Ruby: Assigning array values in a hash

Question

I would like the following hash output:

{ 0 => [0,1,2], 1 => [0,1,2], 2 => [0,1,2],
  3 => [3,4,5], 4 => [3,4,5], 5 => [3,4,5],
  6 => [6,7,8], 7 => [6,7,8], 8 => [6,7,8]
}

Intuitively I can just hard code these values in, but I am struggle to figure out how I can dynamically assign each hash key to the appropriate array value.. For instance perhaps creating a 9.times loop?

{ 0 => (0..2).to_a, 1 => (0..2).to_a, 2 => (0..2).to_a,
  3 => (3..5).to_a, 4 => (3..5).to_a, 5 => (3..5).to_a, 
  6 => (6..8).to_a, 7 => (6..8).to_a, 8 => (6..8).to_a 
}

Thanks for any guidance

Aleksei Matiushkin · Accepted Answer · 2019-02-18 16:18:37Z

2

With a dynamic upper bound:

count, split = 9, 3
0.upto(count - 1).map do |i|
  [i, [*(i / split * split)...(i / split) * split + split]]
end.to_h

answered Feb 18, 2019 at 16:18

Aleksei Matiushkin

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4 Comments

Oleksandr Holubenko Over a year ago

Very nice solution!

Tyl Over a year ago

My first time see [*a..b] usage, tried it and seems wonderful. Tried search but can't find it's document, could you give me a keyword to search it? Thanks :)

rahul mishra Over a year ago

seems to be the clean and optimised solution with O(n) complexity

Aleksei Matiushkin Over a year ago

@Tiw well, * is a splat operator that basically splats the argument. I am not sure there is a specific paper describing it’s usage inside array initialization.

ulferts · Accepted Answer · 2019-02-18 20:56:56Z

2

Using modulo operations (%) you can calculate the values of the subarrays:

(0..8).map { |i| [i, (i - i%3 .. i - i%3 + 2).to_a] }.to_h

# => {0=>[0, 1, 2], 1=>[0, 1, 2], 2=>[0, 1, 2], 3=>[3, 4, 5], 4=>[3, 4, 5], 5=>[3, 4, 5], 6=>[6, 7, 8], 7=>[6, 7, 8], 8=>[6, 7, 8]}

edited Feb 18, 2019 at 20:56

answered Feb 18, 2019 at 16:18

ulferts

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Comments

Oleksandr Holubenko · Accepted Answer · 2019-02-18 16:17:17Z

0

Maybe something like this:

[0..2, 3..5, 6..8].each.with_object({}) do |range, result|
  range.each do |ind|
    result[ind] = range.to_a
  end
end

answered Feb 18, 2019 at 16:17

Oleksandr Holubenko

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Comments

rahul mishra · Accepted Answer · 2019-02-18 16:24:02Z

0

dummy_data = {}
(0..8).to_a.each_slice(3).map{ |m| m.each{|v| dummy_data[v] = m }}

it generates expected dummy data

dummy_data
 => {0=>[0, 1, 2], 1=>[0, 1, 2], 2=>[0, 1, 2], 3=>[3, 4, 5], 4=>[3, 4, 5], 5=>[3, 4, 5], 6=>[6, 7, 8], 7=>[6, 7, 8], 8=>[6, 7, 8]}

answered Feb 18, 2019 at 16:24

rahul mishra

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1 Comment

Aleksei Matiushkin Over a year ago

Don’t abuse each for reducing: (0..8).each_slice(3).each_with_object({}) { |m, acc| m.each {|v| acc[v] = m }}.

Cary Swoveland · Accepted Answer · 2019-02-18 19:32:26Z

0

def make_da_hash(n)
  a = (0..n*n-1).to_a
  a.zip(a.each_slice(n).flat_map { |e| [e]*n }).to_h
end

make_da_hash(3)
  #=> {0=>[0, 1, 2], 1=>[0, 1, 2], 2=>[0, 1, 2],
  #    3=>[3, 4, 5], 4=>[3, 4, 5], 5=>[3, 4, 5],
  #    6=>[6, 7, 8], 7=>[6, 7, 8], 8=>[6, 7, 8]} 

make_da_hash(4)
  #=> {0=>[0, 1, 2, 3], 1=>[0, 1, 2, 3], 2=>[0, 1, 2, 3], 3=>[0, 1, 2, 3],
  #    4=>[4, 5, 6, 7], 5=>[4, 5, 6, 7], 6=>[4, 5, 6, 7], 7=>[4, 5, 6, 7],
  #    8=>[8, 9, 10, 11], 9=>[8, 9, 10, 11], 10=>[8, 9, 10, 11],
  #    11=>[8, 9, 10, 11], 12=>[12, 13, 14, 15], 13=>[12, 13, 14, 15],
  #    14=>[12, 13, 14, 15], 15=>[12, 13, 14, 15]}

answered Feb 18, 2019 at 19:32

Cary Swoveland

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1 Comment

Cary Swoveland Over a year ago

Yes, I could change (0..n*n-1) to (0...n*n), but I never use three dots. ¯\_(ツ)_/¯

sawa · Accepted Answer · 2019-02-19 07:39:38Z

0

{}.merge(*Array.new(3){|i| (0..8).group_by{|n| n / 3 * 3 + i}})

Output:

{
  0=>[0, 1, 2],
  3=>[3, 4, 5],
  6=>[6, 7, 8],
  1=>[0, 1, 2],
  4=>[3, 4, 5],
  7=>[6, 7, 8],
  2=>[0, 1, 2],
  5=>[3, 4, 5],
  8=>[6, 7, 8]
}

answered Feb 19, 2019 at 7:39

sawa

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