How to sort a mixed array by different elements in Ruby?

Question

I have an array similar to this:

a = [
  [0, {:a=>"31", :b=>"21"}],
  [1, {:a=>"32", :b=>"11"}],
  [1, {:a=>"25", :b=>"19"}],
  [0, {:a=>"12", :b=>"10"}]
]

And I want to sort it by the first element of each row or by the various elements of the hash (2nd element in the row).

Tilo · Accepted Answer · 2011-11-17 01:04:54Z

7

to sort by the first item in the Array:

> a.sort_by{|x| x.first}
 => [[0, {:a=>"31", :b=>"21"}], 
     [0, {:a=>"12", :b=>"10"}], 
     [1, {:a=>"32", :b=>"11"}], 
     [1, {:a=>"25", :b=>"19"}]]

to sort by the first item in the Hash:

> a.sort_by{|x| x.last.first}
 => [[0, {:a=>"12", :b=>"10"}],
     [1, {:a=>"25", :b=>"19"}], 
     [0, {:a=>"31", :b=>"21"}], 
     [1, {:a=>"32", :b=>"11"}]]

or you could sort by a given key of the hash:

> sort_key = :b
> a.sort_by{|x| x.last[ sort_key ]}
 => [[0, {:a=>"12", :b=>"10"}],
     [1, {:a=>"32", :b=>"11"}],
     [1, {:a=>"25", :b=>"19"}],
     [0, {:a=>"31", :b=>"21"}]]

If you want to sort by the first array value, and then by the first entry in the hash, as a secondary search criteria, the answer is:

> a.sort_by{|x| [x.first, x.last.first]}
 => [[0, {:a=>"12", :b=>"10"}], 
     [0, {:a=>"31", :b=>"21"}], 
     [1, {:a=>"25", :b=>"19"}], 
     [1, {:a=>"32", :b=>"11"}]]

edited Nov 17, 2011 at 1:04

answered Nov 16, 2011 at 23:26

Tilo

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2 Comments

Andrew Vit Over a year ago

Note that "first entry in the hash" only works in Ruby 1.9. Hashes in ruby 1.8 only have keys, and no implicit order.

shabdar Over a year ago

By "or" I meant to ways of sorting it: 1) by the first element of each row 2) by the first element in the hash (which itself is the second element in the row) P.S. I just added a comment with my solution. It works, but I'm wondering if there's a better way.

steenslag · Accepted Answer · 2011-11-17 01:21:26Z

1

a = [
  [0, {:a=>"31", :b=>"21"}],
  [1, {:a=>"32", :b=>"11"}],
  [1, {:a=>"25", :b=>"19"}],
  [0, {:a=>"12", :b=>"10"}]
]
p a.sort_by{|el| [el.first, *el.last.values]}

Output:

=> [[0, {:a=>"12", :b=>"10"}], 
    [0, {:a=>"31", :b=>"21"}], 
    [1, {:a=>"25", :b=>"19"}], 
    [1, {:a=>"32", :b=>"11"}]]

edited Nov 17, 2011 at 1:21

user979339

answered Nov 16, 2011 at 23:21

steenslag

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4 Comments

user979339 Over a year ago

this doesn't seem to be sorted at all

Andrew Vit Over a year ago

@UnixGuy How is it not sorted? It looks right to me: ordered by the first element, followed by the elements in the hash.

steenslag Over a year ago

@Unix guy : it's sorted by the first element of the arays (so 0 comes before 1); if those are the same the value of key :a of the hash decides who comes first, then the value of key :b. (Added output to the answer).

user979339 Over a year ago

sorry, my bad! I'll take that back. @steenslag posted "a" instead of posting the result, that was confusing ... +1 for his solution! I also fixed his output formatting

shabdar · Accepted Answer · 2011-11-18 00:21:12Z

0

Well, I found an answer to my own question. Please compare & comment.

#sort by first item of each row (number)
a.sort{|x,y| x[0] <=> y[0]}

#sort by the first item in the hash
a.sort{|x,y| x[1][:a] <=> y[1][:a]}

answered Nov 18, 2011 at 0:21

shabdar

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