How to find union between array of first value in ruby - ruby

Question

I have a set of array like this:
[["1","2"],["1","3"],["2","3"],["2","5"]]
I want to find the union of first values like
["1","2"],["1","3"] matches so i need to create new array like ["1","2,3"]
so the resulting array will be like
[["1","2,3"],["2","3,5"]]

Answer 1

Like most problems in Ruby, the Enumerable module does the job:
input = [["1","2"],["1","3"],["2","3"],["2","5"]]
result = input.group_by do |item|
# Group by first element
item[0]
end.collect do |key, items|
# Compose into new format
[
key,
items.collect do |item|
item[1]
end.join(',')
]
end
puts result.inspect
# => [["1", "2,3"], ["2", "3,5"]]
The group_by method comes in very handy when aggregating things like this, and collect is great for rewriting how the elements appear.

Answer 2

What you are asking for is not a true union for a true union of each 2 it would be:
data = [["1","2"],["1","3"],["2","3"],["2","5"]]
data.each_slice(2).map{|a,b| a | b.to_a }
#=> [["1","2","3"],["2","3","5"]]
Here is a very simple solution that modifies this concept to fit your needs:
data = [["1","2"],["1","3"],["2","3"],["2","5"]]
data.each_slice(2).map do |a,b|
unified = (a | b.to_a)
[unified.shift,unified.join(',')]
end
#=> [["1", "2,3"], ["2", "3,5"]]
Added to_a to piped variable b in the event that there are an uneven number of arrays. eg.
data = [["1","2"],["1","3"],["2","3"],["2","5"],["4","7"]]
data.each_slice(2).map do |a,b|
unified = (a | b.to_a)
[unified.shift,unified.join(',')]
end
#=> [["1", "2,3"], ["2", "3,5"], ["4","7"]]
If you meant that you want this to happen regardless of order then this will work but will destroy the data object
data.group_by(&:shift).map{|k,v| [k,v.flatten.join(',')]}
#=> [["1", "2,3"], ["2", "3,5"], ["4","7"]]
Non destructively you could call
data.map(&:dup).group_by(&:shift).map{|k,v| [k,v.flatten.join(',')]}
#=> [["1", "2,3"], ["2", "3,5"], ["4","7"]]

Answer 3

Here's another way.
Code
def doit(arr)
arr.each_with_object({}) { |(i,*rest),h| (h[i] ||= []).concat(rest) }
.map { |i,rest| [i, rest.join(',')] }
end
Examples
arr1 = [["1","2"],["1","3"],["2","3"],["2","5"]]
doit(arr1)
#=> [["1", "2,3"], ["2", "3,5"]]
arr2 = [["1","2","6"],["2","7"],["1","3"],["2","3","9","4","cat"]]
doit(arr2)
# => [["1", "2,6,3"], ["2", "7,3,9,4,cat"]]
Explanation
For arr1 above, we obtain:
enum = arr1.each_with_object({})
#=> #<Enumerator: [["1", "2"], ["1", "3"], ["2", "3"],
# ["2", "5"]]:each_with_object({})>
We can convert enum to an array see its elements:
enum.to_a
#=> [[["1", "2"], {}], [["1", "3"], {}],
# [["2", "3"], {}], [["2", "5"], {}]]
These elements will be passed into the block, and assigned to the block variables, by Enumerator#each, which will invoke Array#each. The first of these elements ([["1", "2"], {}]) can be obtained by invoking Enumerator#next on enum:
(i,*rest),h = enum.next
#=> [["1", "2"], {}]
i #=> "1"
rest #=> ["2"]
h #=> {}
We then execute:
(h[i] ||= []).concat(rest)
#=> (h["1"] ||= []).concat(["2"])
#=> (nil ||= []).concat(["2"])
#=> [].concat(["2"])
#=> ["2"]
each then passes the next element of enum to the block:
(i,*rest),h = enum.next
#=> [["1", "3"], {"1"=>["2"]}]
i #=> "1"
rest #=> ["3"]
h #=> {"1"=>["2"]}
(h[i] ||= []).concat(rest)
#=> (h["1"] ||= []).concat(["3"])
#=> (["2"] ||= []).concat(["3"])
#=> ["2"].concat(["3"])
#=> ["2", "3"]
After passing the last two elements of enum into the block, we obtain:
h=> {"1"=>["2", "3"], "2"=>["3", "5"]}
map creates an enumerator:
enum_h = h.each
#=> > #<Enumerator: {"1"=>["2", "3"]}:each>
and calls Enumerator#each (which calls Hash#each) to pass each element of enum_h into the block:
i, rest = enum_h.next
#=> ["1", ["2", "3"]]
then computes:
[i, rest.join(',')]
#=> ["1", ["2", "3"].join(',')]
#=> ["1", "2,3"]
The other element of enum_h is processed similarly.