Converting array indices to objects in a hash in one line

Question

Given an array

array = ['a','b','c','d','e','f']

and a hash

hash = {"a"=>1,"b"=>2,"c"=>3,"d"=>4,"e"=>5,"f"=>6}

I would like to create a new array which contains the objects corresponding to the value of every index from the third onward. I am currently doing it like this:

newArray = []
array[2..-1].each do |item|
  newArray << hash[item]
end

I feel like this could be done in one line (possibly without creating the newArray beforehand as I would like to put this directly into another object's initialization code). Is this possible?

If you want to do it one line, just remove the linebreaks. In Ruby, you can always remove all linebreaks in all programs, by either replacing them with semicolons, keywords, or sometimes even nothing. In your case: newArray = []; array[2..-1].each do |item| newArray << hash[item] end Voilà: one line. — Jörg W Mittag
– Jörg W Mittag, Commented Apr 12, 2016 at 0:36

Doguita · Accepted Answer · 2016-04-11 22:26:32Z

3

You want to use Hash#values_at

hash.values_at *array[2..-1]
# => [3, 4, 5, 6]

answered Apr 11, 2016 at 22:26

Doguita

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max_ · Accepted Answer · 2016-04-11 22:25:19Z

1

how about this?

array[2..-1].map{|x| hash[x]}

could you post some expected example output please?

edited Apr 11, 2016 at 22:25

max_

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answered Apr 11, 2016 at 22:23

ma489

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patrickmcgraw · Accepted Answer · 2016-04-11 22:20:58Z

0

hash.keep_if {|k,v| array.slice(2, array.length-2).index(k) }.values

or

hash.dup.keep_if {|k,v| array.slice(2, array.length-2).index(k) }.values

if you need to maintain the original hash values

answered Apr 11, 2016 at 22:20

patrickmcgraw

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Schylar · Accepted Answer · 2016-04-11 22:23:46Z

0

array[2..-1].inject([]) {|ary, x| ary << hash[x]}

answered Apr 11, 2016 at 22:23

Schylar

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