Hi Benny,

Your solution is much more elegant and flexible as my hacking of the
core function!

How would you implement 'pad-padding' (I like this name) if every item
of the padding should be repeated in order, not only the last one?

(defn pad-padding [padding]
  (concat padding (repeat padding)))

does not work because of the additional parentheses around 'padding'.

Is there a function that circles through the elements of a coll such
as:

(take 10 (circle [1 2 3]))
;; (1 2 3 1 2 3 1 2 3 1)

Stefan

On Nov 23, 3:17 pm, Benny Tsai <benny.t...@gmail.com> wrote:
> Since (repeat) returns a lazy sequence, and partition will only take
> as many as is needed, I think you don't even have to specify how many
> times to repeat:
>
> user=> (partition 3 3 (repeat "a") [1 2 3 4 5 6 7])
> ((1 2 3) (4 5 6) (7 "a" "a"))
>
> And if the desired behavior is to repeat the last element of provided
> padding as many times as necessary, you could do it with a little
> helper function:
>
> (defn pad-padding [padding]
>   (concat padding (repeat (last padding))))
>
> user=> (partition 4 4 (pad-padding ["a" "b"]) [1 2 3 4 5 6 7 8 9])
> ((1 2 3 4) (5 6 7 8) (9 "a" "b" "b"))
>
> On Nov 23, 12:05 am, Meikel Brandmeyer <m...@kotka.de> wrote:
>
>
>
>
>
>
>
> > Hi,
>
> > while not having looked at your code, partition does what you want:
>
> > user=> (partition 3 3 (repeat 3 "a") [1 2 3 4 5 6 7])
> > ((1 2 3) (4 5 6) (7 "a" "a"))
> > user=> (partition 3 3 (repeat 3 "a") [1 2 3 4 5 6 7 8])
> > ((1 2 3) (4 5 6) (7 8 "a"))
> > user=> (partition 3 3 (repeat 3 "a") [1 2 3 4 5 6 7 8 9])
> > ((1 2 3) (4 5 6) (7 8 9))
> > user=> (partition 3 3 (repeat 3 "a") [1 2 3 4 5 6 7 8 9 10])
> > ((1 2 3) (4 5 6) (7 8 9) (10 "a" "a"))
>
> > (Note: the 3 in repeat is not strictily necessary. 2 would be
> > sufficient)
>
> > Sincerely
> > Meikel

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