根据给定的顺序对数字数组进行排序

Question

I have two arrays. The first array contains the sort order. The second array contains an arbitrary number of elements.

I have the property that all elements (value-wise) from the second array are guaranteed to be in the first array, and I am only working with numbers.

A = [1,3,4,4,4,5,2,1,1,1,3,3]
Order = [3,1,2,4,5]

When I sort A , I would like the elements to appear in the order specified by Order :

[3, 3, 3, 1, 1, 1, 1, 2, 4, 4, 4, 5]

Note that duplicates are fair game. The elements in A should not be altered, only re-ordered. How can I do this?

Answer 1

>> source = [1,3,4,4,4,5,2,1,1,1,3,3]
=> [1, 3, 4, 4, 4, 5, 2, 1, 1, 1, 3, 3]
>> target = [3,1,2,4,5]
=> [3, 1, 2, 4, 5]
>> source.sort_by { |i| target.index(i) }
=> [3, 3, 3, 1, 1, 1, 1, 2, 4, 4, 4, 5]

Answer 2

If (and only if!) @Gareth's answer turns out to be too slow, instead go with:

# Pre-create a hash mapping value to index once only…
index = Hash[ Order.map.with_index.to_a ] #=> {3=>0,1=>1,2=>2,4=>3,5=>4}

# …and then sort using this constant-lookup-time
sorted = A.sort_by{ |o| index[o] } 

---

Benchmarked:

require 'benchmark'

order = (1..50).to_a.shuffle
items = 1000.times.map{ order.sample }
index = Hash[ order.map.with_index.to_a ]

Benchmark.bmbm do |x|
  N = 10_000
  x.report("Array#index"){ N.times{
    items.sort_by{ |n| order.index(n) }
  }}
  x.report("Premade Hash"){ N.times{
    items.sort_by{ |n| index[n] }
  }}
  x.report("Hash on Demand"){ N.times{
    index = Hash[ order.map.with_index.to_a ]
    items.sort_by{ |n| index[n] }
  }}
end

#=>                      user     system      total        real
#=> Array#index     12.690000   0.010000  12.700000 ( 12.704664)
#=> Premade Hash     4.140000   0.000000   4.140000 (  4.141629)
#=> Hash on Demand   4.320000   0.000000   4.320000 (  4.323060)

Answer 3

Another possible solution without explicit sorting:

source = [1,3,4,4,4,5,2,1,1,1,3,3]
target = [3,1,2,4,5]
source.group_by(&lambda{ |x| x }).values_at(*target).flatten(1)