I need this Hash
{"peter" => ["apple", "orange", "mango"], "sandra" => ["flowers", "bike"]}
convert to this Array:
[["peter", "apple"], ["peter", "orange"], ["peter", "mango"], ["sandra", "flowers"], ["sandra", "bike"]]
Now I have got this solution
my_hash.inject([]){|ar, (k,v)| ar << v.map{|c| [k,c]}}.flatten(1)
But I believe here is more elegant solution with those zip or transpose magick :)
You are right to be suspicious about Enumerable#inject solutions. In Ruby, inject/reduce is somewhat abused, we must be careful and choose the right abstraction (map, select, zip, flatten...) if they fit the problem at hand. In this case:
h = {"peter" => ["apple", "orange", "mango"], "sandra" => ["flowers", "bike"]}
h.map { |k, vs| vs.map { |v| [k, v] } }.flatten(1)
#=> [["peter", "apple"], ["peter", "orange"], ["peter", "mango"], ["sandra", "flowers"], ["sandra", "bike"]]
But if you want to use Enumerable#zip don't let anyone stop you ;-)
h.map { |k, vs| [k].cycle(vs.size).zip(vs) }.flatten(1)
And as @steenslag says, also:
h.map { |k, vs| [k].product(vs) }.flatten(1)
So at the end we can write:
h.flat_map { |k, vs| [k].product(vs) }
zipor transpose magick! EDIT: Sorry, you edited it 23 seconds before I wrote this comment.
h.inject([]){|a,(k,vs)| a+vs.map {|v| [k,v]}}
You could also use this version
h.inject([]){|a,(k,vs)| a+=vs.map {|v| [k,v]}}
Which is most efficient because it use the same list rather than creating a new one at each iteration. However it feels wrong (for me) to use inject and modify a variable in place. An each version would do the same.
a = []; h.each {|k,vs| a+=vs.map {|v| [k,v]}}
It's slightly shorter and as expressive.
With zip
hash.inject([]){ |ar, (k,v)| ar << ([k]*v.size).zip(v) }
A plausible solution using transpose too:
[
hash.keys.map{|k| [k]*hash[k].size }.flatten,
hash.keys.map{|k| hash[k] }.flatten
].transpose
Take into account that:
hash.keys should return the keys in the same order in both cases, so don't use it in other language unless you are sure of this.