I am working quite a bit with Maps in javascript. I need the most computationally efficient way to find all items that are in Map a that are not present in Map b. For example,
const a = new Map();
a.set('item1', 'item1value');
a.set('item2', 'item2value');
const b = new Map();
b.set('item1', 'item1value');
The result of the function I'm looking to write would be another Map, with a single entry of key: item2, value: item2value.
I am well aware of the multitude of questions / answers / methods to do this with arrays and objects, however I have not seen such an explanation for maps. I need the absolute most efficient way to do so, as I will need to be call this function up to thousands of times quickly. Is conversion to an array and back to a Map the best way? Are there any tricks with Maps that may help?
No, don't convert the maps to arrays and back. Computing the difference between arrays is slow, in a map you have O(1) lookup. Just loop through the entries of a and put them into the result if you don't find an equivalent entry in b. This will have the optimal time complexity O(n) (where n is the size of the map a).
const result = new Map();
for (const [k, v] of a) {
if (v === undefined && !b.has(k) || b.get(k) !== v) {
result.set(k, v);
}
}
If you know that your map doesn't contain undefined values, you can omit the v === undefined && !b.has(k) || entirely and might get some speedup. Also, notice that if your map can contain NaN values, you'll want to use Object.is instead of ===.
If you want to write it as a single fancy expression, consider a generator:
const result = new Map(function*() {
for (const e of a) {
const [k, v] = e;
if (v === undefined && !b.has(k) || b.get(k) !== v) {
yield e;
}
}
}());
You could iterate the first map and check against the second.
const
a = new Map([['item1', 'item1'], ['item2', 'item2']]),
b = new Map([['item1', 'item1']]),
difference = (a, b) => {
const d = new Map;
a.forEach((v, k) => {
if (!b.has(k) || b.get(k) !== v) d.set(v, k);
});
return d;
}
console.log([...difference(a, b)])Perhaps like this by maintaining a;
var a = new Map(),
b = new Map(),
c;
a.set('item1', 'item1value');
a.set('item2', 'item2value');
b.set('item1', 'item1value');
c = new Map(a);
b.forEach((_,k) => c.delete(k));
console.log(c); // try in dev tools to see result