array x array matrix in ruby

Question

I'd like to convert from

{'key1' => (1..10) ,
 'key2' => (11..20) , 
'key3' => (21..30)}

to

[{'key1' => 1, 'key2' => 11, 'key3' => 21},
{'key1' => 1, 'key2' => 11, 'key3' => 22},...
 .
 .
{'key1' => 10, 'key2' => 20, 'key3' => 30}]

How to solve it?

Freddie, I've edited my answer to address your comment and also added an alternative, which I think is an improvement. — Cary Swoveland
– Cary Swoveland, Commented Sep 29, 2013 at 18:06
What have you tried, they should ask you before they unleashed themselves to answer your otherwise trivial question. — Boris Stitnicky
– Boris Stitnicky, Commented Oct 21, 2013 at 4:52

Arup Rakshit · Accepted Answer · 2013-09-29 15:48:13Z

4

Here it is :

hsh = {'key1' => (1..10) ,'key2' => (11..20) , 'key3' => (21..30)}
keys = hsh.keys
hsh['key1'].to_a.product(hsh['key2'].to_a,hsh['key3'].to_a).map{|a|Hash[keys.zip(a)]}

# => [{'key1' => 1, 'key2' => 11, 'key3' => 21},
#   {'key1' => 1, 'key2' => 11, 'key3' => 22},...
#   .
#   .
#   {'key1' => 10, 'key2' => 20, 'key3' => 30}]

You could also write the above as below,when you have more number of keys:

hsh = {'key1' => (1..10) ,'key2' => (11..20) , 'key3' => (21..30)}
keys = hsh.keys
array = hsh.values_at(*keys[1..-1]).map(&:to_a)
hsh['key1'].to_a.product(*array).map{|a|Hash[keys.zip(a)]}

edited Sep 29, 2013 at 15:48

answered Sep 29, 2013 at 14:29

Arup Rakshit

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2 Comments

freddiefujiwara Over a year ago

GREAT! Solution .. but can I calc it for any size of keys?

Cary Swoveland Over a year ago

Interesting use of splat, Arup. I borrowed it for my second answer.

Cary Swoveland · Accepted Answer · 2013-09-29 19:02:37Z

2

So many ways... A kiss answer (edited to extend to any number of keys):

s = {'key1' => (1..10), 'key2' => (11..20), 'key3' => (21..30)}
r = []
s.each {|k,v| a = []; (v.to_a).each {|i| a << {k=>i}}; r << a}
result = r.shift
r.each {|e| result = result.product(e).map(&:flatten)}
result

edited Sep 29, 2013 at 19:02

answered Sep 29, 2013 at 15:27

Cary Swoveland

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3 Comments

freddiefujiwara Over a year ago

GREAT! Solution! thank you so much. but can I calc it for any size of keys?

Arup Rakshit Over a year ago

@Cary You can write I think result[0].product(result[1]).product(result[2]).flatten(1) as result[0].product(*result[1..-1]).flatten

Cary Swoveland Over a year ago

Thanks, Arup. You spotted my mistake. I've edited the code to address SO's concern.

Stefan · Accepted Answer · 2013-09-30 15:18:24Z

2

h = {
  'key1' => (1..10),
  'key2' => (11..20),
  'key3' => (21..30)
}

h.map { |k,v| [k].product(v.to_a) }.transpose.map { |e| Hash[e] }
#=> [{"key1"=>1, "key2"=>11, "key3"=>21},
#    {"key1"=>2, "key2"=>12, "key3"=>22},
#    {"key1"=>3, "key2"=>13, "key3"=>23},
#    {"key1"=>4, "key2"=>14, "key3"=>24},
#    {"key1"=>5, "key2"=>15, "key3"=>25},
#    {"key1"=>6, "key2"=>16, "key3"=>26},
#    {"key1"=>7, "key2"=>17, "key3"=>27},
#    {"key1"=>8, "key2"=>18, "key3"=>28},
#    {"key1"=>9, "key2"=>19, "key3"=>29},
#    {"key1"=>10, "key2"=>20, "key3"=>30}]

answered Sep 30, 2013 at 15:18

Stefan

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1 Comment

Cary Swoveland Over a year ago

Interesting use of transpose, Stefan. I've tucked that into my grey cells.

steenslag · Accepted Answer · 2013-09-29 14:38:54Z

1

h = {'key1' => (1..10) ,
'key2' => (11..20) , 
'key3' => (21..30)}

arrays = h.values.map(&:to_a).transpose
p arrays.map{|ar| Hash[h.keys.zip(ar)] }
#=> [{"key1"=>1, "key2"=>11, "key3"=>21}, {"key1"=>2, "key2"=>12, "key3"=>22},...

answered Sep 29, 2013 at 14:38

steenslag

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Comments

Community · Accepted Answer · 2017-05-23 11:50:02Z

1

h = {'key1' => (1..10), 'key2' => (11..20), 'key3' => (21..30)}

Edit 1: made some changes, principally the use of inject({}):

f,*r = h.map {|k,v| [k].product(v.to_a)}
f.zip(*r).map {|e| e.inject({}) {|h,a| h[a.first] = a.last; h}}

Edit 2: After seeing the use of Hash[] in @Phrogz's answer to another question:

f,*r = h.map {|k,v| [k].product(v.to_a)}
f.zip(*r).map {|e| Hash[*e.flatten]}

edited May 23, 2017 at 11:50

CommunityBot

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answered Sep 29, 2013 at 19:06

Cary Swoveland

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Comments

Boris Stitnicky · Accepted Answer · 2013-10-21 05:12:38Z

0

Lazier way of doing the same:

h = {
  'key1' => (1..10),
  'key2' => (11..20),
  'key3' => (21..30)
}

result = ( 0...h.values.map( &:to_a ).map( &:size ).max ).map do |i|
  Hash.new { |hsh, k| hsh[k] = h[k].to_a[i] }
end

result[1]['key3'] #=> 22

answered Oct 21, 2013 at 5:12

Boris Stitnicky

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