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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