ruby one-liner from two hashes - ruby

Question

a = {"rows" => [{"id" => "231-z", "name" => 'jon', "age"=> 27, "state" => 'AL'},
{"id" => "4121-x", "name" => 'ton', "age"=> 37, "state" => 'VA'}
]
}
b = {"rows" => [{"key" => ["xyz","4121-x"], "value" =>{"sum" => 12312, "realage" => 29}},
{"key" => ["xyz","231-z"], "value" =>{"sum" => 1212, "realage" => 33}}
]
}
In hash a, age is incorrect
In hash b, realage is correct. Also in hash b id is the second value in the first array that maps to id of hash a . Those are 4121-x, 231-z correspond to hash a
I want to correct the age in hash a and swap it with the realage of hash b
I can do it in multiple steps, but is it possible to do it in one liner or very short? So finally correct hash a should look like
a = {"rows" => [{"id" => "231-z", "name" => 'jon', "age"=> 33, "state" => 'AL'},
{"id" => "4121-x", "name" => 'ton', "age"=> 29, "state" => 'VA'}
]
}

Answer 1

does this look reasonable?
a['rows'].each_with_index do |ah, i|
(bh = b['rows'].select {|h| h['key'].last == ah['id'] }.first) &&
a['rows'][i] = ah.update('age' => bh['value']['realage'])
end
p a
{
"rows" => [
[0] {
"id" => "231-z",
"name" => "jon",
"age" => 33,
"state" => "AL"
},
[1] {
"id" => "4121-x",
"name" => "ton",
"age" => 29,
"state" => "VA"
}
]
}
Please note it will update a only if corresponding id found in b.
Also, the rows order does not matter, nor matter the rows number, it is only important b to have a row with same id as processed row in a
Here is a Working Demo