Sorting: Sort array based on multiple conditions in Ruby

Question

I have a mulitdimensional array like so:

[
  [name, age, date, gender]
  [name, age, date, gender]
  [..]
]

I'm wondering the best way to sort this array based on multiple conditions...For instance, how would I sort based on age first then by name?

I was messing around with the sort method like so:

array.sort { |a,b| [ a[1], a[0] ] <=> [ b[1], b[0] ] }

Besides that I don't really understand this syntax, I'm not getting the results I would expect. Should I be using the sort method? Should I be individually comparing results by mapping the array?

possible duplicate of Sort a collection of objects by number (highest first) then by letter (alphabetical) — Andrew Grimm
– Andrew Grimm, Commented Jan 17, 2012 at 2:15
@pruett: no disrespect for robbrit's answer, but you should consider the selected answer, there is nothing wrong with using Enumerable#sort except when Enumerable#sort_by does the job. This can be misleading for people landing here. — tokland
– tokland, Commented Jan 17, 2012 at 8:11

Iulian Onofrei · Accepted Answer · 2020-10-20 06:35:06Z

51

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] }

edited Oct 20, 2020 at 6:35

Iulian Onofrei

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answered Jan 17, 2012 at 2:56

Jörg W Mittag

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Comments

robbrit · Accepted Answer · 2012-01-17 00:12:03Z

17

This should do the trick:

array.sort { |a,b| [ a[1], a[0] ] <=> [ b[1], b[0] ] }

So what does this do? It uses a lot of Ruby idioms.

First is blocks, which are sort of like callbacks or anonymous functions/classes in other languages. The sort method of Array uses them to compare two elements based on the return value of the block. You can read all about them here.
Next is the <=> operator. It returns -1 if the first argument is less than the second, 0 if they are equal, and 1 if the first is greater than the second. When you use it with arrays, it will compare the arrays element-wise until one of them returns -1 or 1. If the arrays are equal, you will get 0.

answered Jan 17, 2012 at 0:12

robbrit

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1 Comment

mu is too short Over a year ago

You could also use sort_by { |a| [a[1], a[0]] }, constructing all those arrays doesn't come for free.

Fab · Accepted Answer · 2012-01-17 00:26:28Z

7

As I understand it you want to order by age first, and then if more than one record has the same age, arrange that subset by name.

This works for me

people = [
      ["bob", 15, "male"], 
      ["alice", 25, "female"], 
      ["bob", 56, "male"], 
      ["dave", 45, "male"], 
      ["alice", 56, "female"], 
      ["adam", 15, "male"]
    ]

people.sort{|a,b| (a[1] <=> b[1]) == 0 ? (a[0] <=> b[0]) : (a[1] <=> b[1]) }

# The sorted array is

[["adam", 15, "male"], 
 ["bob", 15, "male"], 
 ["alice", 25, "female"], 
 ["dave", 45, "male"], 
 ["alice", 56, "female"], 
 ["bob", 56, "male"]]

What this is doing is comparing by age first, and if the age is the same (<=> returs 0) it comparing the name.

answered Jan 17, 2012 at 0:26

Fab

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3 Comments

pruett Over a year ago

very nice! i like that syntax

inquisitive Over a year ago

what if the age is same and then we want to sort the name in descending order?

user1515295 Over a year ago

@Inquisitive then in that case I think you would switch the "a" and the "b" in the "(a[1] <=> b[1])" on the right.

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