All possible combinations from a hash of arrays in Ruby
Asked Answered
T

4

22

What I have:

Let's say I have a hash like this, with various values belonging to one parameter.

a = {}
a[:bitrate] = ["100", "500", "1000"]
a[:fps] = ["15", "30"]
a[:qp] = ["20", "30"]

What I need:

I need some way to iteratively get all the possible combinations of these values, so, with all the parameter/value pairs:

  • bitrate = 100, fps = 15, qp = 20
  • bitrate = 500, fps = 15, qp = 30
  • ...

The number of parameters (i.e. the keys) and the number of values (i.e. the length of the value arrays) are not known beforehand. Ideally, I'd do something like:

a.foo do |ret|
  puts ret.keys   # => ["bitrate", "fps", "qp"]
  puts ret.values # => ["100", "15", "20"]
end

… where the block is called for each possible combination. How can I define foo?


What I (probably) don't need:

Now, I know this: Combine array of array into all possible combinations, forward only, in Ruby, suggesting something like:

a.first.product(*a[1..-1]).map(&:join)

But this operates on values and arrays in arrays only, and I need the original reference to the parameter's name.

Tater answered 20/3, 2012 at 11:54 Comment(0)
M
32
a = {}
a[:bitrate] = ["100", "500", "1000"]
a[:fps] = ["15", "30"]
a[:qp] = ["20", "30"]

def product_hash(hsh)
  attrs   = hsh.values
  keys    = hsh.keys
  product = attrs[0].product(*attrs[1..-1])
  product.map{ |p| Hash[keys.zip p] }
end

product_hash(a)

you'll get

[{:bitrate=>"100", :fps=>"15", :qp=>"20"},
 {:bitrate=>"100", :fps=>"15", :qp=>"30"},
 {:bitrate=>"100", :fps=>"30", :qp=>"20"},
 {:bitrate=>"100", :fps=>"30", :qp=>"30"},
 {:bitrate=>"500", :fps=>"15", :qp=>"20"},
 {:bitrate=>"500", :fps=>"15", :qp=>"30"},
 {:bitrate=>"500", :fps=>"30", :qp=>"20"},
 {:bitrate=>"500", :fps=>"30", :qp=>"30"},
 {:bitrate=>"1000", :fps=>"15", :qp=>"20"},
 {:bitrate=>"1000", :fps=>"15", :qp=>"30"},
 {:bitrate=>"1000", :fps=>"30", :qp=>"20"},
 {:bitrate=>"1000", :fps=>"30", :qp=>"30"}]

You can also add new key to your hash.

a = {}
a[:bitrate] = ["100", "500", "1000"]
a[:fps] = ["15", "30"]
a[:qp] = ["20", "30"]
a[:bw] = [true, false]

product_hash(a)

#=>
[{:bitrate=>"100", :fps=>"15", :qp=>"20", :bw=>true},
 {:bitrate=>"100", :fps=>"15", :qp=>"20", :bw=>false},
 {:bitrate=>"100", :fps=>"15", :qp=>"30", :bw=>true},
 {:bitrate=>"100", :fps=>"15", :qp=>"30", :bw=>false},
 {:bitrate=>"100", :fps=>"30", :qp=>"20", :bw=>true},
 {:bitrate=>"100", :fps=>"30", :qp=>"20", :bw=>false},
 {:bitrate=>"100", :fps=>"30", :qp=>"30", :bw=>true},
 {:bitrate=>"100", :fps=>"30", :qp=>"30", :bw=>false},
 {:bitrate=>"500", :fps=>"15", :qp=>"20", :bw=>true},
 {:bitrate=>"500", :fps=>"15", :qp=>"20", :bw=>false},
 {:bitrate=>"500", :fps=>"15", :qp=>"30", :bw=>true},
 {:bitrate=>"500", :fps=>"15", :qp=>"30", :bw=>false},
 {:bitrate=>"500", :fps=>"30", :qp=>"20", :bw=>true},
 {:bitrate=>"500", :fps=>"30", :qp=>"20", :bw=>false},
 {:bitrate=>"500", :fps=>"30", :qp=>"30", :bw=>true},
 {:bitrate=>"500", :fps=>"30", :qp=>"30", :bw=>false},
 {:bitrate=>"1000", :fps=>"15", :qp=>"20", :bw=>true},
 {:bitrate=>"1000", :fps=>"15", :qp=>"20", :bw=>false},
 {:bitrate=>"1000", :fps=>"15", :qp=>"30", :bw=>true},
 {:bitrate=>"1000", :fps=>"15", :qp=>"30", :bw=>false},
 {:bitrate=>"1000", :fps=>"30", :qp=>"20", :bw=>true},
 {:bitrate=>"1000", :fps=>"30", :qp=>"20", :bw=>false},
 {:bitrate=>"1000", :fps=>"30", :qp=>"30", :bw=>true},
 {:bitrate=>"1000", :fps=>"30", :qp=>"30", :bw=>false}]
Maze answered 20/3, 2012 at 12:49 Comment(0)
T
2

Just FYI I took fl00r's approach and monkey-patched it. I like it a bit better.

class Hash
  def product
    product = values[0].product(*values[1..-1])
    product.map{|p| Hash[keys.zip p]}
  end
end
Typebar answered 14/2, 2013 at 18:19 Comment(0)
M
1

Please try OCG options combination generator.

require "ocg"

generator = OCG.new(
  :bitrate => %w[100 500 1000],
  :fps => %w[15 30],
  :qp => %w[20 30]
)

puts generator.next until generator.finished?

Generator includes much more functionality that will help you to deal with other options.

Magdamagdaia answered 1/11, 2019 at 10:57 Comment(0)
S
0

I believe fl00r's answer is almost perfect but has a drawback. It assumes that hsh.values and hsh.keys will have a matching order, which as far as I know, is not warrantied. So you will probably need an extra step to ensure that. Maybe something like:

def product_hash(hsh)
  keys  = hsh.keys
  attrs = keys.map { |key| hsh[key] }
  product = attrs[0].product(*attrs[1..-1])
  product.map{ |p| Hash[keys.zip p] }
end

But fl00r can correct me if I'm wrong.

Sunburn answered 25/5, 2016 at 10:49 Comment(3)
For Ruby >= 1.9 they are in the same orderMaze
Is that warrantied by the language specification or something that only applies to the MRI? In both cases, can you please point me to the documentation ensuring that? I never know where to search for such stuff. Thanks.Sunburn
ruby-doc.org/core-2.4.1/Hash.html Hashes enumerate their values in the order that the corresponding keys were inserted.Maze

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