From: Jeremy Bopp <jeremy@...>
Date: 2010-10-14T14:50:49+09:00
Subject: Re: sort_by: multiple fields with reverse sort

On 10/13/2010 09:50 PM, Steve Howell wrote:
> It would be nice if Ruby supported a sort_by on steroids.
> 
>   sorted_list = list.multi_field_sort_by(
>     { |x| x.department.name },
>     { |x| x.position.name },
>     desc { |x| x.level },
>     desc { |x| x.salary ? x.salary : x.rate * db_lookup(x,
> 'hours_worked') }
>   )
> 
> I believe you could write a decent multi_field_sort_by in Ruby that
> would be efficient for large enough lists to outperform more tedious
> approaches, but it would be even better if Ruby natively supported it.
> 
> My proposed syntax might be slightly off, but you get the idea.  You'd
> pass a list of blocks that represent the successive tiebreakers, and
> multi_field_sort_by would presumably cache the results from each
> transformation, evaluating the blocks only as necessary.  The "desc"
> thingy would actually produce some kind of wrapper that
> multi_field_sort_by could introspect to know that it needs to apply a
> particular tiebreaker in reverse order.

How about something like this:

module Enumerable
  # sort_by will take a key generator and an optional comparator
  # and perform a Schwartzian Transform over the data.
  #
  # This is a general solution which is relatively inefficient than
  # a purpose-built sorting function if the keys are trivial to
  # generate.
  def sort_by(cmp = lambda { |a, b| a <=> b }, &key)
    collect do |item|   # Generate keys from the list items.
      [key[item], item]
    end.sort do |a, b|  # Sort the keys.
      cmp[a[0], b[0]]
    end.collect do |kv| # Return the items in key sort order.
      kv[1]
    end
  end
end

# This will cause the sort to operate over the second and then
# the first column.
key = lambda { |item| [item[1], item[0]] }

# This will cause the sort to operate normally on the second
# column and in reverse on the first column.
cmp = lambda do |a, b|
        res = a[0] <=> b[0]
        break res unless res == 0
        b[1] <=> a[1]
      end

# Sample data from Ryan's post.
a = [
  ["radio", 30, 5],
  ["radio", 20, 5],
  ["archie", 20, 5],
  ["newton", 10, 3]
]

# Sort over the second and then first columns each in normal order.
p a.sort_by(&key)
# => [["newton", 10, 3], ["archie", 20, 5], ["radio", 20, 5],
#                                                  ["radio", 30, 5]]

# Sort over the second column in normal order and then the first
# column in reverse order.
p a.sort_by(cmp, &key)
# => [["newton", 10, 3], ["radio", 20, 5], ["archie", 20, 5],
#                                                  ["radio", 30, 5]]