He's another (longer) way
(vec (filter identity
(map
(fn [x idx] (if (zero? (mod 3 idx)) x))
[2 4 5 8 6 4]
(iterate inc 0))))
I'd use a form of partition production, though. Some thing like this:
(apply concat (partition-all 2 3 [2 4 5 8 6 4]))
The concat should be faster than flatten, because it doesn't have to
do a seqable? check.
Sean
On Dec 3, 12:42 pm, Mark N <nels2...@gmail.com> wrote:
> You might want to use partition-all from seq-utils, since partition
> may drop the end elements if it can't make a full size partition.
>
> For example
> user=> (flatten (partition 2 3 [2 4 5 8]))
> (2 4)
>
> But
> user=> (flatten (partition-all 2 3 [2 4 5 8]))
> (2 4 8)
>
> On Dec 1, 9:31 pm, Wilson MacGyver <wmacgy...@gmail.com> wrote:
>
> > you can do it using partition and flatten from clojure.contrib.seq-utils
>
> > (use 'clojure.contrib.seq-utils)
> > (flatten (partition 2 3 [2 4 5 8 6 4]))
>
> > this yields (2 4 8 6)
>
> > On Tue, Dec 1, 2009 at 10:06 PM, Don <josereyno...@gmail.com> wrote:
> > > I have a vector [2 4 5 8 6 4]
>
> > > And I want to remove a value based on index. Specifically, I want to
> > > remove every third item.
>
> > > So my new vector would be [2 4 8 6]
>
> > > Thank you.
>
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