Reputation: 7280
I have an array of hashes, id
being one of the keys in the hashes. I want to sort the array elements according to a given order of ID
values.
Suppose my array(size=5) is:
[{"id"=>1. ...}, {"id"=>4. ...}, {"id"=>9. ...}, {"id"=>2. ...}, {"id"=>7. ...}]
I want to sort the array elements such that their id
s are in the following order:
[1,3,5,7,9,2,4,6,8,10]
So the expected result is:
[{'id' => 1},{'id' => 7},{'id' => 9},{'id' => 2},{'id' => 4}]
Upvotes: 3
Views: 2678
Reputation: 419
a= [{"id"=>1}, {"id"=>4}, {"id"=>9}, {"id"=>2}, {"id"=>7}]
b=[1,3,5,7,9,2,4,6,8,10]
a.sort_by{|x| b.index (x['id'])}
Upvotes: 0
Reputation: 110755
Why not just sort
?
def doit(arr, order)
arr.sort { |h1,h2| order.index(h1['id']) <=> order.index(h2['id']) }
end
order = [1,3,5,7,9,2,4,6,8,10]
arr = [{'id' => 1}, {'id' => 4}, {'id' => 9}, {'id' => 2}, {'id' => 7}]
doit(arr, order)
#=> [{'id' => 1}, {'id' => 7}, {'id' => 9}, {'id' => 2}, {'id' => 4}]
Upvotes: 0
Reputation: 13921
Here is a solution for any custom index:
def my_index x
# Custom code can be added here to handle items not in the index.
# Currently an error will be raised if item is not part of the index.
[1,3,5,7,9,2,4,6,8,10].index(x)
end
my_collection = [{"id"=>1}, {"id"=>4}, {"id"=>9}, {"id"=>2}, {"id"=>7}]
p my_collection.sort_by{|x| my_index x['id'] } #=> [{"id"=>1}, {"id"=>7}, {"id"=>9}, {"id"=>2}, {"id"=>4}]
Then you can format it in any way you want, maybe this is prettier:
my_index = [1,3,5,7,9,2,4,6,8,10]
my_collection.sort_by{|x| my_index.index x['id'] }
Upvotes: 7
Reputation: 16514
General note on sorting. Use #sort_by
method of the ruby's array class:
[{'id' => 1},{'id'=>3},{'id'=>2}].sort_by {|x|x['id'] }
# => [{"id"=>1}, {"id"=>2}, {"id"=>3}]
Or with usage #values
method as a callback:
[{'id' => 1},{'id'=>3},{'id'=>2}].sort_by(&:values)
# => [{"id"=>1}, {"id"=>2}, {"id"=>3}]
or you can use more obvious version with #sort
method:
[{'id' => 1},{'id'=>3},{'id'=>2}].sort {|x,y| x['id'] <=> y['id'] }
# => [{"id"=>1}, {"id"=>2}, {"id"=>3}]
For your case, to sort with extended condition use #%
to split even and odd indexes:
[{'id' => 1},{'id'=>4},{'id'=>9},{'id'=>2},{'id'=>7}].sort do |x,y|
u = y['id'] % 2 <=> x['id'] % 2
u == 0 && y['id'] <=> x['id'] || u
end
# => [{"id"=>1}, {"id"=>7}, {"id"=>9}, {"id"=>2}, {"id"=>4}]
For your case, to sort with extended condition use #%
to split according the index, even id
value is absent in the index array:
index = [1,3,5,7,4,2,6,8,10] # swapped 2 and 4, 9 is absent
[{'id' => 1},{'id'=>4},{'id'=>9},{'id'=>2},{'id'=>7}].sort do |x,y|
!index.rindex( x[ 'id' ] ) && 1 || index.rindex( x[ 'id' ] ) <=> index.rindex( y[ 'id' ] ) || -1
end
# => [{"id"=>1}, {"id"=>7}, {"id"=>4}, {"id"=>2}, {"id"=>9}]
Upvotes: 1
Reputation: 4417
I would map the hash based on the values like so:
a = [{"id"=>1}, {"id"=>4}, {"id"=>9}, {"id"=>2}, {"id"=>7}]
[1,3,5,7,9,2,4,6,8,10].map{|x| a[a.index({"id" => x})] }.compact
#=> [{"id"=>1}, {"id"=>7}, {"id"=>9}, {"id"=>2}, {"id"=>4}]
Upvotes: 1