Sort the array with reference to another array

Question

I have two different arrays. Let's say:

a = [1, 2, 13, 4, 10, 11, 43]
b = [44, 23, 1, 4, 10, 2, 55, 13]

Now I have to sort the array b by referring to the array a. I tried the following solution:

lookup = {}
a.each_with_index do |item, index|
  lookup[item] = index
end

b.sort_by do |item|
  lookup.fetch(item)
end

But I'm getting the KeyError: key not found: 44 error. Can anyone help me find a solution?

Expected output is [1, 2, 13, 4, 10, 23, 44, 55].

I (and I expect others) cannot infer your question from an example. Please edit to state your question in words, then, if necessary, give one or more examples, each with the desired result. Also, please double-check your values. — Cary Swoveland
– Cary Swoveland, Commented Jun 14, 2017 at 6:02
This seems clumsy and not sure if it's robust ((a & b).sort_by {|e| a.index(e)}) + (b - (a & b)).sort #=> [1, 2, 13, 4, 10, 23, 44, 55] — Sagar Pandya
– Sagar Pandya, Commented Jun 14, 2017 at 6:10
Please provide more information on your logic for sorting. It's not clear as to why expected output has to be [1, 2, 13, 4, 10, 23, 44, 55]? — Surya
– Surya, Commented Jun 14, 2017 at 6:25

ndnenkov · Accepted Answer · 2017-06-14 06:55:40Z

4

Comparing arrays checks the first value, if it's equal goes to the second value and so on. Hence this will compare by the order of occurrence in a and then by the actual value for the ones not in a:

b.sort_by { |e| [a.index(e) || a.size, e] }

To keep O(nlogn), you could:

ai = a.each_with_index.to_h
b.sort_by { |e| [ai[e] || a.size, e] }

edited Jun 14, 2017 at 6:55

answered Jun 14, 2017 at 6:47

ndnenkov

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1 Comment

Sergio Tulentsev Over a year ago

I feel that complexity can be improved here. Perhaps by hashifying the order array, as attempted in the question. But this, as it is, certainly should work, yes.

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