I agree with number 2: it would be very nice to have this in contrib. I needed it last month and rolled my own (less clean) version.
Thomas On Sep 10, 10:26 pm, Matt Smith <m0sm...@gmail.com> wrote: > problem: convert a collection [1 2 0 1 2 3 0 1 2 3 0 0 1 2] into > partitions like: > ((1 2) (0 1 2 3) (0 1 2 3) (0) (0 1 2)) > In this case, start each partition on a 0. > > I looked at the various partition functions but none of them would do > the trick without adding unnecessary complexity. Instead I wrote a > new function based on partition-by: > > Solution: > (defn partition-starting-every > "Partition the sequence starting each partition when the f is true." > [f coll] > (if-let [coll (seq coll)] > (let [p (cons (first coll) (take-while (complement f) (rest > coll)))] > (lazy-seq (cons p (partition-starting-every f (drop (count p) > coll))))))) > > user=>(partition-starting-every zero? [1 2 0 1 2 3 0 1 2 3 0 0 1 2]) > ((1 2) (0 1 2 3) (0 1 2 3) (0) (0 1 2)) > > Questions: > 1 - Is there a simpler way to do this using existing partition > functions? > 2 - If not, is this something people are interested in having > contributed? > > In looking at the partition functions they are very similar. Maybe > there is a why to combine them in to a single function. -- 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
