If I want merge in an algorithm where it is often the case that the new map has the same keys as the old map, I'd assume using the new map as a replacement is faster in those cases.
Of course comparing the keys costs and I'd like to know which is the fastest way. Can I do (= (keys old-map) (keys new-map)) or are the orders not guaranteed? I see an alternative in (every? new-map (keys old-map)) which would also account for cases where the keysets are different but old-map would be completely replaced by the merge. Which one would you recommend, performance wise? On a side note: I noticed Clojures merge merges a map onto nil by replacing nil with an empty map. Wouldn't it be faster if it returned the map right away? Should I write a ticket? Kind regards, Leon. -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en --- You received this message because you are subscribed to the Google Groups "Clojure" group. To unsubscribe from this group and stop receiving emails from it, send an email to clojure+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
