)), then the outmost 'chain' will remove 'Task(User)'

Now if we want to find User's best friend id:

Db.find() // Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)
    .chain(user => Db.find(user.best_friend_id)) // Task(User) --Db.find--> Task(Task(Right(User))) --chain--> Task(Right(User))

We know that 'Db.find(user.best_friend_id)' returns 'Task(Right(User))', inner most function call return 'Task(Task(Right(User)))', after outter most 'chain', we got 'Task(Right(User))'.

Now we can apply the trick again:

Db.find() // Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)
    .chain(user => Db.find(user.best_friend_id)) // Task(User) --Db.find--> Task(Task(Right(User))) --chain--> Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)

We only got 'Task(User)' in the end, ready to be forked!

Db.find() // Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)
    .chain(user => Db.find(user.best_friend_id)) // Task(User) --Db.find--> Task(Task(Right(User))) --chain--> Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)
    .fork(console.error, console.log); // { id: 5, name: 'user1', best_friend_id: 6 }

--

const Either = require('data.either');
const Task = require('data.task');
const {List, Map} = require('immutable-ext');
const {Right, Left, fromNullable} = Either;

const log = console.log;

// F a -> G a
// nt(F).map(f) === nt(F.map(f))

//////Exp1/////
const res = List(['Hello', 'wolrd']).chain(x => x.split(''));
console.log(res.toJS()); // [ 'H', 'e', 'l', 'l', 'o', 'w', 'o', 'l', 'r', 'd' ]

//////Exp2/////
const first = xs => fromNullable(xs[]);
const largeNumbers = xs => xs.filter(x => x > );
const doulbe = x => x * ;

// We first run though the whole array to doulbe the number
// Then apply first to either transform -> Proformance cost
// Since nt(F).map(f) === nt(F.map(f))
const transform1 = xs => first(largeNumbers(xs).map(doulbe));
// Now we get first of array and transform to either, then apply our doulbe function
const transform2 = xs => first(largeNumbers(xs)).map(doulbe);
const app = xs => transform2(xs);
console.log(app([,,,])); // Either { value: 800 }

//////Exp3/////const fake = id =>
    ({id, name: 'user1', best_friend_id: id + });
// Db :: find :: id -> Task(Either)
const Db = ({
    find: id => new Task((rej, res) => {
        res(id >  ? Right(fake(id)): Left('not found'))
    })
});

const eitherToTask = e =>
    e.fold(Task.rejected, Task.of);

Db.find() // Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)
    .chain(user => Db.find(user.best_friend_id)) // Task(User) --Db.find--> Task(Task(Right(User))) --chain--> Task(Right(User))
    .chain(eitherToTask) // Task(Right(User)) --EtT--> Task(Task(User)) --chain--> Task(User)
    .fork(console.error, console.log); // { id: 5, name: 'user1', best_friend_id: 6 }