Modify hashes in an array based on another array - arrays

Question

I have two arrays like this:
a = [{'one'=>1, 'two'=>2},{'uno'=>1, 'dos'=>2}]
b = ['english', 'spanish']
I need to add a key-value pair to each hash in a to get this:
a = [{'one'=>1, 'two'=>2, 'language'=>'english'},{'uno'=>1, 'dos'=>2, 'language'=>'spanish'}]
I attempted this:
(0..a.length).each {|c| a[c]['language']=b[c]}
and it does not work. With this:
a[1]['language']=b[1]
(0..a.length).each {|c| puts c}
an error is shown:
NoMethodError (undefined method '[]=' for nil:NilClass)
How can I fix this?

Answer 1

a.zip(b){|h, v| h["language"] = v}
a # => [
# {"one"=>1, "two"=>2, "language"=>"english"},
# {"uno"=>1, "dos"=>2, "language"=>"spanish"}
# ]

Answer 2

When the each iterator over your Range reaches the last element (i.e. a.length), you will attempt to access a nonexisting element of a.
In your example, a.length is 2, so on the last iteration of your each, you will attempt to access a[2], which doesn't exist. (a only contains 2 elements wich indices 0 and 1.) a[2] evaluates to nil, so you will now attempt to call nil['language']=b[2], which is syntactic sugar for nil.[]=('language', b[2]), and since nil doesn't have a []= method, you get a NoMethodError.
The immediate fix is to not iterate off the end of a, by using an exclusive Range:
(0...a.length).each {|c| a[c]['language'] = b[c] }
By the way, the code you posted:
(0..a.length).each {|c| puts c }
should clearly have shown you that you iterate till 2 instead of 1.
That's only the immediate fix, however. The real fix is to simply never iterate over a datastructure manually. That's what iterators are for.
Something like this, where Ruby will keep track of the index for you:
a.each_with_index do |hsh, i| hsh['language'] = b[i] end
Or, without fiddling with indices at all:
a.zip(b.zip(['language'].cycle).map(&:reverse).map(&Array.method(:[])).map(&:to_h)).map {|x, y| x.merge!(y) }
[Note: this last one doesn't mutate the original Arrays and Hashes unlike the other ones.]

Answer 3

The problem you're having is that your (0..a.length) is inclusive. a.length = 2 so you want to modify it to be 0...a.length which is exclusive.
On a side note, you could use Array#each_with_index like this so you don't have to worry about the length and so on.
a.each_with_index do |hash, index|
hash['language'] = b[index]
end

Answer 4

Here is another method you could use
b.each_with_index.with_object(a) do |(lang,i),obj|
obj[i]["language"] = lang
obj
end
#=>[
{"one"=>1, "two"=>2, "language"=>"english"},
{"uno"=>1, "dos"=>2, "language"=>"spanish"}
]
What this does is creates an Enumerator for b with [element,index] then it calls with_object using a as the object. It then iterates over the Enumerator passing in each language and its index along with the a object. It then uses the index from b to find the proper index in a and adds a language key to the hash that is equal to the language.
Please know this is a destructive method where the objects in a will mutate during the process. You could make it non destructive using with_object(a.map(&:dup)) this will dup the hashes in a and the originals will remain untouched.
All that being said I think YAML would be better suited for a task like this but I am not sure what your constraints are. As an example:
yml = <<YML
-
one: 1
two: 2
language: "english"
-
uno: 1
dos: 2
language: "spanish"
YML
require 'yaml'
YAML.load(yml)
#=>[
{"one"=>1, "two"=>2, "language"=>"english"},
{"uno"=>1, "dos"=>2, "language"=>"spanish"}
]
Although using YAML I would change the structure for numbers to be more like language => Array of numbers by index e.g. {"english" => ["zero","one","two"]}. That way you can can access them like ["english"][0] #=> "zero"