Sort a List with Multiple Conditions Ruby on Rails

Sort a list with multiple conditions Ruby on Rails

Use sort_by and supply an array of values. It will compare them in order from left to right. Default sort is ascending, so if you want the opposite (e.g. matches_won should sort descendingly so that most matches one comes first, as opposed to matches_lost, which should sort lowest to highest), you need to negate the value.

Here's an example

require 'pp'
Team = Struct.new :won, :lost, :demerits, :style

teams = Array.new(20) { Team.new rand(5), rand(5), rand(5), rand(5) }
puts "Before sort:"
pp teams

puts "", "After sort:"
pp teams
teams.sort_by! { |team| [-team.won, team.lost, team.demerits, -team.style] }
# >> Before sort:
# >> [#<struct Team won=1, lost=2, demerits=4, style=3>,
# >>  #<struct Team won=0, lost=4, demerits=2, style=2>,
# >>  #<struct Team won=4, lost=1, demerits=2, style=3>,
# >>  #<struct Team won=1, lost=4, demerits=1, style=4>,
# >>  #<struct Team won=3, lost=1, demerits=1, style=4>,
# >>  #<struct Team won=3, lost=4, demerits=1, style=2>,
# >>  #<struct Team won=4, lost=0, demerits=4, style=0>,
# >>  #<struct Team won=3, lost=4, demerits=1, style=2>,
# >>  #<struct Team won=1, lost=4, demerits=1, style=2>,
# >>  #<struct Team won=3, lost=0, demerits=1, style=1>,
# >>  #<struct Team won=3, lost=4, demerits=3, style=4>,
# >>  #<struct Team won=1, lost=4, demerits=0, style=0>,
# >>  #<struct Team won=3, lost=4, demerits=2, style=0>,
# >>  #<struct Team won=3, lost=0, demerits=3, style=0>,
# >>  #<struct Team won=1, lost=2, demerits=0, style=1>,
# >>  #<struct Team won=3, lost=0, demerits=1, style=0>,
# >>  #<struct Team won=0, lost=4, demerits=1, style=4>,
# >>  #<struct Team won=1, lost=4, demerits=3, style=0>,
# >>  #<struct Team won=3, lost=3, demerits=2, style=3>,
# >>  #<struct Team won=0, lost=4, demerits=4, style=4>]
# >> 
# >> After sort:
# >> [#<struct Team won=4, lost=0, demerits=4, style=0>,
# >>  #<struct Team won=4, lost=1, demerits=2, style=3>,
# >>  #<struct Team won=3, lost=0, demerits=1, style=1>,
# >>  #<struct Team won=3, lost=0, demerits=1, style=0>,
# >>  #<struct Team won=3, lost=0, demerits=3, style=0>,
# >>  #<struct Team won=3, lost=1, demerits=1, style=4>,
# >>  #<struct Team won=3, lost=3, demerits=2, style=3>,
# >>  #<struct Team won=3, lost=4, demerits=1, style=2>,
# >>  #<struct Team won=3, lost=4, demerits=1, style=2>,
# >>  #<struct Team won=3, lost=4, demerits=2, style=0>,
# >>  #<struct Team won=3, lost=4, demerits=3, style=4>,
# >>  #<struct Team won=1, lost=2, demerits=0, style=1>,
# >>  #<struct Team won=1, lost=2, demerits=4, style=3>,
# >>  #<struct Team won=1, lost=4, demerits=0, style=0>,
# >>  #<struct Team won=1, lost=4, demerits=1, style=4>,
# >>  #<struct Team won=1, lost=4, demerits=1, style=2>,
# >>  #<struct Team won=1, lost=4, demerits=3, style=0>,
# >>  #<struct Team won=0, lost=4, demerits=1, style=4>,
# >>  #<struct Team won=0, lost=4, demerits=2, style=2>,
# >>  #<struct Team won=0, lost=4, demerits=4, style=4>]

How to sort array of strings with multiple conditions?

If these attributes are database columns, you can use:

Organization.order(attribute_a: :desc, attribute_b: :asc)

Otherwise, use sort with an array:

Arrays are compared in an “element-wise” manner; the first two elements that are not equal will determine the return value for the whole comparison.

Exchanging the first elements sorts them in descending order:

array.sort { |x, y| [y.attribute_a, x.attribute_b] <=> [x.attribute_a, y.attribute_b] }
#                    ^                                  ^
#                    |                                  |
#                    +-------- x and y exchanged -------+

To generate a list as mentioned in your comment, you can use group_by:

<% sorted_array.group_by(&:attribute_a).each do |attr, group| %>
  <%= attr %> #=> "Type z"
  <% group.each do |item| %>
    <%= item.attribute_b %> #=> "name a", "name b", ...
  <% end %>
<% end %>

Sorting: Sort array based on multiple conditions in Ruby

You should always use sort_by for a keyed sort. Not only is it much more readable, it is also much more efficient. In addition, I would also prefer to use destructuring bind, again, for readability:

array.sort_by {|name, age| [age, name] }

Sort an array and make it unique on multiple conditions - Ruby

you can use sort_by method and uniq and values_at

hashForAnimals.sort_by{ |a| a[:sortOrder] }.uniq{ |k| k.values_at(:animalCd, :animalType) }

# => [{:animalCd=>"Cow", :animalType=>"Carnivore", :sortOrder=>1}, {:animalCd=>"Rabbit", :animalType=>"Herbivore", :sortOrder=>2}, {:animalCd=>"Tiger", :animalType=>"Carnivore", :sortOrder=>3}, {:animalCd=>"Shark", :animalType=>"Carnivore", :sortOrder=>4}, {:animalCd=>"Cow", :animalType=>"Herbivore", :sortOrder=>5}, {:animalCd=>"Bear", :animalType=>"Omnivore", :sortOrder=>7}]

Sorting By Multiple Conditions in Ruby

You should always sort by the same criteria to insure a meaningful order. If comparing two nil dates, it is fine that the position will judge of the order, but if comparing one nil date with a set date, you have to decide which goes first, irrespective of the position (for example by mapping nil to a day way in the past).

Otherwise imagine the following:

a.date = nil                   ; a.position = 1
b.date = Time.now - 1.day      ; b.position = 2
c.date = Time.now              ; c.position = 0

By your original criteria, you would have: a < b < c < a. So, which one is the smallest??

You also want to do the sort at once. For your <=> implementation, use #nonzero?:

def <=>(other)
  return nil unless other.is_a?(Post)
  (self.category <=> other.category).nonzero? ||
  ((self.date || AGES_AGO) <=> (other.date || AGES_AGO)).nonzero? ||
  (self.position <=> other.position).nonzero? ||
  0
end

If you use your comparison criteria just once, or if that criteria is not universal and thus don't want to define <=>, you could use sort with a block:

post_ary.sort{|a, b| (a.category <=> ...).non_zero? || ... }

Better still, there is sort_by and sort_by! which you can use to build an array for what to compare in which priority:

post_ary.sort_by{|a| [a.category, a.date || AGES_AGO, a.position] }

Besides being shorter, using sort_by has the advantage that you can only obtain a well ordered criteria.

Notes:

• sort_by! was introduced in Ruby 1.9.2. You can require 'backports/1.9.2/array/sort_by' to use it with older Rubies.
• I'm assuming that Post is not a subclass of ActiveRecord::Base (in which case you'd want the sort to be done by the db server).

How can I sort by multiple conditions with different orders?

How about:

ordered_list = [[1, "b"], [1, "a"], [2, "a"]]
ordered_list.sort! do |a,b|
  [a[0],b[1]] <=> [b[0], a[1]]
end

sorting and rearranging an array of hashes based on multiple conditions

Here is another option using what you already have as a base (Since you were basically all the way there)

a = [
  { "name" => "X", "year" => "2013-08"},
  { "name" => "A", "year" => "2017-01"},
  { "name" => "X", "year" => "2000-08"},
  { "name" => "B", "year" => "2018-05"},
  { "name" => "D", "year" => "2016-04"},
  { "name" => "C", "year" => "2016-04"}
]

a.sort do  |a,b| 
  a_ord, b_ord = [a,b].map {|e| e["name"] == "X" ? 0 : 1 }
  [a_ord,b["year"],a["name"] ] <=> [b_ord, a["year"],b["name"]]
end

Here we just make sure that "X" is always in front by assigning it a 0 and everything else a 1. Then since 0 and 0 would be equivalent X will fall back to the same logic you already have applied as will all the others. We can make this a bit fancier as:

a.sort do  |a,b| 
  [a,b].map {|e| e["name"] == "X" ? 0 : 1 }.zip(
    [b["year"],a["year"]],[a["name"],b["name"]]
  ).reduce(:<=>)
end

How to sort a ruby array by two conditions

This should do it:

require 'ostruct'
arr = [
  OpenStruct.new(percent: 73, type: 1),
  OpenStruct.new(percent: 70, type: 4),
  OpenStruct.new(percent: 60, type: 4),
  OpenStruct.new(percent: 50, type: 4),
  OpenStruct.new(percent: 64, type: 1),
  OpenStruct.new(percent: 74, type: 2)
]

puts arr.sort_by { |a| [a.type % 4, -a.percent] }

output:

#<OpenStruct percent=70, type=4>
#<OpenStruct percent=60, type=4>
#<OpenStruct percent=50, type=4>
#<OpenStruct percent=73, type=1>
#<OpenStruct percent=64, type=1>
#<OpenStruct percent=74, type=2>

Ruby Sort an array of arrays of numbers based on multiple conditions

My understanding is that when a[2] >= 0, sorting is to on the array [a[1], a[2]], and elements for which a[2] < 0 are to be at the end of the sorted array and sorted by [-a[1], -a[2]].

biggest_plus_1 = to_sort.map { |a| a[2] }.max + 1
  #=> 3
to_sort.sort_by { |a| a[2] >= 0 ? [0, a[1], a[2]] : [biggest_plus_1, -a[1], -a[2]] }
  #=> [[6, 27, 1, 11.0], [7, 27, 1, 12.0], [8, 27, 1, 13.0], [9, 27, 2, 14.0],
  #    [5, 27, -2, 5.0], [2, 27, -2, 2.0], [3, 27, -2, 3.0], [4, 27, -2, 4.0],
  #    [1, 27, -3, 1.0]] 

Array#sort and Enumerable#sort_by rely on the method Array#<=> for determining the ordering of each pair of arrays being sorted. Two arrays, a and b are ordered lexicographically, meaning the following. If a[0] < b[0] then a is less than b (a < b), or equivalently, a <=> b #=> -1. Similarly, if a[0] > b[0] then a is greater than b (a > b) and a <=> b #=> 1. If a[0] == b[0], the tie is broken by the comparing the second elements in the same way, and so on. If a is smaller than b (a.size < b.size), and the first a.size elements of each array are equal, a < b. a and b are equal if and only if a <=> b #=> 0.

Since elements a for which a[2] < 0 are to be placed at the end of the sorted array, we need to sort by arrays whose first elements place the array at the front or back of the sorted array. It is for that reason that I made the first element of the sort-by array zero when a[2] >= 0 and biggest_plus_1 when a[2] < 0, where biggest_plus_1 is the largest value of a[2] plus 1.

The remaining elements of the sort-by arrays determine how each of the two groups of arrays are to be sorted.

Note that biggest_plus_1 will be non-positive if all a[2] < 0, but that doesn't matter, as no element will be sorted by an array whose first element is zero.

Sort array of active records by multiple columns

A good way is to use sort and <=> and nonzero? like this:

jobs.sort{|a,b| 
  (a.organization <=> b.organization).nonzero? ||
  (b.created_at <=> a.created_at)
}

This code says:

1. Compare A and B by organization.
2. If they differ, then we have our answer.
3. If they are the same, then we need to do more.
4. Compare B and A by time. (Note B & A are in reverse order)
5. If they differ, then we have our answer.
6. If they are the same, then the sort order doesn't matter. (Ruby sort is "unstable")

Example code independent of ActiveRecord:

require 'time'
require 'ostruct'

jobs = [
  OpenStruct.new(_id: 1, created_at: Time.parse("2014-07-15 19:18:40 UTC"), organization: "Acme Inc"),
  OpenStruct.new(_id: 3, created_at: Time.parse("2014-05-20 09:27:38 UTC"), organization: "Baxter"), 
  OpenStruct.new(_id: 2, created_at: Time.parse("2014-11-25 12:21:00 UTC"), organization: "Wizard"), 
  OpenStruct.new(_id: 3, created_at: Time.parse("2015-01-15 07:20:10 UTC"), organization: "Baxter")
]

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