On 29 March 2012 21:41, Cedric Greevey <cgree...@gmail.com> wrote:
>
> On Thu, Mar 29, 2012 at 4:18 PM, David Jagoe <davidja...@gmail.com> wrote:
> > Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2]
> >
> > partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)]
> >!
>
> (defn partition-distinct [s]
> (lazy-seq
> (loop [seen #{} s (seq s) part []]
> (if s
> (let [[f & r] s]
> (if (seen f)
> (cons (seq part) (partition-distinct s))
> (recur (conj seen f) r (conj part f))))
> (if-let [part (seq part)]
> (list part))))))
>
Thanks! I had a nasty feeling that it could be done in a one-liner
using partition-by or something similar. But of course you need to be
able to look at previous elements...
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