Andy
Thanks. Makes complete sense when explained!
Thanks for the speedy reply.
On Oct 2, 6:27 pm, Andy Fingerhut <andy.finger...@gmail.com> wrote:
> user> (def a ["a" "a" "a" "b" "c" "c" "a"])
> #'user/a
> user> (pack a)
> (("a" "a" "a") ("b") ("c" "c") ("a"))
>
> Now calling (map f (pack a)) will cause the function you give as the
> first argument of map to be called with each of the elements of the
> collection (pack a) in turn. Those elements are, in order:
>
> ("a" "a" "a") a list of 3 elements
> ("b") a list of 1 element (note: it is not the same as "b" by itself)
> ("c" "c") a list of 2 elements
> ("a") a list of 1 element
>
> Then each of the return values of those function calls will be put
> into a list (or more precisely, a lazily generated list).
>
> With your failing version of encodemodifed, when it is called with the
> list ("b"), (count %) is equal to 1, so it returns its argument, which
> is the list ("b").
>
> Andy
>
> On Oct 2, 2010, at 9:26 AM, swheeler wrote:
>
>
>
>
>
>
>
> > (defn pack
> > [col]
> > (partition-by identity
> > col))
>
> > (defn encodemodified
> > [col]
> > (map #(if (== (count %) 1) % (list (count %) (first
> > %)))
> > (pack col)))
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