CODEWITHSUNDEEP
ruby
MikeHolford
MikeHolford

Reputation: 1931

Ruby loop through each element in an array n times

I've had a tough time finding an exact example of this.

If I have an array that contains 5 elements. E.g. 

list = [5, 8, 10, 11, 15]

I would like to fetch what would be the 8th (for example) element of that array if it were to be looped. I don't want to just duplicate the array and fetch the 8th element because the nth element may change

Essentially the 8th element should be the number 10.

Any clean way to do this?

Upvotes: 1

Views: 219

Answers (5)

Stefan
Stefan

Reputation: 114138

You could use rotate:

[5, 8, 10, 11, 15].rotate(7).first
#=> 10

It's 7 because arrays are zero based.

Upvotes: 2

Cryptex Technologies
Cryptex Technologies

Reputation: 1163

I ran these in my irb to get the output,

irb(main):006:0> list = [5, 8, 10, 11, 15]
=> [5, 8, 10, 11, 15]

irb(main):007:0> list[(8 % list.length) - 1]
=> 10

hope it will help you.

Upvotes: 0

Aleksei Matiushkin
Aleksei Matiushkin

Reputation: 121000

Just out of curiosity using Array#cycle:

[5, 8, 10, 11, 15].cycle.take(8).last

This is quite inefficient but fancy.

Upvotes: 1

Ursus
Ursus

Reputation: 30056

Math to the rescue

list[(8 % list.length) - 1]

A link about this modulo operator that we love

Upvotes: 7

Jagdeep Singh
Jagdeep Singh

Reputation: 4920

This should do:

def fetch_cycled_at_position(ary, num)
  ary[(num % ary.length) - 1]
end

ary = _
 => [5, 8, 10, 11, 15]

fetch_cycled_at_position(ary, 1)   # Fetch first element
 => 5

fetch_cycled_at_position(ary, 5)   # Fetch 5th element
 => 15

fetch_cycled_at_position(ary, 8)   # Fetch 8th element
 => 10

Upvotes: 2

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