Questions on implementing hashes in ruby - ruby

I'm new to ruby, I am solving a problem that involves hashes and key. The problem asks me to Implement a method, #pet_types, that accepts a hash as an argument. The hash uses people's # names as keys, and the values are arrays of pet types that the person owns. My question is about using Hash#each method to iterate through each num inside the array. I was wondering if there's any difference between solving the problem using hash#each or hash.sort.each?
I spent several hours coming up different solution and still to figure out what are the different approaches between the 2 ways of solving the problem below.
I include my code in repl.it: https://repl.it/H0xp/6 or you can see below:
# Pet Types
# ------------------------------------------------------------------------------
# Implement a method, #pet_types, that accepts a hash as an argument. The hash uses people's
# names as keys, and the values are arrays of pet types that the person owns.
# Example input:
# {
# "yi" => ["dog", "cat"],
# "cai" => ["dog", "cat", "mouse"],
# "venus" => ["mouse", "pterodactyl", "chinchilla", "cat"]
# }
def pet_types(owners_hash)
results = Hash.new {|h, k| h[k] = [ ] }
owners_hash.sort.each { |k, v| v.each { |pet| results[pet] << k } }
results
end
puts "-------Pet Types-------"
owners_1 = {
"yi" => ["cat"]
}
output_1 = {
"cat" => ["yi"]
}
owners_2 = {
"yi" => ["cat", "dog"]
}
output_2 = {
"cat" => ["yi"],
"dog" => ["yi"]
}
owners_3 = {
"yi" => ["dog", "cat"],
"cai" => ["dog", "cat", "mouse"],
"venus" => ["mouse", "pterodactyl", "chinchilla", "cat"]
}
output_3 = {
"dog" => ["cai", "yi"],
"cat" => ["cai", "venus", "yi"],
"mouse" => ["cai", "venus"],
"pterodactyl" => ["venus"],
"chinchilla" => ["venus"]
}
# method 2
# The 2nd and 3rd method should return a hash that uses the pet types as keys and the values should
# be a list of the people that own that pet type. The names in the output hash should
# be sorted alphabetically
# switched_hash = Hash.new()
# owners_hash.each do |owner, pets_array|
# pets_array.each do |pet|
# select_owners = owners_hash.select { |owner, pets_array|
owners_hash[owner].include?(pet) }
# switched_hash[pet] = select_owners.keys.sort
# end
# end
# method 3
#switched_hash
# pets = Hash.new {|h, k| h[k] = [ ] } # WORKS SAME AS: pets = Hash.new( Array.new )
# owners = owners_hash.keys.sort
# owners.each do |owner|
# owners_hash[owner].each do |pet|
# pets[pet] << owner
# end
# end
# pets
# Example output:
# output_3 = {
# "dog" => ["cai", "yi"],
# "cat" => ["cai", "venus", "yi"], ---> (sorted alphabetically!)
# "mouse" => ["cai", "venus"],
# "pterodactyl" => ["venus"],
# "chinchilla" => ["venus"]
# }
I used a hash data structure in my program to first solve this problem. Then I tried to rewrite it using the pet_hash. And my final codes is the following:
def pet_types(owners_hash)
pets_hash = Hash.new { |k, v| v = [] }
owners_hash.each do |owner, pets|
pets.each do |pet|
pets_hash[pet] += [owner]
end
end
pets_hash.values.each(&:sort!)
pets_hash
end
puts "-------Pet Types-------"
owners_1 = {
"yi" => ["cat"]
}
output_1 = {
"cat" => ["yi"]
}
owners_2 = {
"yi" => ["cat", "dog"]
}
output_2 = {
"cat" => ["yi"],
"dog" => ["yi"]
}
owners_3 = {
"yi" => ["dog", "cat"],
"cai" => ["dog", "cat", "mouse"],
"venus" => ["mouse", "pterodactyl", "chinchilla", "cat"]
}
output_3 = {
"dog" => ["cai", "yi"],
"cat" => ["cai", "venus", "yi"],
"mouse" => ["cai", "venus"],
"pterodactyl" => ["venus"],
"chinchilla" => ["venus"]
}
puts pet_types(owners_1) == output_1
puts pet_types(owners_2) == output_2
puts pet_types(owners_3) == output_3

Hash#sort has the same effect (at least for my basic test) as Hash#to_a followed by Array#sort.
hash = {b: 2, a: 1}
hash.to_a.sort # => [[:a, 1, [:b, 2]]
hash.sort # => the same
Now let's look at #each, both on Hash and Array.
When you provide two arguments to the block, that can handle both cases. For the hash, the first argument will be the key and the second will be the value. For the nested array, the values essentially get splatted out to the args:
[[:a, 1, 2], [:b, 3, 4]].each { |x, y, z| puts "#{x}-#{y}-#{z}" }
# => a-1-2
# => b-3-4
So basically, you should think of Hash#sort to be a shortcut to Hash#to_a followed by Array#sort, and recognize that #each will work the same on a hash as a hash converted to array (a nested array). In this case, it doesn't matter which approach you take. Clearly if you need to sort iteration by the keys then you should use sort.

Related

How to find the largest value of a hash in an array of hashes

In my array, I'm trying to retrieve the key with the largest value of "value_2", so in this case, "B":
myArray = [
"A" => {
"value_1" => 30,
"value_2" => 240
},
"B" => {
"value_1" => 40,
"value_2" => 250
},
"C" => {
"value_1" => 18,
"value_2" => 60
}
]
myArray.each do |array_hash|
array_hash.each do |key, value|
if value["value_2"] == array_hash.values.max
puts key
end
end
end
I get the error:
"comparison of Hash with Hash failed (ArgumentError)".
What am I missing?
Though equivalent, the array given in the question is generally written:
arr = [{ "A" => { "value_1" => 30, "value_2" => 240 } },
{ "B" => { "value_1" => 40, "value_2" => 250 } },
{ "C" => { "value_1" => 18, "value_2" => 60 } }]
We can find the desired key as follows:
arr.max_by { |h| h.values.first["value_2"] }.keys.first
#=> "B"
See Enumerable#max_by. The steps are:
g = arr.max_by { |h| h.values.first["value_2"] }
#=> {"B"=>{"value_1"=>40, "value_2"=>250}}
a = g.keys
#=> ["B"]
a.first
#=> "B"
In calculating g, for
h = arr[0]
#=> {"A"=>{"value_1"=>30, "value_2"=>240}}
the block calculation is
a = h.values
#=> [{"value_1"=>30, "value_2"=>240}]
b = a.first
#=> {"value_1"=>30, "value_2"=>240}
b["value_2"]
#=> 240
Suppose now arr is as follows:
arr << { "D" => { "value_1" => 23, "value_2" => 250 } }
#=> [{"A"=>{"value_1"=>30, "value_2"=>240}},
# {"B"=>{"value_1"=>40, "value_2"=>250}},
# {"C"=>{"value_1"=>18, "value_2"=>60}},
# {"D"=>{"value_1"=>23, "value_2"=>250}}]
and we wish to return an array of all keys for which the value of "value_2" is maximum (["B", "D"]). We can obtain that as follows.
max_val = arr.map { |h| h.values.first["value_2"] }.max
#=> 250
arr.select { |h| h.values.first["value_2"] == max_val }.flat_map(&:keys)
#=> ["B", "D"]
flat_map(&:keys) is shorthand for:
flat_map { |h| h.keys }
which returns the same array as:
map { |h| h.keys.first }
See Enumerable#flat_map.
Code
p myArray.pop.max_by{|k,v|v["value_2"]}.first
Output
"B"
I'd use:
my_array = [
"A" => {
"value_1" => 30,
"value_2" => 240
},
"B" => {
"value_1" => 40,
"value_2" => 250
},
"C" => {
"value_1" => 18,
"value_2" => 60
}
]
h = Hash[*my_array]
# => {"A"=>{"value_1"=>30, "value_2"=>240},
# "B"=>{"value_1"=>40, "value_2"=>250},
# "C"=>{"value_1"=>18, "value_2"=>60}}
k = h.max_by { |k, v| v['value_2'] }.first # => "B"
Hash[*my_array] takes the array of hashes and turns it into a single hash. Then max_by will iterate each key/value pair, returning an array containing the key value "B" and the sub-hash, making it easy to grab the key using first:
k = h.max_by { |k, v| v['value_2'] } # => ["B", {"value_1"=>40, "value_2"=>250}]
I guess the idea of your solution is looping through each hash element and compare the found minimum value with hash["value_2"].
But you are getting an error at
if value["value_2"] == array_hash.values.max
Because the array_hash.values is still a hash
{"A"=>{"value_1"=>30, "value_2"=>240}}.values.max
#=> {"value_1"=>30, "value_2"=>240}
It should be like this:
max = nil
max_key = ""
myArray.each do |array_hash|
array_hash.each do |key, value|
if max.nil? || value.values.max > max
max = value.values.max
max_key = key
end
end
end
# max_key #=> "B"
Another solution:
myArray.map{ |h| h.transform_values{ |v| v["value_2"] } }.max_by{ |k| k.values }.keys.first
You asked "What am I missing?".
I think you are missing a proper understanding of the data structures that you are using. I suggest that you try printing the data structures and take a careful look at the results.
The simplest way is p myArray which gives:
[{"A"=>{"value_1"=>30, "value_2"=>240}, "B"=>{"value_1"=>40, "value_2"=>250}, "C"=>{"value_1"=>18, "value_2"=>60}}]
You can get prettier results using pp:
require 'pp'
pp myArray
yields:
[{"A"=>{"value_1"=>30, "value_2"=>240},
"B"=>{"value_1"=>40, "value_2"=>250},
"C"=>{"value_1"=>18, "value_2"=>60}}]
This helps you to see that myArray has only one element, a Hash.
You could also look at the expression array_hash.values.max inside the loop:
myArray.each do |array_hash|
p array_hash.values
end
gives:
[{"value_1"=>30, "value_2"=>240}, {"value_1"=>40, "value_2"=>250}, {"value_1"=>18, "value_2"=>60}]
Not what you expected? :-)
Given this, what would you expect to be returned by array_hash.values.max in the above loop?
Use p and/or pp liberally in your ruby code to help understand what's going on.

Convert array into hash and add a counter value to the new hash

I have the following array of hashes:
[
{"BREAD" => {:price => 1.50, :discount => true }},
{"BREAD" => {:price => 1.50, :discount => true }},
{"MARMITE" => {:price => 1.60, :discount => false}}
]
And I would like to translate this array into a hash that includes the counts for each item:
Output:
{
"BREAD" => {:price => 1.50, :discount => true, :count => 2},
"MARMITE" => {:price => 1.60, :discount => false, :count => 1}
}
I have tried two approaches to translate the array into a hash.
new_cart = cart.inject(:merge)
hash = Hash[cart.collect { |item| [item, ""] } ]
Both work but then I am stumped at how to capture and pass the count value.
Expected output
{
"BREAD" => {:price => 1.50, :discount => true, :count => 2},
"MARMITE" => {:price => 1.60, :discount => false, :count => 1}
}
We are given the array:
arr = [
{"BREAD" => {:price => 1.50, :discount => true }},
{"BREAD" => {:price => 1.50, :discount => true }},
{"MARMITE" => {:price => 1.60, :discount => false}}
]
and make the assumption that each hash has a single key and if two hashes have the same (single) key, the value of that key is the same in both hashes.
The first step is create an empty hash to which will add key-value pairs:
h = {}
Now we loop through arr to build the hash h. I've added a puts statement to display intermediate values in the calculation.
arr.each do |g|
k, v = g.first
puts "k=#{k}, v=#{v}"
if h.key?(k)
h[k][:count] += 1
else
h[k] = v.merge({ :count => 1 })
end
end
displays:
k=BREAD, v={:price=>1.5, :discount=>true}
k=BREAD, v={:price=>1.5, :discount=>true}
k=MARMITE, v={:price=>1.6, :discount=>false}
and returns:
#=> [{"BREAD" =>{:price=>1.5, :discount=>true}},
# {"BREAD" =>{:price=>1.5, :discount=>true}},
# {"MARMITE"=>{:price=>1.6, :discount=>false}}]
each always returns its receiver (here arr), which is not what we want.
h #=> {"BREAD"=>{:price=>1.5, :discount=>true, :count=>2},
# "MARMITE"=>{:price=>1.6, :discount=>false, :count=>1}}
is the result we need. See Hash#key? (aka, has_key?), Hash#[], Hash#[]= and Hash#merge.
Now let's wrap this in a method.
def hashify(arr)
h = {}
arr.each do |g|
k, v = g.first
if h.key?(k)
h[k][:count] += 1
else
h[k] = v.merge({ :count=>1 })
end
end
h
end
hashify(arr)
#=> {"BREAD"=>{:price=>1.5, :discount=>true, :count=>2},
# "MARMITE"=>{:price=>1.6, :discount=>false, :count=>1}}
Rubyists would often use the method Enumerable#each_with_object to simplify.
def hashify(arr)
arr.each_with_object({}) do |g,h|
k, v = g.first
if h.key?(k)
h[k][:count] += 1
else
h[k] = v.merge({ :count => 1 })
end
end
end
Compare the two methods to identify their differences. See Enumerable#each_with_object.
When, as here, the keys are symbols, Ruby allows you to use the shorthand { count: 1 } for { :count=>1 }. Moreover, she permits you to write :count = 1 or count: 1 without the braces when the hash is an argument. For example,
{}.merge('cat'=>'meow', dog:'woof', :pig=>'oink')
#=> {"cat"=>"meow", :dog=>"woof", :pig=>"oink"}
It's probably more common to see the form count: 1 when keys are symbols and for the braces to be omitted when a hash is an argument.
Here's a further refinement you might see. First create
h = arr.group_by { |h| h.keys.first }
#=> {"BREAD" =>[{"BREAD"=>{:price=>1.5, :discount=>true}},
# {"BREAD"=>{:price=>1.5, :discount=>true}}],
# "MARMITE"=>[{"MARMITE"=>{:price=>1.6, :discount=>false}}]}
See Enumerable#group_by. Now convert the values (arrays) to their sizes:
counts = h.transform_values { |arr| arr.size }
#=> {"BREAD"=>2, "MARMITE"=>1}
which can be written in abbreviated form:
counts = h.transform_values(&:size)
#=> {"BREAD"=>2, "MARMITE"=>1}
See Hash#transform_values. We can now write:
uniq_arr = arr.uniq
#=> [{"BREAD"=>{:price=>1.5, :discount=>true}},
#= {"MARMITE"=>{:price=>1.6, :discount=>false}}]
uniq_arr.each_with_object({}) do |g,h|
puts "g=#{g}"
k,v = g.first
puts " k=#{k}, v=#{v}"
h[k] = v.merge(counts: counts[k])
puts " h=#{h}"
end
which displays:
g={"BREAD"=>{:price=>1.5, :discount=>true}}
k=BREAD, v={:price=>1.5, :discount=>true}
h={"BREAD"=>{:price=>1.5, :discount=>true, :counts=>2}}
g={"MARMITE"=>{:price=>1.6, :discount=>false}}
k=MARMITE, v={:price=>1.6, :discount=>false}
h={"BREAD"=>{:price=>1.5, :discount=>true, :counts=>2},
"MARMITE"=>{:price=>1.6, :discount=>false, :counts=>1}}
and returns:
#=> {"BREAD"=>{:price=>1.5, :discount=>true, :counts=>2},
# "MARMITE"=>{:price=>1.6, :discount=>false, :counts=>1}}
See Array#uniq.
This did the trick:
arr = [
{ bread: { price: 1.50, discount: true } },
{ bread: { price: 1.50, discount: true } },
{ marmite: { price: 1.60, discount: false } }
]
Get the count for each occurrence of hash, add as key value pair and store:
h = arr.uniq.each { |x| x[x.first.first][:count] = arr.count(x) }
Then convert hashes into arrays, flatten to a single array then construct a hash:
Hash[*h.collect(&:to_a).flatten]
#=> {:bread=>{:price=>1.50, :discount=>true, :count=>2}, :marmite=>{:price=>1.60, :discount=>false, :count=>1}}
Combined a couple of nice ideas from here:
https://raycodingdotnet.wordpress.com/2013/08/05/array-of-hashes-into-single-hash-in-ruby/
and here:
http://carol-nichols.com/2015/08/07/ruby-occurrence-couting/

Initializing a hash with an empty array keyed to an array of strings - Ruby

I have:
people=["Bob","Fred","Sam"]
holidays = Hash.new
people.each do |person|
a=Array.new
holidays[person]=a
end
gifts = Hash.new
people.each do |person|
a=Array.new
gifts[person]=a
end
Feels clunky. I can't seem to figure a more streamline way with an initialization block or somesuch thing. Is there an idiomatic approach here?
Ideally, I'd like to keep an array like:
lists["holidays","gifts",...]
... and itterate through it to initialize each element in the lists array.
people = %w|Bob Fred Sam|
data = %w|holidays gifts|
result = data.zip(data.map { people.zip(people.map { [] }).to_h }).to_h
result['holidays']['Bob'] << Date.today
#⇒ {
# "holidays" => {
# "Bob" => [
# [0] #<Date: 2016-11-04 ((2457697j,0s,0n),+0s,2299161j)>
# ],
# "Fred" => [],
# "Sam" => []
# },
# "gifts" => {
# "Bob" => [],
# "Fred" => [],
# "Sam" => []
# }
# }
More sophisticated example would be:
result = data.map do |d|
[d, Hash.new { |h, k| h[k] = [] if people.include?(k) }]
end.to_h
The latter produces the “lazy initialized nested hashes.” It uses the Hash#new with a block constructor for nested hashes.
Play with it to see how it works.
A common way of doing that would be to use Enumerable#each_with_objrect.
holidays = people.each_with_object({}) { |p,h| h[p] = [] }
#=> {"Bob"=>[], "Fred"=>[], "Sam"=>[]}
gifts is the same.
If you only want a number of such hashes then, the following should suffice:
count_of_hashes = 4 // lists.count; 4 is chosen randomly by throwing a fair die
people = ["Bob", "Fred", "Sam"]
lists = count_of_hashes.times.map do
people.map {|person| [person, []]}.to_h
end
This code also ensures the arrays and the hashes all occupy their own memory. As can be verified by the following code:
holidays, gifts, *rest = lists
holidays["Bob"] << "Rome"
And checking the values of all the other hashes:
lists
=> [
{"Bob"=>["Rome"], "Fred"=>[], "Sam"=>[]},
{"Bob"=>[], "Fred"=>[], "Sam"=>[]},
{"Bob"=>[], "Fred"=>[], "Sam"=>[]},
{"Bob"=>[], "Fred"=>[], "Sam"=>[]}
]

visiting hash with keys from array

I have a big hash with lots of nested key value pairs.
Eg.
h = {"foo" => {"bar" => {"hello" => {"world" => "result" } } } }
Now I want to access result and I have keys for that in array in proper sequence.
keys_arr = ["foo", "bar", "hello", "world"]
The motive is clear, I want to do following:
h["foo"]["bar"]["hello"]["world"]
# => "result"
But I don't know how to do this. I am currently doing:
key = '["' + keys_arr.join('"]["') + '"]'
eval("h"+key)
# => "result"
Which looks like a hack. Also it greatly reduces my ability to work with hash in real environment.
Please suggest alternate and better ways.
Using Enumerable#inject (or Enumerable#reduce):
h = {"foo" => {"bar" => {"hello" => {"world" => "result" } } } }
keys_arr = ["foo", "bar", "hello", "world"]
keys_arr.inject(h) { |x, k| x[k] }
# => "result"
UPDATE
If you want to do something like: h["foo"]["bar"]["hello"]["world"] = "ruby"
innermost = keys_arr[0...-1].inject(h) { |x, k| x[k] } # the innermost hash
innermost[keys_arr[-1]] = "ruby"
keys_arr.inject(h, :[])
will do
Another way:
h = {"foo" => {"bar" => {"hello" => {"world" => 10 } } } }
keys = ["foo", "bar", "hello", "world"]
result = h
keys.each do |key|
result = result[key]
end
puts result #=>10
If the key may not exist, see here:
Dealing with many [...] in Ruby

Ruby: Link two arrays of objects by attribute value

I'm pretty new in Ruby programming. In Ruby there are plenty ways to write elegant code. Is there any elegant way to link two arrays with objects of the same type by attribute value?
It's hard to explain. Let's look at the next example:
a = [ { :id => 1, :value => 1 }, { :id => 2, :value => 2 }, { :id => 3, :value => 3 } ]
b = [ { :id => 1, :value => 2 }, { :id => 3, :value => 4 } ]
c = link a, b
# Result structure after linkage.
c = {
"1" => {
:a => { :id => 1, :value => 1 },
:b => { :id => 1, :value => 1 }
},
"3" => {
:a => { :id => 3, :value => 3 },
:b => { :id => 3, :value => 4 }
}
}
So the basic idea is to get pairs of objects from different arrays by their common ID and construct a hash, which will give this pair by ID.
Thanks in advance.
If you want to take an adventure through Enumerable, you could say this:
(a.map { |h| [:a, h] } + b.map { |h| [:b, h] })
.group_by { |_, h| h[:id] }
.select { |_, a| a.length == 2 }
.inject({}) { |h, (n, v)| h.update(n => Hash[v]) }
And if you really want the keys to be strings, say n.to_s => Hash[v] instead of n => Hash[v].
The logic works like this:
We need to know where everything comes from we decorate the little hashes with :a and :b symbols to track their origins.
Then add the decorated arrays together into one list so that...
group_by can group things into almost-the-final-format.
Then find the groups of size two since those groups contain the entries that appeared in both a and b. Groups of size one only appeared in one of a or b so we throw those away.
Then a little injection to rearrange things into their final format. Note that the arrays we built in (1) just somehow happen to be in the format that Hash[] is looking for.
If you wanted to do this in a link method then you'd need to say things like:
link :a => a, :b => b
so that the method will know what to call a and b. This hypothetical link method also easily generalizes to more arrays:
def link(input)
input.map { |k, v| v.map { |h| [k, h] } }
.inject(:+)
.group_by { |_, h| h[:id] }
.select { |_, a| a.length == input.length }
.inject({}) { |h, (n, v)| h.update(n => Hash[v]) }
end
link :a => [...], :b => [...], :c => [...]
I assume that, for any two elements h1 and h2 of a (or of b), h1[:id] != h2[:id].
I would do this:
def convert(arr) Hash[arr.map {|h| [h[:id], h]}] end
ah, bh = convert(a), convert(b)
c = ah.keys.each_with_object({}) {|k,h|h[k]={a: ah[k], b: bh[k]} if bh.key?(k)}
# => {1=>{:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}},
# 3=>{:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}}
Note that:
ah = convert(a)
# => {1=>{:id=>1, :value=>1}, 2=>{:id=>2, :value=>2}, 3=>{:id=>3, :value=>3}}
bh = convert(b)
# => {1=>{:id=>1, :value=>2}, 3=>{:id=>3, :value=>4}}
Here's a second approach. I don't like it as well, but it represents a different way of looking at the problem.
def sort_by_id(a) a.sort_by {|h| h[:id]} end
c = Hash[*sort_by_id(a.select {|ha| b.find {|hb| hb[:id] == ha[:id]}})
.zip(sort_by_id(b))
.map {|ha,hb| [ha[:id], {a: ha, b: hb}]}
.flatten]
Here's what's happening. The first step is to select only the elements ha of a for which there is an element hb of b for which ha[:id] = hb[id]. Then we sort both (what's left of) a and b on h[:id], zip them together and then make the hash c.
r1 = a.select {|ha| b.find {|hb| hb[:id] == ha[:id]}}
# => [{:id=>1, :value=>1}, {:id=>3, :value=>3}]
r2 = sort_by_id(r1)
# => [{:id=>1, :value=>1}, {:id=>3, :value=>3}]
r3 = sort_by_id(b)
# => [{:id=>1, :value=>2}, {:id=>3, :value=>4}]
r4 = r2.zip(r3)
# => [[{:id=>1, :value=>1}, {:id=>1, :value=>2}],
# [{:id=>3, :value=>3}, {:id=>3, :value=>4}]]
r5 = r4.map {|ha,hb| [ha[:id], {a: ha, b: hb}]}
# => [[1, {:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}}],
# [3, {:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}]]
r6 = r5.flatten
# => [1, {:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}},
# 3, {:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}]
c = Hash[*r6]
# => {1=>{:a=>{:id=>1, :value=>1}, :b=>{:id=>1, :value=>2}},
# 3=>{:a=>{:id=>3, :value=>3}, :b=>{:id=>3, :value=>4}}}
Ok, I've found the answer by myself. Here is a quite short line of code, which should do the trick:
Hash[a.product(b)
.select { |pair| pair[0][:id] == pair[1][:id] }
.map { |pair| [pair[0][:id], { :a => pair[0], :b => pair[1] }] }]
The product method gives us all possible pairs, then we filter them by equal IDs of pair elements. And then we map pairs to the special form, which will produce a Hash we are looking for.
So Hash[["key1", "value1"], ["key2", "value2"]] returns { "key1" => "value1", "key2" => "value2" }. And I use this to get the answer on my question.
Thanks.
P.S.: you can use pair.first instead of pair[0] and pair.last instead of pair[1] for better readability.
UPDATE
As Cary pointed out, it is better to replace |pair| with |ha, hb| to avoid these ugly indices:
Hash[a.product(b)
.select { |ha, hb| ha[:id] == hb[:id] }
.map { |ha, hb| [ha[:id], { :a => ha, :b => hb }] }]

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