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  1. Practice
  2. Functional Programming
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  4. Filter Array
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Filter Array

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  • Darshit_Shah + 0 comments

    Another solution approach in Scala using for expression:

    for (i <- arr if i < delim) yield i

  • PitrDubovich + 0 comments

    In Scala, assuming we're not able to use filter, we could instead use the fact that flatMap will filter out values of None.

    arr.flatMap(a => if (a < delim) Some(a) else None)

  • nevertiree + 4 comments

    Haskell

    f n arr = filter (< n) arr
    • har33sh + 0 comments

      f n arr = [ a | a <- arr , a < n ]

    • gabrielhuff + 0 comments

      f = filter . (>)

    • matharumanpreet + 1 comment

      anytime you have same dangling arguments at the end on both sides, you can get rid of it and be point free, like this

      f n = filter (< n)

      • M_F_H + 0 comments

        matter of taste, whether you prefer "f(x) = sin(x)" or "f = sin"... But anyway, the exercise is to write your own filter function and not use the built-in one.

    • M_F_H + 0 comments
      [deleted]

  • metaculus + 0 comments

    Here is how to read the stdin for this question in Ocaml:

    let f n arr = 
    (* implementation *)
    
let () =
    (* first line is parsed as the first arg to f *)
    let num = int_of_string (read_line ()) in
    
    (* the rest until EOF are parsed into an array for second arg *)
    f num (
        let rec aux arr =
            try
                let input = read_line () in
                    match input with
                    | "" -> arr
                    | a  -> aux (arr @ [int_of_string a])
            with End_of_file -> arr
        in aux [] )

  • jb_gn + 0 comments

    I decided I didn't want to use the Enum library at all and implemented my Elixir solution using pure naive comprehensions. One of the interesting things about recursion is that the list is represented backwards in memory immediately after make_list is run, but since filter walks the list again and places the last element in the first position, everything comes back out in the correct order once it is time to print :D

    defmodule NaiveList do
  def print([]), do:
    nil
  def print([value | rest]) do
    IO.puts(value)
    print(rest)
  end

  def filter(limit, list), do:
    filter(limit, list, [])
  def filter(_limit, [], output), do:
    output
  def filter(limit, [value | rest], output) when value < limit, do:
    filter(limit, rest, [value | output])
  def filter(limit, [_value| rest], output), do:
    filter(limit, rest, output)

  def make_list(), do:
    make_list(IO.gets(""), [])
  defp make_list(""<>number, list), do:
    make_list(IO.gets(""), [(number |> String.trim |> String.to_integer) | list])
  defp make_list(:eof, list), do:
    list
end

n = IO.gets("") |> String.trim |> String.to_integer
list = NaiveList.make_list()
NaiveList.filter(n, list) |> NaiveList.print
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