On Thu, Oct 21, 2021 at 5:26 AM George Neuner <[email protected]> wrote:
> > On 10/20/2021 5:53 PM, unlimitedscolobb wrote: > > > You can get a lot of mileage out of the 'set' datatype, which removes ordering from the equation, especially since lists will act as sets in a pinch. (When you want to improve performance, use 'in-set' and/or 'list->set'. To see if 2 hashes contain the same set of keys: > > (and (= (hash-count hash1) (hash-count hash2)) > (for/and ([k (in-list (hash-keys hash1))]) > (hash-has-key? hash2 k))) > Alternatively: (set=? (hash-keys hash1) (hash-keys hash2)) ; Return an unordered list of the keys that are in hash1 but not in hash2 (set-subtract (hash-keys hash1) (hash-keys hash2)) ; Get a new hash consisting of the key/values that are in hash1 but not in hash2 (for/hash ([k (set-subtract (hash-keys hash1) (hash-keys hash2))]) (values k (hash-ref hash1 k))) ; Get a ore detailed breakdown: (require handy) (define hash1 (for/hash ([k '(a b c d e f g)] [v 10]) (values k v))) (define hash2 (for/hash ([k '(a b c d e z y)] [v 10]) (values k v))) (define hash3 (hash-set* hash2 'c 111 'd 184)) (disjunction hash1 hash3) Result: (dict-disjunction '#hash((c . (2 111)) (d . (3 184))); values that differ between the hashes '#hash((f . 5) (g . 6)) ; key/values that exist only in hash1 '#hash((y . 6) (z . 5)) ; key/values that exist only in hash3 '#hash((a . 0) (b . 1) (c . 2) (d . 3) (e . 4) (f . 5) (g . 6)) ; hash1 '#hash((a . 0) (b . 1) (c . 111) (d . 3) (e . 4) (y . 6) (z . 5))) ; hash3 > Unfortunately, there is no variant of "for" that creates mutable hashes. > But the general form works for anything. > If you don't mind inefficiency then handy can be, well, handy: (define imm-h (for/hash ([k '(a b c)][v 3]) (values k v))) (immutable? imm-h) (immutable? (hash->mutable imm-h)) hash->mutable takes an existing hash, which can be either immutable or mutable, and adds its key/values to a new mutable hash one by one, then returns that hash. The handy module is a bit of a Fibber McGee that really needs to be broken out. It's thoroughly documented, but unfortunately only in comments. Converting that to proper scribble is one of my Copious Free Times projects. -- You received this message because you are subscribed to the Google Groups "Racket Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/racket-users/CAE8gKoeZWCPg4yS41L9jPeL7gZzgW1iLusr3xJJYQTsj6491Xg%40mail.gmail.com.
