Yet another way:
(reduce #(let [[k [v1 v2]] %2] (vec (concat % [[k v1]] [[k v2]]))) [] [[:a [1
2]] [:b [3 4]]])
On Jun 8, 2014, at 3:43 PM, boz wrote:
> Thanks for the replies!!
>
> I goofed and only gave part of my loopy solution ...
> here's the whole thing
>
> (defn do-one [id v]
Wow! 'for' is a powerful thing!!
This is perfect!
On Sunday, June 8, 2014 12:32:44 PM UTC-7, François Rey wrote:
>
> (for [[k v] [[:a [1 2]] [:b [3 4]]]
> n v]
> [k n])
>
>
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Thanks for the replies!!
I goofed and only gave part of my loopy solution ...
here's the whole thing
(defn do-one [id v]
(loop [a (first v) r (rest v) result []]
(if a
(recur (first r) (rest r) (conj result [id a]))
result)))
(defn do-all [x]
(loop [a (first x) b (rest x) re
If you really want a vector, just wrap my answer with (into [] ...):
(into[]
(for[[k v] [[:a [1 2]] [:b [3 4]]]
n v]
[k n]))
=> [[:a 1] [:a 2] [:b 3] [:b 4]]
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On Jun 8, 2014, at 3:12 PM, boz wrote:
> Is there a better way to take this...
>
> [[:a [1 2]]
>[:b [3 4]]]
>
> and convert it to this...
>
> [[:a 1]
>[:a 2]
>[:b 3]
>[:b 4]]
>
> than using a loop like this...
>
> (defn doit [id v]
>
(mapcat (fn [[k coll]] (map vector (repeat k) coll))
[[:a [1 2]] [:b [3 4]]])
=> '([:a 1] [:a 2] [:b 3] [:b 4])
On Sunday, June 8, 2014 2:12:19 PM UTC-5, boz wrote:
>
> Is there a better way to take this...
>
> [[:a [1 2]]
> [:b [3 4]]]
>
> and convert it to this...
>
> [[:a 1]
>
(for[[k v] [[:a [1 2]] [:b [3 4]]]
n v]
[k n])
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