Any way to handle duplicates?

[barbazul]

I have a set of elements with possible duplication.

I need to get all possible permutations of combinations but removing duplicates.

Like so:

const all = ['a', 'a', 'b'];
const cmb = Combinatorics.permutationCombination(all);
console.log(cmb.toArray());

Expected output is:

[ [ 'a' ],
  [ 'b' ],
  [ 'a', 'a' ],
  [ 'a', 'b' ],
  [ 'b', 'a' ],
  [ 'a', 'a', 'b' ],
  [ 'a', 'b', 'a' ],
  [ 'b', 'a', 'a' ] ]

but instead, I am getting

[ [ 'a' ],
  [ 'a' ],
  [ 'b' ],
  [ 'a', 'a' ],
  [ 'a', 'a' ],
  [ 'a', 'b' ],
  [ 'b', 'a' ],
  [ 'a', 'b' ],
  [ 'b', 'a' ],
  [ 'a', 'a', 'b' ],
  [ 'a', 'b', 'a' ],
  [ 'a', 'a', 'b' ],
  [ 'a', 'b', 'a' ],
  [ 'b', 'a', 'a' ],
  [ 'b', 'a', 'a' ] ]

So far I have been able to work around this with a control list and checking while iterating. I could refactor this into a lazyFilter that checks such list, but it would pretty much be the same as I still need to externally keep track of the already visited permutations.

Any way to keep this contained within the library?

[barbazul]

An example of external control list would be something like this:

const checked = [];
let controlHash = '';
let element;

while (element = cmb.next()) {
    controlHash = element.join('');
    if (checked.indexOf(controlHash) === -1) {
        console.log(element);
        checked.push(controlHash);
    }
}

As you can see it adds quite a bit to a simple code that should simply iterate a precalculated list

[blanchg]

Why don't you create your input as a <code>Set</code>?

…

On Sat., 29 Dec. 2018, 6:04 am Barbazul ***@***.*** wrote: An example of external control list would be something like this: const checked = [];let controlHash = '';let element; while (element = cmb.next()) { controlHash = element.join(''); if (checked.indexOf(controlHash) === -1) { console.log(element); checked.push(controlHash); } } As you can see it adds quite a bit to a simple code that should simply iterate a precalculated list — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#56 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AAFR38akwu0E0fHKufbcHUJ1OWzvPs6Pks5u9nlTgaJpZM4ZkLXD> .