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Hash destructuring in Ruby 3

Tags:

ruby

I Ruby 2 you could do the following:

my_hash = {a: {aa: 1, ab: 2, ac: 3}}

my_hash.each do |key, aa:, ab: 4, **|
  puts key
  puts aa
  puts ab
end

In Ruby 3 this now results in missing keywords :aa, :ab. What would be the best way to refactor code like this in Ruby 3?

Something like the following would not work because it doesn't support setting default values:

my_hash.each do |key, values|
  values in {aa: aa, ab: ab}
end

The best way I can think of is putting the existing code in a wrapper:

lambda = ->(key, aa:, ab: 4, **) do
  puts key
  puts aa
  puts ab
end

my_hash.each do |key, values|
  lambda.call(key, **values)
end

Any better options?

like image 811
Albert Avatar asked May 31 '21 13:05

Albert


Video Answer


3 Answers

I can't think of a way to convert the hash to keyword arguments.

But why not use the hash the way it is, i.e. without treating its keys like keywords? Unless your actual code is more complex, fetch seems to do what you want:

my_hash = {a: {aa: 1, ab: 2, ac: 3}}

my_hash.each do |key, values|
  puts key
  puts values.fetch(:aa)
  puts values.fetch(:ab, 4)
end

The above works fine in both, Ruby 2 and 3. Just like your example, it will raise an error if :aa is missing and it will use a default value of 4 if :ab is missing.

like image 55
Stefan Avatar answered Oct 28 '22 00:10

Stefan


with ||

As stated by @Stefan it will not works if the value can contains nil or false

my_hash = {a: {aa: 1, ac: 3}}
 
my_hash.each do |key, values|
  puts key
  puts values[:aa]
  puts values[:ab] || 4
end

with key?

my_hash = {a: {aa: 1, ac: 3}}
default_values_hash = { :ab => 4 }
 
my_hash.each do |key, values|
  puts key
  puts values[:aa]
  puts values.key?(:ab) ? values[:ab] : default_values_hash[:ab]
end

with default

my_hash = {a: {aa: 1, ab: 2, ac: 3}}

my_hash.each do |key, values|
  values.default = :notfound
  puts key
  puts values[:aa]
  puts values[:ab] == :notfound ? 4 : values[:ab]
end
my_hash = {a: {aa: 1, ac: 3}}
 
my_hash.each do |key, values|
  values.default = :notfound
  puts key
  puts values[:aa]
  puts values[:ab] == :notfound ? 4 : values[:ab]
end

with default-proc

my_hash = {a: {aa: 1, ab: 2, ac: 3}}
default_values_hash = { :ab => 4 }

my_hash.each do |key, values|
  values.default_proc = proc { |hash, key| default_values_hash[key] || :notfound }
  puts key
  puts values[:aa]
  puts values[:ab]
end

with merge

my_hash = {a: {aa: 1, ac: 3}}
default_values_hash = { :ab => 4 }
 
my_hash.each do |key, values|
  values = default_values_hash.merge(values)
  puts key
  puts values[:aa]
  puts values[:ab]
end

All above code works in both Ruby 2 and 3

Note: I would suggest the key? method to test and check if key exists and handle the case accordingly as it would be much faster as comparing to other cases which require additional steps internally.

like image 37
Chandan Avatar answered Oct 27 '22 23:10

Chandan


If you don't need to use default value but need just destruct hash, you can use => operator for pattern matching

my_hash = { a: { aa: 1, ab: 2, ac: 3 } }

my_hash.each do |key, values|
  values => { aa:, ab:, ac: }

  puts key, aa, ab, ac
end

This code will raise NoMatchingPatternKeyError if values don't include all keys

like image 1
mechnicov Avatar answered Oct 27 '22 22:10

mechnicov