Proper way to make array from integer division in Ruby












1














tl;dr I want to make an array from the division by 5 results:



20 => [5,5,5,5]

16 => [5,5,5,1]

7  => [5,2]


My current implementation is straightforward yet too large. How can I make it simpler and shorter? 



  max_count = 5

  total_count = input_value



  count_array = 

  div = total_count / max_count

  mod = total_count % max_count

  div.times { count_array << max_count }

  count_array << mod unless mod == 0





arrays ruby integer







share|improve this question




















  • 2




    16 => [5,5,5,1]
    – Sagar Pandya
    Nov 22 '18 at 21:27






  • 1




    20.divmod(5) #=> [4, 0]. Contains the all the information you require i.e. the quotient and the remainder.
    – Sagar Pandya
    Nov 22 '18 at 21:35
















1














tl;dr I want to make an array from the division by 5 results:



20 => [5,5,5,5]

16 => [5,5,5,1]

7  => [5,2]


My current implementation is straightforward yet too large. How can I make it simpler and shorter? 



  max_count = 5

  total_count = input_value



  count_array = 

  div = total_count / max_count

  mod = total_count % max_count

  div.times { count_array << max_count }

  count_array << mod unless mod == 0





arrays ruby integer







share|improve this question




















  • 2




    16 => [5,5,5,1]
    – Sagar Pandya
    Nov 22 '18 at 21:27






  • 1




    20.divmod(5) #=> [4, 0]. Contains the all the information you require i.e. the quotient and the remainder.
    – Sagar Pandya
    Nov 22 '18 at 21:35














1












1








1


1





tl;dr I want to make an array from the division by 5 results:



20 => [5,5,5,5]

16 => [5,5,5,1]

7  => [5,2]


My current implementation is straightforward yet too large. How can I make it simpler and shorter? 



  max_count = 5

  total_count = input_value



  count_array = 

  div = total_count / max_count

  mod = total_count % max_count

  div.times { count_array << max_count }

  count_array << mod unless mod == 0





arrays ruby integer







share|improve this question















tl;dr I want to make an array from the division by 5 results:



20 => [5,5,5,5]

16 => [5,5,5,1]

7  => [5,2]


My current implementation is straightforward yet too large. How can I make it simpler and shorter? 



  max_count = 5

  total_count = input_value



  count_array = 

  div = total_count / max_count

  mod = total_count % max_count

  div.times { count_array << max_count }

  count_array << mod unless mod == 0





arrays ruby integer




arrays ruby integer






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 23 '18 at 10:22









Stefan

75.2k894141




75.2k894141










asked Nov 22 '18 at 20:45









laechoppe

14110




14110








  • 2




    16 => [5,5,5,1]
    – Sagar Pandya
    Nov 22 '18 at 21:27






  • 1




    20.divmod(5) #=> [4, 0]. Contains the all the information you require i.e. the quotient and the remainder.
    – Sagar Pandya
    Nov 22 '18 at 21:35














  • 2




    16 => [5,5,5,1]
    – Sagar Pandya
    Nov 22 '18 at 21:27






  • 1




    20.divmod(5) #=> [4, 0]. Contains the all the information you require i.e. the quotient and the remainder.
    – Sagar Pandya
    Nov 22 '18 at 21:35








2




2




16 => [5,5,5,1]
– Sagar Pandya
Nov 22 '18 at 21:27




16 => [5,5,5,1]
– Sagar Pandya
Nov 22 '18 at 21:27




1




1




20.divmod(5) #=> [4, 0]. Contains the all the information you require i.e. the quotient and the remainder.
– Sagar Pandya
Nov 22 '18 at 21:35




20.divmod(5) #=> [4, 0]. Contains the all the information you require i.e. the quotient and the remainder.
– Sagar Pandya
Nov 22 '18 at 21:35












4 Answers
4






active

oldest

votes


















4















  1. You don't need total_count.


  2. div.times { count_array << max_count } is [max_count] * count_array

  3. Using splat, we can simplify it further




max_count = 5



[*[max_count] * (input_value / max_count), input_value % max_count] - [0]


Alternatively, using divmod



max_count = 5



n, mod = input_value.divmod(max_count)

[*[max_count] * n, mod] - [0]


Last line can also be written as:



(Array.new(n) { max_count } << mod) - [0]


or as Stefan suggested in the comment, using Numeric#nonzero?:



Array.new(n, max_count).push(*mod.nonzero?)





share|improve this answer























  • ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
    – Cary Swoveland
    Nov 22 '18 at 22:11






  • 1




    I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
    – Stefan
    Nov 23 '18 at 9:09












  • @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
    – Marcin Kołodziej
    Nov 23 '18 at 9:12



















0














One option more:



d = 5

n = 24

Array.new(n/d){d}.tap{ |a| a << n%d if (n%d).nonzero? }

#=> [5, 5, 5, 5, 4]





share|improve this answer



















  • 1




    @MarcinKołodziej, fixed thanks.
    – iGian
    Nov 23 '18 at 9:14



















0














You can try this as well.



max=5

num=48

q, r=num.divmod(max) # => [9, 3]

Array.new.fill(max, 0, q).push(r.nonzero?).compact

# => [5, 5, 5, 5, 5, 5, 5, 5, 5, 3]





share|improve this answer



















  • 1




    This might result in an array ending with 0 (which the OP doesn't want)
    – Stefan
    Nov 23 '18 at 10:21










  • Just realized that :P.Thanks @Stefan.
    – Wysiati
    Nov 23 '18 at 10:23



















0














What about this?



[20].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 5]

[16].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 1]

[7] .tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 2]





share|improve this answer





















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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4















    1. You don't need total_count.


    2. div.times { count_array << max_count } is [max_count] * count_array

    3. Using splat, we can simplify it further




    max_count = 5
    

    
[*[max_count] * (input_value / max_count), input_value % max_count] - [0]


    Alternatively, using divmod



    max_count = 5
    

    
n, mod = input_value.divmod(max_count)
    
[*[max_count] * n, mod] - [0]


    Last line can also be written as:



    (Array.new(n) { max_count } << mod) - [0]


    or as Stefan suggested in the comment, using Numeric#nonzero?:



    Array.new(n, max_count).push(*mod.nonzero?)





    share|improve this answer























    • ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
      – Cary Swoveland
      Nov 22 '18 at 22:11






    • 1




      I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
      – Stefan
      Nov 23 '18 at 9:09












    • @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
      – Marcin Kołodziej
      Nov 23 '18 at 9:12
















    4















    1. You don't need total_count.


    2. div.times { count_array << max_count } is [max_count] * count_array

    3. Using splat, we can simplify it further




    max_count = 5
    

    
[*[max_count] * (input_value / max_count), input_value % max_count] - [0]


    Alternatively, using divmod



    max_count = 5
    

    
n, mod = input_value.divmod(max_count)
    
[*[max_count] * n, mod] - [0]


    Last line can also be written as:



    (Array.new(n) { max_count } << mod) - [0]


    or as Stefan suggested in the comment, using Numeric#nonzero?:



    Array.new(n, max_count).push(*mod.nonzero?)





    share|improve this answer























    • ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
      – Cary Swoveland
      Nov 22 '18 at 22:11






    • 1




      I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
      – Stefan
      Nov 23 '18 at 9:09












    • @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
      – Marcin Kołodziej
      Nov 23 '18 at 9:12














    4












    4








    4







    1. You don't need total_count.


    2. div.times { count_array << max_count } is [max_count] * count_array

    3. Using splat, we can simplify it further




    max_count = 5
    

    
[*[max_count] * (input_value / max_count), input_value % max_count] - [0]


    Alternatively, using divmod



    max_count = 5
    

    
n, mod = input_value.divmod(max_count)
    
[*[max_count] * n, mod] - [0]


    Last line can also be written as:



    (Array.new(n) { max_count } << mod) - [0]


    or as Stefan suggested in the comment, using Numeric#nonzero?:



    Array.new(n, max_count).push(*mod.nonzero?)





    share|improve this answer















    1. You don't need total_count.


    2. div.times { count_array << max_count } is [max_count] * count_array

    3. Using splat, we can simplify it further




    max_count = 5
    

    
[*[max_count] * (input_value / max_count), input_value % max_count] - [0]


    Alternatively, using divmod



    max_count = 5
    

    
n, mod = input_value.divmod(max_count)
    
[*[max_count] * n, mod] - [0]


    Last line can also be written as:



    (Array.new(n) { max_count } << mod) - [0]


    or as Stefan suggested in the comment, using Numeric#nonzero?:



    Array.new(n, max_count).push(*mod.nonzero?)






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 23 '18 at 9:14

























    answered Nov 22 '18 at 21:05









    Marcin Kołodziej

    4,256315




    4,256315












    • ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
      – Cary Swoveland
      Nov 22 '18 at 22:11






    • 1




      I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
      – Stefan
      Nov 23 '18 at 9:09












    • @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
      – Marcin Kołodziej
      Nov 23 '18 at 9:12


















    • ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
      – Cary Swoveland
      Nov 22 '18 at 22:11






    • 1




      I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
      – Stefan
      Nov 23 '18 at 9:09












    • @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
      – Marcin Kołodziej
      Nov 23 '18 at 9:12
















    ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
    – Cary Swoveland
    Nov 22 '18 at 22:11




    ...or ([max_count] * n).tap { |arr| arr << mod if mod > 0 }. rem may be a better name than mod.
    – Cary Swoveland
    Nov 22 '18 at 22:11




    1




    1




    I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
    – Stefan
    Nov 23 '18 at 9:09






    I'd use Array.new(n, max_count).push(*mod.nonzero?) – nonzero? returns the receiver if it's not zero or nil otherwise.
    – Stefan
    Nov 23 '18 at 9:09














    @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
    – Marcin Kołodziej
    Nov 23 '18 at 9:12




    @Stefan that's really nice, I did not know about nonzero?, today I learned :) I'll edit and add it.
    – Marcin Kołodziej
    Nov 23 '18 at 9:12













    0














    One option more:



    d = 5
    
n = 24
    
Array.new(n/d){d}.tap{ |a| a << n%d if (n%d).nonzero? }
    
#=> [5, 5, 5, 5, 4]





    share|improve this answer



















    • 1




      @MarcinKołodziej, fixed thanks.
      – iGian
      Nov 23 '18 at 9:14
















    0














    One option more:



    d = 5
    
n = 24
    
Array.new(n/d){d}.tap{ |a| a << n%d if (n%d).nonzero? }
    
#=> [5, 5, 5, 5, 4]





    share|improve this answer



















    • 1




      @MarcinKołodziej, fixed thanks.
      – iGian
      Nov 23 '18 at 9:14














    0












    0








    0






    One option more:



    d = 5
    
n = 24
    
Array.new(n/d){d}.tap{ |a| a << n%d if (n%d).nonzero? }
    
#=> [5, 5, 5, 5, 4]





    share|improve this answer














    One option more:



    d = 5
    
n = 24
    
Array.new(n/d){d}.tap{ |a| a << n%d if (n%d).nonzero? }
    
#=> [5, 5, 5, 5, 4]






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 23 '18 at 9:11

























    answered Nov 23 '18 at 4:53









    iGian

    3,2702622




    3,2702622








    • 1




      @MarcinKołodziej, fixed thanks.
      – iGian
      Nov 23 '18 at 9:14














    • 1




      @MarcinKołodziej, fixed thanks.
      – iGian
      Nov 23 '18 at 9:14








    1




    1




    @MarcinKołodziej, fixed thanks.
    – iGian
    Nov 23 '18 at 9:14




    @MarcinKołodziej, fixed thanks.
    – iGian
    Nov 23 '18 at 9:14











    0














    You can try this as well.



    max=5
    
num=48
    
q, r=num.divmod(max) # => [9, 3]
    
Array.new.fill(max, 0, q).push(r.nonzero?).compact
    
# => [5, 5, 5, 5, 5, 5, 5, 5, 5, 3]





    share|improve this answer



















    • 1




      This might result in an array ending with 0 (which the OP doesn't want)
      – Stefan
      Nov 23 '18 at 10:21










    • Just realized that :P.Thanks @Stefan.
      – Wysiati
      Nov 23 '18 at 10:23
















    0














    You can try this as well.



    max=5
    
num=48
    
q, r=num.divmod(max) # => [9, 3]
    
Array.new.fill(max, 0, q).push(r.nonzero?).compact
    
# => [5, 5, 5, 5, 5, 5, 5, 5, 5, 3]





    share|improve this answer



















    • 1




      This might result in an array ending with 0 (which the OP doesn't want)
      – Stefan
      Nov 23 '18 at 10:21










    • Just realized that :P.Thanks @Stefan.
      – Wysiati
      Nov 23 '18 at 10:23














    0












    0








    0






    You can try this as well.



    max=5
    
num=48
    
q, r=num.divmod(max) # => [9, 3]
    
Array.new.fill(max, 0, q).push(r.nonzero?).compact
    
# => [5, 5, 5, 5, 5, 5, 5, 5, 5, 3]





    share|improve this answer














    You can try this as well.



    max=5
    
num=48
    
q, r=num.divmod(max) # => [9, 3]
    
Array.new.fill(max, 0, q).push(r.nonzero?).compact
    
# => [5, 5, 5, 5, 5, 5, 5, 5, 5, 3]






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 23 '18 at 10:28

























    answered Nov 23 '18 at 10:16









    Wysiati

    1044




    1044








    • 1




      This might result in an array ending with 0 (which the OP doesn't want)
      – Stefan
      Nov 23 '18 at 10:21










    • Just realized that :P.Thanks @Stefan.
      – Wysiati
      Nov 23 '18 at 10:23














    • 1




      This might result in an array ending with 0 (which the OP doesn't want)
      – Stefan
      Nov 23 '18 at 10:21










    • Just realized that :P.Thanks @Stefan.
      – Wysiati
      Nov 23 '18 at 10:23








    1




    1




    This might result in an array ending with 0 (which the OP doesn't want)
    – Stefan
    Nov 23 '18 at 10:21




    This might result in an array ending with 0 (which the OP doesn't want)
    – Stefan
    Nov 23 '18 at 10:21












    Just realized that :P.Thanks @Stefan.
    – Wysiati
    Nov 23 '18 at 10:23




    Just realized that :P.Thanks @Stefan.
    – Wysiati
    Nov 23 '18 at 10:23











    0














    What about this?



    [20].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 5]
    
[16].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 1]
    
[7] .tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 2]





    share|improve this answer


























      0














      What about this?



      [20].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 5]
      
[16].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 1]
      
[7] .tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 2]





      share|improve this answer
























        0












        0








        0






        What about this?



        [20].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 5]
        
[16].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 1]
        
[7] .tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 2]





        share|improve this answer












        What about this?



        [20].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 5]
        
[16].tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 5, 5, 1]
        
[7] .tap{|a| a.push(5, a.pop - 5) while a.last > 5} # => [5, 2]






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 26 '18 at 6:54









        sawa

        129k27197299




        129k27197299






























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            Sphinx de Gizeh

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            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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