Maybe not what you're looking for, but this is my take on this particular
problem:
(defn rem-dup [str]
(->>
str
(partition-by (partial = \space))
(map #(if (= (first %) \space) \space %))
flatten))
I find partition-by the 'weapon of choice' when a sequence contains values
that needs to be distinguished from the rest (such as spaces). This makes
the code pretty simple to understand. The partition-by statement clearly
says that spaces should be in one group, and the rest should be in another
group.
Jonathan
On Tue, May 31, 2011 at 12:15 AM, iz bash <[email protected]>wrote:
> so clojures like my first programming language. most of the time
> now ,i use map /reduce/.. with lazy sequences but sometimes i just
> cant seem to find a solution other than using loop-recur. then i read
> somewhere almost all loop recur situations can be translated into
> reduce. so ive posted a function below...how'd you implement it using
> only higher order functions?
>
>
> (defn rem-dup [stri]
> (loop [[x y & z] (seq stri) ,bui []]
> (cond (nil? x) bui
> (= x y \space) (recur (cons y z) bui)
> :else (recur (cons y z) (conj bui x)))))
>
>
> (rem-dup "aaaa bb cc") => [\a \a \a \a \space \b
> \b \space \c \c]
>
> also itd be great if you guys could give some pointers on using higher
> order functions and lazy sequences in place of a loop recur.
>
> Thanks!!
> and btw clojures totally awesome :)
>
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