On Wed, Nov 10, 2010 at 3:28 PM, David Jacobs
<develo...@allthingsprogress.com> wrote:
> I have a sorted list, and I'd like to derive from that a list
> containing all "ties" in the original.
>
> That is, if I have (1 2 3 4 4 5 5 5 5 5 6 9 12 12), I want to get back
> the sequence (4 4 5 5 5 5 5 12 12).
>
> My first thought was to try to filter down the list, but filter takes
> each element of a list in step. One hack might be to pass the original
> list through a partition function[0], then filter that list using
> (filter #(> (count %) 1)). Seems messy, though.
>
> Is there a standard solution to this problem that I'm not aware of?
If the sequence is always sorted as in the example:
(loop [s the-input-seq o [] last nil]
(if (empty? s)
o
(let [n (first s)]
(if (or (= n (second s)) (= n last))
(recur (rest s) (conj o n) n)
(recur (rest s) o n)))))
It's non-lazy but oughta do the job. :)
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