Sort array by other array

Question

I have two arrays:

a = [ 1, 0, 2, 1, 0]
b = ['a', 'b', 'c', 'd', 'e']

I want to order the b array according to a's elements values. I can make this by merging the two arrays into a Hash and the order by key:

h = Hash[b.zip a]
=> {"a"=>1, "b"=>0, "c"=>2, "d"=>1, "e"=>0}

h2 = Hash[h.sort_by{|k, v| v}]
=> {"b"=>0, "e"=>0, "a"=>1, "d"=>1, "c"=>2}

array = h2.keys
=> ["b", "e", "a", "d", "c"]

Where there is a tie the order may be chosen arbitrary.

Is there a way (maybe more compact), I can achieve this without using the hash.

Doguita · Accepted Answer · 2015-04-02 17:41:27Z

3

a.zip(b).sort.map(&:last)

In parts:

p a.zip(b) # => [[1, "a"], [0, "b"], [2, "c"], [1, "d"], [0, "e"]] 
p a.zip(b).sort # => [[0, "b"], [0, "e"], [1, "a"], [1, "d"], [2, "c"]] 
p a.zip(b).sort.map(&:last) # => ["b", "e", "a", "d", "c"]

answered Apr 2, 2015 at 17:41

Doguita

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3 Comments

Cary Swoveland Over a year ago

...which can alternatively be written: [a,b].transpose.sort.map(&:last).

Mark Thomas Over a year ago

@CarySwoveland It's not an exact equivalent, as the error behavior is different. If the arrays are different sizes, zip will pad with nil but transpose will throw an error.

Cary Swoveland Over a year ago

Thanks, @MarkThomas. Yes, but in this case I thought it was reasonable to assume the arrays are of the same size, both because they are in the example and because no rule was given for the case where they are unequal in size. But you do make a good good point about zip vs. transpose.

steenslag · Accepted Answer · 2015-04-02 17:55:44Z

2

a = [ 1, 0, 2, 1, 0]
b = ['a', 'b', 'c', 'd', 'e']

p b.sort_by.each_with_index{|el,i| a[i]}
# => ["b", "e", "a", "d", "c"]

answered Apr 2, 2015 at 17:55

steenslag

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