.map() a Javascript ES6 Map?

Question

So .map itself only offers one value you care about... That said, there are a few ways of tackling this:

// instantiation
const myMap = new Map([
  [ "A", 1 ],
  [ "B", 2 ]
]);

// what's built into Map for you
myMap.forEach( (val, key) => console.log(key, val) ); // "A 1", "B 2"

// what Array can do for you
Array.from( myMap ).map(([key, value]) => ({ key, value })); // [{key:"A", value: 1}, ... ]

// less awesome iteration
let entries = myMap.entries( );
for (let entry of entries) {
  console.log(entry);
}

Note, I'm using a lot of new stuff in that second example... ...Array.from takes any iterable (any time you'd use [].slice.call( ), plus Sets and Maps) and turns it into an array... ...Maps, when coerced into an array, turn into an array of arrays, where el[0] === key && el[1] === value; (basically, in the same format that I prefilled my example Map with, above).

I'm using destructuring of the array in the argument position of the lambda, to assign those array spots to values, before returning an object for each el.

If you're using Babel, in production, you're going to need to use Babel's browser polyfill (which includes "core-js" and Facebook's "regenerator").
I'm quite certain it contains Array.from.

You should just use Spread operator:

var myMap = new Map([["thing1", 1], ["thing2", 2], ["thing3", 3]]);

var newArr = [...myMap].map(value => value[1] + 1);
console.log(newArr); //[2, 3, 4]

var newArr2 = [for(value of myMap) value = value[1] + 1];
console.log(newArr2); //[2, 3, 4]

Just use Array.from(iterable, [mapFn]).

var myMap = new Map([["thing1", 1], ["thing2", 2], ["thing3", 3]]);

var newEntries = Array.from(myMap, ([key, value]) => [key, value + 1]);
var newMap = new Map(newEntries);

You can use this function:

function mapMap(map, fn) {
  return new Map(Array.from(map, ([key, value]) => [key, fn(value, key, map)]));
}

usage:

var map1 = new Map([["A", 2], ["B", 3], ["C", 4]]);

var map2 = mapMap(map1, v => v * v);

console.log(map1, map2);
/*
Map { A → 2, B → 3, C → 4 }
Map { A → 4, B → 9, C → 16 }
*/

Using Array.from I wrote a Typescript function that maps the values:

function mapKeys<T, V, U>(m: Map<T, V>, fn: (this: void, v: V) => U): Map<T, U> {
  function transformPair([k, v]: [T, V]): [T, U] {
    return [k, fn(v)]
  }
  return new Map(Array.from(m.entries(), transformPair));
}

const m = new Map([[1, 2], [3, 4]]);
console.log(mapKeys(m, i => i + 1));
// Map { 1 => 3, 3 => 5 }

Actually you can still have a Map with the original keys after converting to array with Array.from. That's possible by returning an array, where the first item is the key, and the second is the transformed value.

const originalMap = new Map([
  ["thing1", 1], ["thing2", 2], ["thing3", 3]
]);

const arrayMap = Array.from(originalMap, ([key, value]) => {
    return [key, value + 1]; // return an array
});

const alteredMap = new Map(arrayMap);

console.log(originalMap); // Map { 'thing1' => 1, 'thing2' => 2, 'thing3' => 3 }
console.log(alteredMap);  // Map { 'thing1' => 2, 'thing2' => 3, 'thing3' => 4 }

If you don't return that key as the first array item, you loose your Map keys.