I really like the 'partition' technique. That said, as a non-expert, I find
the recursive approach marginally easier to read:
(defn has22 [coll]
(when-let [s (seq coll)]
(or (= 2 (first s) (second s)) (recur (rest s)))))
In my microbenchmarks, the above technique runs about 5-10x faster for all
sequences. Close enough that I suspect it is largely a matter of personal
preference, but still slightly faster on my machine.
As a footnote, using destructuring instead ran only about half as fast as
the above code:
(defn has22-destructing [coll]
(when-let [[x & xs] (seq coll)]
(or (= 2 x (first xs)) (recur xs))))
Not entirely sure why that would be.
Best regards,
-DeWitt
On Sun, Jul 29, 2012 at 8:22 PM, Yoshinori Kohyama <[email protected]>wrote:
> Hi John,
>
> 'partition' will be useful for you, as Moritz pointed out.
>
> (partition 2 1 [1 2 3 4]) -> ((1 2) (2 3) (3 4))
> (partition 2 1 [1 2 2 4]) -> ((1 2) (2 2) (2 4))
> (partition 2 1 [1 2 2 2]) -> ((1 2) (2 2) (2 2))
>
> (some #(= % [2 2]) (partition 2 1 [1 2 3 4])) -> nil
> (some #(= % [2 2]) (partition 2 1 [1 2 2 4])) -> true
> (some #(= % [2 2]) (partition 2 1 [1 2 2 2])) -> true
>
> (filter #(= % [2 2]) (partition 2 1 [1 2 3 4])) -> ()
> (filter #(= % [2 2]) (partition 2 1 [1 2 2 4])) -> ((2 2))
> (filter #(= % [2 2]) (partition 2 1 [1 2 2 2])) -> ((2 2) (2 2))
>
> I'm sorry I can't recognize whether you need a pair of 2s or two pairs of
> 2s.
>
> If you need one or more pairs of 2s, do
> (defn has22 [coll] (boolean (some #(= % [2 2]) (partition 2 1 coll))))
> (has22 [1 2 3 4]) -> false
> (has22 [1 2 2 4]) -> true
> (has22 [1 2 2 2]) -> true
>
> If you need two or more pairs of 2s, do
> (defn has222 [coll] (< 1 (count (filter #(= % [2 2]) (partition 2 1
> coll)))))
> (has222 [1 2 3 4]) -> false
> (has222 [1 2 2 4]) -> false
> (has222 [1 2 2 2]) -> true
>
> Regards,
> Yoshinori Kohyama
>
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