I don't know about a standard solution, but my take is that you can
use the built-in 'frequencies' function to build up a frequency table
for the list, then filter on anything that occurs more than once:
(defn get-ties [coll]
(let [freqs (frequencies coll)]
(filter #(> (freqs %) 1) coll)))
On Nov 10, 1:28 pm, David Jacobs <develo...@allthingsprogress.com>
wrote:
> I have a sorted list, and I'd like to derive from that a list
> containing all "ties" in the original.
>
> That is, if I have (1 2 3 4 4 5 5 5 5 5 6 9 12 12), I want to get back
> the sequence (4 4 5 5 5 5 5 12 12).
>
> My first thought was to try to filter down the list, but filter takes
> each element of a list in step. One hack might be to pass the original
> list through a partition function[0], then filter that list using
> (filter #(> (count %) 1)). Seems messy, though.
>
> Is there a standard solution to this problem that I'm not aware of?
>
> Thanks,
> David
>
> [0]:http://groups.google.com/group/clojure/browse_thread/thread/f1593f492...
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