Creating two arrays containing an original array with different values added at the end

Question

So I am trying to build a multidimensional array by starting with a single array and splitting it into separate arrays to account for possible added values.

Example:

Original Array: [2,3]

adding either 4 or 5

New array: [[2,3,4],[2,3,5]]

I have tried the following:

array=[2,3]
array1=array<<4
array2=array<<5

array=[2,3]
array1=array<<4
array.pop
array2=array<<5

array=[2,3]
array1=array.push 4
array.pop
array2=array.push 5

The results I get are:

[[2,3,4,5],[2,3,4,5]]
[[2,3,5],[2,3,5]]
[[2,3,5],[2,3,5]]

Is there a way to alter the original array only in the new variables so that the variables don't end up equal when I combine them?

array1 = array + [4]; array2 = array + [5]

ptd
– ptd

2014-12-08 19:37:10 +00:00
Commented Dec 8, 2014 at 19:37 — ptd
– ptd, Commented Dec 8, 2014 at 19:37

tadman · Accepted Answer · 2014-12-08 19:40:22Z

1

There are a number of methods on Array that are in-place modifiers, that is they don't make copies, and << is one of them.

What you might find easier is this:

array = [ 2, 3 ]
array1 = array + [ 4 ]
array2 = array + [ 5 ]

The result in this case is two independent arrays.

answered Dec 8, 2014 at 19:40

tadman

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Comments

Uri Agassi · Accepted Answer · 2014-12-08 20:20:34Z

1

Another interesting way to do this is using the splat operator:

array = [2, 3]
array1 = [*array, 4]
# => [2, 3, 4]
array2 = [*array, 5]
# => [2, 3, 5]

answered Dec 8, 2014 at 20:20

Uri Agassi

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1 Comment

Cary Swoveland Over a year ago

...which lends itself nicely to mapping an array of values to add to array. Nice.

Cary Swoveland · Accepted Answer · 2014-12-08 20:04:17Z

0

If you have several times to add to:

array = [1, 2, 3]

say:

b = [*(4..20)]
  #=> [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]

you could use the method Array#product:

[array].product(b).map(&:flatten)
  #=> [[1, 2, 3,  4], [1, 2, 3,  5], [1, 2, 3,  6], [1, 2, 3,  7],
  #    [1, 2, 3,  8], [1, 2, 3,  9], [1, 2, 3, 10], [1, 2, 3, 11],
  #    [1, 2, 3, 12], [1, 2, 3, 13], [1, 2, 3, 14], [1, 2, 3, 15],
  #    [1, 2, 3, 16], [1, 2, 3, 17], [1, 2, 3, 18], [1, 2, 3, 19],
  #    [1, 2, 3, 20]]

answered Dec 8, 2014 at 20:04

Cary Swoveland

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1 Comment

jasmine noack Over a year ago

Thanks, that may be helpful I will have to play with the #product method to make sure I understand it.

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