Assume I have the following storage map:

ZAccounts: StorageMap(AccountId -> AccountInfoOf<T>)

Is there a simple way to draw n AccountId's randomly. I assume that I already have a secure random source (BABE or VRF).

I mean if I use let accounts: Vec<_> = ZAccounts::<T>::iter().collect(); (to select a random subset later) I have the flaw of an unbounded vector. And additionally I have a storage read for every iteration (pretty expensive).

Then there is the possibility to have a BoundedVec. That would be easy, but I have the constraint that there is no limit (unbounded).

So maybe you know a possibility to know beforehand the key to query for a storage map based on internal storage structure with hasher.

Can I somehow calculate a key from storage and then map some index to it. Then I could map an index to a storage key. Like for example:

StorageMap(u128 -> AccountId)

It could be done like that, but what if one (index -> AccountId) is deleted. Then there are missing indices. At the end this leads to a list of active indices again (like [0, 2, 3, 4 ...] where 1 got removed for some dynamic use-case in which an AccountId should be deleted).

On Zeitgeist a court system needs to get implemented. This requires to select n random jurors from an unbounded list of possible jurors. What would you propose?

1 Answer 1


Maybe you need a StorageValue<Vec<(Index, AccountId)>>.

let v = StorageValue::get();
let index = v[random % v.len()];
let account_id = v[i];

Or StorageMap<Index, AccountId> + StorageValue<Count>.

let count = StorageValue::get();
let index = random % count;
let account_id = StorageMap::get(index);

Note you must not delete the middle item in this map(actually, it looks like a link list structure). You should use swap and delete the item from the tail (also, update the counter correctly).

  • Good thought. I would like to have an unlimited number of AccountId's. So there should be for example a million and more juror account ids.
    – Chralt
    Feb 2, 2023 at 16:24
  • How many elements in this vector could be possible? I mean there is some weight limitation with this operation, right? As soon as I access the content of the StorageValue I need to decode a large vector.
    – Chralt
    Feb 2, 2023 at 16:27
  • How about my second option? Feb 2, 2023 at 16:36
  • That's better. The interesting part is to swap and deleting the element from the tail. I mean at the moment when removing from the middle all indices need to shift (minus one for following elements).
    – Chralt
    Feb 2, 2023 at 16:41
  • You shouldn't remove it. You need to use a new account to swap the account you want to remove. Feb 2, 2023 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.