I thought I would have a go myself without copying (although having read
them earlier) the other functions and this is what I came up with:
(first (reduce (fn [[results seen] item]
(let [occurrences ((fnil identity 0) (get seen item))
seen (assoc seen item (inc occurrences))
item (if (> occurrences 0) (format-fn item
occurrences) item)]
[(conj results item) seen])) [[] {}] (seq items)))
This doesn't solve the problem you mention, but baby steps.
Being really anal I could claim the original a_2 should remain a_2 and the
third instance of a jump to being a_3. I thought about this and couldn't
see how to do this with reduce because I really want to say "oh, I have a
new name, recurse into the function again with the new proposed name", i.e.
loop the generation of the proposed name until it is unique, but I haven't
got that far yet (without potentially blowing the stack!)
Then I saw your 'recur' used outside of a loop which points the way...
Thanks!
On Friday, 10 January 2014 20:16:28 UTC, puzzler wrote:
>
>
> On Fri, Jan 10, 2014 at 11:52 AM, Colin Yates <[email protected]<javascript:>
> > wrote:
> > way to take the wind out of our sails! Well spotted :).
>
>
> It's not too hard to fix. Here's an adapted version of Jonas' solution
> that should do the trick:
>
> (defn uniqueify [items]
> (first
> (reduce (fn [[results count-map] item]
> (let [n (count-map item 0)]
> (if (zero? n)
> [(conj results item) (assoc count-map item (inc n))]
> (recur [results (assoc count-map item (inc n))]
> (str item \_ n)))))
> [[] {}]
> items)))
>
> => (uniqueify ["a" "a" "a" "a" "b" "a_2" "a_3" "a_3_1" "a_3_1" "a"])
> ["a" "a_1" "a_2" "a_3" "b" "a_2_1" "a_3_1" "a_3_1_1" "a_3_1_2" "a_4"]
>
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