Skip to content

Instantly share code, notes, and snippets.

Embed
What would you like to do?
// Replace items in `as` with a matching `id` with the
// respective item in `bs`, otherwise append new items.
const abs = Object.fromEntries([
...as.map(a => [a.id, a]),
...bs.map(b => [b.id, b])
]);
// The 'trick' is that if you have two entries with the same id
// (in this case, implying that the value was updated within `bs`)
// then with this method the value of the right-most entry will be used.
@sebinsua
Copy link
Author

sebinsua commented Nov 6, 2022

In general, Object.fromEntries seems useful in situations in which you want to deduplicate while considering items that are right-most as superseding those to the left. We could sort with this in mind...

@sebinsua
Copy link
Author

sebinsua commented Nov 7, 2022

Another idea, might be to group by values other than id in order to "select" more aggressively (e.g. instead of removing items with the same id, we could key by something else like a hash of some of the internal values).

@sebinsua
Copy link
Author

sebinsua commented Nov 8, 2022

As for grouping:

function append(previousItems = [], element, index, array) {
  return previousItems.concat([element]);
}

function groupToMap(items, callbackFn, reduceFn = append) {
  if (!Array.isArray(items)) {
    throw new Error(`groupToMap must receive an array as its first argument.`);
  }
  if (typeof callbackFn !== "function") {
    throw new Error(
      `groupToMap must receive a function as its second argument.`
    );
  }
  if (typeof reduceFn !== "function") {
    throw new Error(
      `groupToMap must receive a function as its third argument.`
    );
  }
  return items.reduce((map, element, index, array) => {
    const key = callbackFn(element, index, array);
    return map.set(
      key,
      map.has(key) ? reduceFn(map.get(key), element, index, array) : reduceFn(undefined, element, index, array)
    );
  }, new Map());
}

function group(items, callbackFn, appendFn = append) {
  return Object.fromEntries(groupToMap(items, callbackFn, appendFn).entries());
}

Create two partitions for odd and even values containing a count of the number of items belonging to these partitions:

groupToMap(
  [1, 2, 3, 4, 5],
  (v) => (v % 2 === 0 ? "even" : "odd"),
  (previous = 0) => previous + 1
);
// Map {"odd" => 3, "even" => 2}

Create two partitions for odd and even values containing the sum of the values within these partitions:

groupToMap(
  [1, 2, 3, 4, 5],
  (v) => (v % 2 === 0 ? "even" : "odd"),
  (previous = 0, v) => previous + v
);
// Map(2) {"odd" => 9, "even" => 6}

@sebinsua
Copy link
Author

sebinsua commented Nov 9, 2022

If I get time I will implement an immutable version of Counter.

function total(map) {
  return map.values().reduce((acc, count) => acc + count, 0);
}

function mostCommon(map, n) {
  // ...
  // Not trivial to do well, so I'll come back to this... :)
}

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment