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I have an array and a hash and what I want to achieve is to sort the hash (with each id inside each hash) based on the sort that is in the array and if the id (in hash) doesn't exist in my_array, they should not be deleted and just be pushed down in the sort.

my_array = [4, 2, 5, 3, 1]
hash = [
  {"id" => 1, "field_name" => "foo"},
  {"id" => 2, "field_name" => "bar"},
  {"id" => 3, "field_name" => "abc"},
  {"id" => 4, "field_name" => "zsh"},
  {"id" => 5, "field_name" => "kql"},
  {"id" => 6, "field_name" => "plo"},
  {"id" => 7, "field_name" => "cde"}
]

Needed output

[
  {"id" => 4, "field_name" => "zsh"},
  {"id" => 2, "field_name" => "bar"},
  {"id" => 5, "field_name" => "kql"},
  {"id" => 3, "field_name" => "abc"},
  {"id" => 1, "field_name" => "foo"},
  {"id" => 6, "field_name" => "plo"},
  {"id" => 7, "field_name" => "cde"}
]

Appreciate any help and thanks in advance!

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5
  • 2
    Can you explain that "let them be as they were" part a little further? What if "id"=>6 and "id"=>7 were at positions 1 and 2? Those places will be taken by "id"=>4 and "id"=>2 during the sort, so they have to move somewhere else. Commented Jun 16, 2022 at 14:32
  • Thanks @Stefan by that I man not to remove them since they are not in the array list. But yeah of course, they need to be moved down. I was talking more about removing them from the list Commented Jun 16, 2022 at 15:22
  • So they should just move to the end but keep their relative position (6 before 7), right? Commented Jun 16, 2022 at 15:29
  • Thanks. @Stefan the position is only important for the ones in the array basically so that they come at top (based on their position in the array) and the rest come just after those sorted one Commented Jun 16, 2022 at 15:33
  • It would be more correct to say that you have an array of hashes. Commented Jun 17, 2022 at 2:11

2 Answers 2

Reset to default
2 
my_array = [4, 7, 5, 3, 1]

arr = [
  {"id" => 1, "field_name" => "foo"},
  {"id" => 2, "field_name" => "bar"},
  {"id" => 3, "field_name" => "abc"},
  {"id" => 4, "field_name" => "zsh"},
  {"id" => 5, "field_name" => "kql"},
  {"id" => 6, "field_name" => "plo"},
  {"id" => 7, "field_name" => "cde"}
]

Note that I've changed my_array from that given in the question and have renamed the array to something more appropriate.

arr.sort_by.with_index do |h,i|
  my_array.include?(h["id"]) ? [0, h["id"]] : [1,i]
end
  #=> [{"id"=>1, "field_name"=>"foo"},
  #    {"id"=>3, "field_name"=>"abc"},
  #    {"id"=>4, "field_name"=>"zsh"},
  #    {"id"=>5, "field_name"=>"kql"},
  #    {"id"=>7, "field_name"=>"cde"},
  #    {"id"=>2, "field_name"=>"bar"},
  #    {"id"=>6, "field_name"=>"plo"}]

See Enumerable#sort_by. That method uses Array#<=> to order elements. See especially the third paragraph at the doc for <=>.

Below I've listed the arrays used by sort_by for each element of arr.

{"id" => 1, "field_name" => "foo"}  ->  [0, 1]
{"id" => 2, "field_name" => "bar"}  ->  [1, 1]
{"id" => 3, "field_name" => "abc"}  ->  [0, 3]   
{"id" => 4, "field_name" => "zsh"}  ->  [0, 4]
{"id" => 5, "field_name" => "kql"}  ->  [0, 5]
{"id" => 6, "field_name" => "plo"}  ->  [1, 5]
{"id" => 7, "field_name" => "cde"}  ->  [0, 7]

If [type, x] is the array sort_by uses for sorting, it will put all elements for which type 0 before those for which type = 1. To break ties it orders elements by x. For type = 1, x is the index of the element, which keeps them in order at the end.

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4 Comments

thank you so much @Cary. I just has a question regarding the sorting. In my_array the sorting is 4, 7, 5, 3, 1 whereas the output of the example code has the sorting (ids) as: 4, 2, 7, 5, 3, 6, 1 and they don't match 4, 7, 5, 3, 1, 4, 6 or am I misunderstanding something :)
arild, I will elaborate within the next couple of hours. Also, I believe I can simplify the calculations some. In the meantime please address @engineersmnky's comment, as the solution is greatly simplified if the elements not to be sorted are to be placed at the end; just partition the array into two groups, sort one (no need for an actual sort) and then tack the other group on the end of the sorted array.
You are the best @CarySwoveland. Appreciate your help!
Note that all of the comments above pertained to my original answer, which reflected a misunderstanding of the question that the OP has clarified by editing the question.
2

Map Over Your ID Array to Select Matching Hashes in Order

While there may be a more elegant way to do this, I think the solution below is quite clear. It uses your Array of id elements as an Enumerable object to map over, returning the matching Hash object with a matching value for each "id" key encountered during iteration (if found), and then removes any elements where no match was found, e.g. the Array returned by #map returns nil.

Using Ruby 3.1.2:

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

array_of_hashes = [
  {"id" => 1, "field_name" => "foo"},
  {"id" => 2, "field_name" => "bar"},
  {"id" => 3, "field_name" => "abc"},
  {"id" => 4, "field_name" => "zsh"},
  {"id" => 5, "field_name" => "kql"},
  {"id" => 6, "field_name" => "plo"},
  {"id" => 7, "field_name" => "cde"}
]

id_order.map { |id| array_of_hashes.detect { |h| h["id"] == id } }.compact

This correctly returns your ordered results as specified in the first Array:

#=> 
[{"id"=>4, "field_name"=>"zsh"},                                                                                         
 {"id"=>2, "field_name"=>"bar"},                                                                                         
 {"id"=>5, "field_name"=>"kql"},                                                                                         
 {"id"=>3, "field_name"=>"abc"},                                                                                         
 {"id"=>1, "field_name"=>"foo"}]

Note that it doesn't return Hash objects for IDs 6 or 7 because they aren't present in your Array of IDs to search for. You can adjust the #map to treat those as pass-throughs (although you haven't explained how or why they should appear at the end of your resulting Array) or add them to the Array of elements to search for, which would be the preferred approach. However, since they weren't addressed in your original post, this is in fact the correct output if no matching ID is found.

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1 Comment

thank you so much @Todd. I appreciate it. Any way you could also tell me how I can make it so that IDs of 6 & 7 doesn't get removed and they can just be at the end of the sort? Thanks in advance!

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