How to expand elements of an array into sub-arrays?

Question

I have a huge array

huge = 1000
huge_array = (1..huge).to_a

How to best "expand" this array so that each element becomes a sub-array of format [original_element, "default value"], preferably in a memory-friendly way (without an explicit #map loop?)

expanded_huge_array = huge_array.some_magic
#=> [[1, "default value"],[2, "default value"], ... [1000, "default value"]]

What does map have to do with it? If you need an array filled with values then you need an array filled with values. Why have the array index (plus one) be in the array at all? Etc. — Dave Newton
– Dave Newton, Commented Dec 7, 2015 at 15:40
Always assume the question is a watered down, easy-to-ask and easy-to-answer version of a non-trivial problem. The actual "huge_array" people may have will contain anything but an index. — Epigene
– Epigene, Commented Dec 8, 2015 at 7:30
No, assume the question is complete, and provides enough information to answer it in the most reasonable way possible. If there are hidden criteria then it's difficult to reason about, and answers may not suit what your'e actually asking. — Dave Newton
– Dave Newton, Commented Dec 8, 2015 at 11:50
Good point about hidden criteria. However, I rarely miss a chance to slightly over-interpret a question to provide a better learning and problem-solving opportunity. — Epigene
– Epigene, Commented Dec 8, 2015 at 12:18
Right--and without actual context, you can't provide a better learning and problem-solving opportunity, because you don't know what's really being asked. You introduced criteria without any explanation (e.g., why no map? Both map and zip make new arrays, but in different ways. Both iterate over the original array, the zip answer creates a new array of the huge array's size then iterates, etc) — Dave Newton
– Dave Newton, Commented Dec 8, 2015 at 12:37

Aleksei Matiushkin · Accepted Answer · 2015-12-07 15:54:12Z

2

huge_array.zip(['default value'] * huge_array.size)

BTW, you might simulate this behaviour with Hash with default:

arr = Hash.new { |h, key| huge_array.include?(key) ? [key, 'default value'] : nil }
arr[1]
#⇒ [1, 'default value']
arr[10000]
#⇒ nil

edited Dec 7, 2015 at 15:54

answered Dec 7, 2015 at 15:46

Aleksei Matiushkin

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2 Comments

Epigene Over a year ago

Looks like #zip is the precision tool for this. Plus points for hash-with-block idea. (I wonder, what happens if such a hash is returned to a context where huge_array is not defined.)

Aleksei Matiushkin Over a year ago

No surprises, as the hash is called on an undefined key, the “local variable is not defined error” is raised.

Roman Kiselenko · Accepted Answer · 2015-12-07 15:53:09Z

1

Try Array#product:

Returns an array of all combinations of elements from all arrays.

>> [1,2,3].product(["a"])
=> [[1, "a"], [2, "a"], [3, "a"]]

edited Dec 7, 2015 at 15:53

answered Dec 7, 2015 at 15:41

Roman Kiselenko

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1 Comment

Aleksei Matiushkin Over a year ago

> without an explicit #map loop?

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