Y'know, I'm guilty of not looking into flatland/useful enough. I'll have to
try out codox on it, and see if that makes it easier for me to get
acquainted with it.
Thanks, Alex.
On Wednesday, March 19, 2014 6:57:27 AM UTC-4, Alex Nixon wrote:
Or use glue from flatland.useful.seq
Brilliant. Thanks, Shantanu!
On Wednesday, March 19, 2014 12:09:56 AM UTC-4, Shantanu Kumar wrote:
Something like this?
(defn x [1 3 4 5 7 9 10 13])
(reduce (fn [a i] (let [y (last a) z (last y)] (if (and z (= (inc z) i))
(conj (pop a) (conj y i)) (conj a [i] [] x)
Shantanu
On
Or use glue from flatland.useful.seq https://github.com/flatland/useful
and:
user= (glue conj [] (fn [s n] (= (inc (last s)) n)) [1 3 4 5 7 9 10 13])
([1] [3 4 5] [7] [9 10] [13])
On 19 March 2014 04:10, Shantanu Kumar kumar.shant...@gmail.com wrote:
On Wednesday, 19 March 2014 09:39:56
If you've got a sorted list of numbers, for example:
[1 3 4 5 7 9 10 13]
where some are consecutive, how can you pull out the consecutive runs? That
is, either produce
[1 [3 4 5] 7 [9 10] 13]; or maybe something like
[[1 7 13] [3 4 5] [9 10]] ; (the first vec is the elements
Something like this?
(defn x [1 3 4 5 7 9 10 13])
(reduce (fn [a i] (let [y (last a) z (last y)] (if (and z (= (inc z) i))
(conj (pop a) (conj y i)) (conj a [i] [] x)
Shantanu
On Wednesday, 19 March 2014 08:26:43 UTC+5:30, John Gabriele wrote:
If you've got a sorted list of numbers, for
On Wednesday, 19 March 2014 09:39:56 UTC+5:30, Shantanu Kumar wrote:
Something like this?
(defn x [1 3 4 5 7 9 10 13])
Sory for the typo. Should be (def x [1 3 4 5 7 9 10 13])
(reduce (fn [a i] (let [y (last a) z (last y)] (if (and z (= (inc z) i))
(conj (pop a) (conj y i)) (conj a
