Iterating hash of arrays to create an array of array

Question

Suppose I have a hash

@attribute_type = {
  typeA: ['a', 'b', 'c'],
  typeB: ['1', '2', '3'],
  typeC: ['9', '8', '7']
}

I want to iterate over the values so I can create an array having all distinct possible combinations of the three arrays, for example:

['a', '1', '9'], ['a', '1', '8'], ['a', '1', '7'], ['a', '2', '9'], ...

Is this possible?

The answer you selected offers the possible combinations of type[:A] in combination with types B and C. However your question states "all possible combinations" in which case h.values.flatten.combination(3).to_a is potentially more correct depending on if you meant distinct values or permutations. Right now based on the chosen answer ["a","b","c"] is not considered a valid "combination" so "all" might be misleading — engineersmnky
– engineersmnky, Commented May 25, 2018 at 17:44
I meant all distinct possible combinations. I'll edit the question to reflect this. Thank you. — IBG
– IBG, Commented May 29, 2018 at 0:40
So would ["a", "b", 1] be considered valid? If so the selected answer does not cover this. Nor does your example to be honest — engineersmnky
– engineersmnky, Commented May 29, 2018 at 0:54
No I just want all distinct possible combinations from the three arrays. I don't know how to word it. Can you please help me with that? What I want is the combination of 1 value per set inside the array, so like {typeA * typeB * typeC}, but I want the values for each combination. — IBG
– IBG, Commented May 29, 2018 at 2:26
Your description seems sufficient based on the answer you selected I was just looking clarification to make sure. — engineersmnky
– engineersmnky, Commented May 29, 2018 at 12:56

Cary Swoveland · Accepted Answer · 2018-05-25 04:02:24Z

11

h = { :typeA=>['a','b','c'], :typeB=>['1','2','3'], :typeC=>['9','8','7'] }

first, *rest = h.values
  #=> [["a", "b", "c"], ["1", "2", "3"], ["9", "8", "7"]]
first.product(*rest)
  #=> [["a", "1", "9"], ["a", "1", "8"], ["a", "1", "7"],
  #    ["a", "2", "9"], ["a", "2", "8"], ["a", "2", "7"],
  #    ["a", "3", "9"], ["a", "3", "8"], ["a", "3", "7"],
  #    ["b", "1", "9"], ["b", "1", "8"], ["b", "1", "7"],
  #    ["b", "2", "9"], ["b", "2", "8"], ["b", "2", "7"],
  #    ["b", "3", "9"], ["b", "3", "8"], ["b", "3", "7"],
  #    ["c", "1", "9"], ["c", "1", "8"], ["c", "1", "7"],
  #    ["c", "2", "9"], ["c", "2", "8"], ["c", "2", "7"],
  #    ["c", "3", "9"], ["c", "3", "8"], ["c", "3", "7"]]

See Array#product.

answered May 25, 2018 at 4:02

Cary Swoveland

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4 Comments

IBG Over a year ago

If you don't mind me asking, what does first, *rest = h.values mean? I don't get the first part, and the comma as well.

Cary Swoveland Over a year ago

Suppose arr = [[1,2,3], [4,5,6], [7,8,9]]. Let's try it: first, *rest = arr; first #=> [1,2,3]; rest #=> [[4,5,6], [7,8,9]]. The asterisk is called a splat. This article provides a good explanation, especially the section Array Destruction.

mu is too short Over a year ago

I think your autocorrect meant to let you say Destructuring in that comment.

Stefan Over a year ago

@IBG you can also target the values explicitly, i.e. h[:typeA].product(h[:typeB], h[:typeC])

iGian · Accepted Answer · 2018-05-25 05:22:41Z

1

My 2 cents.

Pretty the same as Cary Swoveland, but just one line:

h.values.first.product(*h.values.drop(1))

answered May 25, 2018 at 5:22

iGian

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