Removing the 5th, 10th, 15th ... element of a list

[Cecil Westerhof]

What is the best way to remove all elements which position (counting from 1) is a multiply of five out of a list?

So the list:
    (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21)

​becomes:
    (1 2 3 4 6 7 8 9 11 12 13 14 16 17 18 19 21)​

--

Cecil Westerhof

[Timothy Baldridge]

Tweak as needed:

(keep-indexed 

              (fn [i v]

                (if (= 0 (mod i 5))

                  nil

                  v))

              (range 30))

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(Robert Firth)

[Ivan L]

this works

=> (filter #(not= (mod % 5) 0) (range 22))

(1 2 3 4 6 7 8 9 11 12 13 14 16 17 18 19 21)

[Timothy Baldridge]

True, that works, but he wants to remove based on element position, not element value. 

Timothy

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[Cecil Westerhof]

2015-02-17 20:26 GMT+01:00 Timothy Baldridge <tbald...@gmail.com>:

Tweak as needed:

(keep-indexed 

              (fn [i v]

                (if (= 0 (mod i 5))

                  nil

                  v))

              (range 30))

​I made the following:
​

 

​    (defn indexed-sieve [index this-list]
      (keep-indexed
        (fn [i v]
          (if (= 0 (mod (inc i) index))
              nil
            v))
        this-list))

The first element should not be filtered (counting from 1) and because I will use it more often I created a function.​

 

On Tue, Feb 17, 2015 at 12:21 PM, Cecil Westerhof <cldwes...@gmail.com> wrote:

What is the best way to remove all elements which position (counting from 1) is a multiply of five out of a list?

So the list:
    (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21)

​becomes:
    (1 2 3 4 6 7 8 9 11 12 13 14 16 17 18 19 21)​

--

Cecil Westerhof

[Colin Yates]

Style police would point out zero? and when-not :).

--

[Timothy Baldridge]

Clearly the answer is a program that runs every snippet posted to this mailing list through kibit (https://github.com/jonase/kibit). 

[Colin Yates]

+1

And don't forget Eastwood as well :).

On a more serious note, I found http://blog.mattgauger.com/blog/2014/09/15/clojure-code-quality-tools/ has some other great tips related to this.

[Colin Jones]

Sounds almost like a mirror image of `clojure.core/take-nth`, so something like this is kind of fun:

(defn drop-nth [n coll] 

  (lazy-seq 

    (when-let [s (seq coll)] 

      (concat (take (dec n) s) (drop-nth n (drop n s))))))

[Fluid Dynamics]

On Tuesday, February 17, 2015 at 2:21:20 PM UTC-5, Cecil Westerhof wrote:

The correct answer, obviously, is

(->> s
  (cons ::sentinel)
  (partition n n nil)
  (map next)
  (mapcat identity))

with, in this case, s = (range 1 22) and n = 5

[Ben Wolfson]

why not (mapcat next)?

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--

Ben Wolfson
"Human kind has used its intelligence to vary the flavour of drinks, which may be sweet, aromatic, fermented or spirit-based. ... Family and social life also offer numerous other occasions to consume drinks for pleasure." [Larousse, "Drink" entry]

[Fluid Dynamics]

On Tuesday, February 17, 2015 at 6:47:59 PM UTC-5, Ben wrote:

why not (mapcat next)?

Yeah, that should work too, though conceptually trimming each part and then reflattening the partition are two separate steps of the computation. :)

[Udayakumar Rayala]

Then probably (apply concat) is better than (mapcat identity) isnt it?

What is (cons ::sentinel)? Why do you need it here?

--

[Fluid Dynamics]

On Tuesday, February 17, 2015 at 10:24:29 PM UTC-5, Udayakumar Rayala wrote:

Then probably (apply concat) is better than (mapcat identity) isnt it?

What is (cons ::sentinel)? Why do you need it here?

It'll drop the wrong elements otherwise:

=> (->> (range 1 22)
     (partition 5 5 nil)
     (mapcat next))
(2 3 4 5 7 8 9 10 12 13 14 15 17 18 19 20)

All the partitions need to be shifted by one position to the left, so the elements to drop are at their left edges, if we want to use the efficient "next" instead of the evil linear-time "butlast" in the later step. But that can be done by prepending a dummy element to the input, which will conveniently go away as one of the elided elements.

[Udayakumar Rayala]

Ah ok. Got it.

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