Something like this?
(defn partition-when [f l]
(reduce #(if (f %2)
(conj %1 (vector %2))
(conj (butlast %1)
(conj (last %1) %2)))
[] l))
On Sat, Nov 27, 2010 at 4:15 PM, Ken Wesson <kwess...@gmail.com> wrote:
> On Sat, Nov 27, 2010 at 4:00 PM, rob levy <r.p.l...@gmail.com> wrote:
> > partition-by does exactly what you need.
>
> Not quite.
>
> user=> (take 10 (partition-by #(= 0 (rem % 3)) (iterate inc 1)))
> ((1 2)
> (3)
> (4 5)
> (6)
> (7 8)
> (9)
> (10 11)
> (12)
> (13 14)
> (15))
>
> At first it seems you can fix this as follows:
>
> user=> (defn questionable-split-when [pred coll]
> (let [p (partition-by pred coll)]
> (cons (first p) (map #(apply concat %) (partition 2 2 []
> (rest p))))))
> #'user/questionable-split-when
> user=> (take 10 (questionable-split-when #(= 0 (rem % 3)) (iterate inc 1)))
> ((1 2)
> (3 4 5)
> (6 7 8)
> (9 10 11)
> (12 13 14)
> (15 16 17)
> (18 19 20)
> (21 22 23)
> (24 25 26)
> (27 28 29))
>
> So far, so good. But:
>
> user=> (questionable-split-when #(= 0 (rem % 3)) [1 2 3 6 7 8 9])
> ((1 2)
> (3 6 7 8)
> (9))
>
> Whereas:
>
> user=> (split-when #(= 0 (rem % 3)) [1 2 3 6 7 8 9])
> ((1 2)
> (3)
> (6 7 8)
> (9))
>
> The latter correctly starts a new partition each time the pred is
> true; the former fails if the pred is ever true twice in a row.
>
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