问题
arr = [1,3,2,4]
arr.sort #=> [1,2,3,4]
I would like an array [0, 2, 1, 3] (original indexes in arr.sort order)
Is there a simple way to do that with Ruby 1.9.3?
thank you
回答1:
xs = [1, 3, 2, 4]
original_indexes = xs.map.with_index.sort.map(&:last)
#=> [0, 2, 1, 3]
回答2:
arr=[1,3,2,4]
p arr.map{|e| arr.sort.index(e)}
to avoid sorting each time, better is:
arr=[1,3,2,4]
arr_s = arr.sort
p arr.map{|e| arr_s.index(e)}
UPDATED
arr=[1,3,2,4]
start_time = Time.now
(1..100000).each do |i|
arr.map{|e| arr.sort.index(e)}
end
elapsed = Time.now - start_time
p elapsed
xs = [1, 3, 2, 4]
start_time = Time.now
(1..100000).each do |i|
xs.map.with_index.sort.map(&:last)
end
elapsed = Time.now - start_time
p elapsed
and got the result:
0.281736
0.504314
回答3:
I tested on MRI Ruby 2.2.1p85 (on both Mac and CentOS), tokland's solution return a wrong result:
xs = [8,3,2,7,5]
xs.map.with_index.sort.map(&:last)
#=> [2, 1, 4, 3, 0] # wrong
Yevgeniy Anfilofyev solution works but does not support non-unique array:
arr = [8,3,2,7,5]
arr_s = arr.sort
arr.map{|e| arr_s.index(e)}
#=> [4, 1, 0, 3, 2] # correct
arr = [8,3,5,2,8,8,7,5]
arr_s = arr.sort
arr.map{|e| arr_s.index(e)}
#=> [5, 1, 2, 0, 5, 5, 4, 2]
I come up this:
arr = [8,3,5,2,8,8,7,5]
index_order = []
arr.uniq.sort.each do |a|
index_order += arr.each_index.select{|i| arr[i] == a }
end
r = []
index_order.each_with_index do |a, i|
r[a] = i
end
r
#=> [5, 1, 2, 0, 6, 7, 4, 3]
回答4:
array = [6, 20, 12, 2, 9, 22, 17]
sorted = array.sort
indices = []
array.each do |n|
index = (0...sorted.length).bsearch { |x| n <=> sorted[x] }
indices << index
end
indices
This solution works with O(nlogn)
回答5:
(0..arr.size - 1).sort_by { |i| arr[i] }
来源:https://stackoverflow.com/questions/14446181/originals-indexes-of-sorted-elements-in-ruby