Another option is: ((fn pascal ([n] (pascal n [1M])) ([n row] (if (= n 1) row (recur (dec n) (mapv (partial reduce +) (partition 2 1 (cons 0 (conj row 0)))))))) 500)
Because row is a vector you can conj 0 to the end (quickly) and cons 0 to the front (quickly) and then mapv makes sure you get a vector back. Sean On Feb 4, 2014, at 2:39 AM, Andy Smith <the4thamig...@googlemail.com> wrote: > Ok thanks, thats really helpful. The second link suggests using doall, which > seems to do the trick : > > ((fn pascal ([n] (pascal n [1M])) ([n row] (if (= n 1) row (recur (dec n) > (map (partial reduce +) (doall (partition 2 1 (concat [0] row [0])))))))) 500) > > However you do lose the laziness, but here the laziness is not needed ... >
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