Changing a value of a key

Question

Is there a shorter way to change the value of a key?

 my_array.each do |x| 
    if my_hash.key?(x) 
      my_hash[x] += 1
    else
      my_hash[x] = 1
    end
  end

possible duplicate of [Ruby: What is the easiest method to update Hash values?](http://stackoverflow.com/questions/5215713/ruby-what-is-the-easiest-method-to-update-hash-values) — Roman Kiselenko, Nov 20 '14 at 09:11
@Зелёный, what if a key doesn't exist, will it still work? — Incerteza, Nov 20 '14 at 09:27
@AlexanderSupertramp : Check my answer if this is what you are looking for then. otherwise let me know so I can update my answer accordingly — Hetal Khunti, Nov 20 '14 at 09:32

score 1 · Answer 1 · answered Nov 20 '14 at 09:25

1

A bit shorter:

my_array.each do |x| 
  my_hash[x] ||= 0
  my_hash[x] += 1
end

answered Nov 20 '14 at 09:25

Thomas

but I don't want it to be 0 by default. – Incerteza Nov 20 '14 at 09:27
1

It won't be zero since you add 1 right after initialisation – Thomas Nov 20 '14 at 09:27
I don't want it to have any default value. – Incerteza Nov 20 '14 at 13:38

score 1 · Answer 2 · answered Nov 20 '14 at 09:27

1

In your specific case it's probably easiest to give the hash a default value:

my_hash = Hash.new(1)
=> {}
my_hash[:x]
=> 1
my_hash[:y] += 1
=> 2

Please note that this only makes sense if the default value is immutable, otherwise the reference will be shared between all keys.

answered Nov 20 '14 at 09:27

Michael Kohl

how can y be equal to 2? – Incerteza Nov 20 '14 at 13:38
Because it had a default value of 1. If you don't want that, initialize the hash with `Hash.new(0)`. – Michael Kohl Nov 21 '14 at 03:44
I don't want it to have a default value at all. I want to upsert a value. – Incerteza Nov 21 '14 at 03:51

score 0 · Answer 3 · answered Nov 20 '14 at 17:01

0

You can use the ternary operator to write your statement on one line.

my_array.each {|x| my_hash.key?(x) ? my_hash[x].next : my_hash[x] = 1}

You can also use the next method if you are increasing a value by just 1.

answered Nov 20 '14 at 17:01

Darkmouse

3 Answers3