How do I create the intersection of two hashes?

Question

I have two hashes:

hash1 = {1 => "a" , 2 => "b" , 3 => "c" , 4 => "d"} 
hash2 = {3 => "hello", 4 => "world" , 5 => "welcome"} 

I need a hash which contains common keys in both hashes:

hash3 = {3 => "hello" , 4 => "world"}

Is it possible to do it without any loop?

Answer 1

Ruby 2.5 has added Hash#slice, which allows a compact code like:

hash3 = hash1.slice(*hash2.keys)

In older rubies this was possible in rails or projects using active support's hash extensions.

Answer 2

hash2.select {|key, value| hash1.has_key? key }
# => {3=>"hello", 4=>"world"}

Answer 3

I'd go with this:

hash1 = {1 => "a" , 2 => "b" , 3 => "c" , 4 => "d"} 
hash2 = {3 => "hello", 4 => "world" , 5 => "welcome"} 

Hash[(hash1.keys & hash2.keys).zip(hash2.values_at(*(hash1.keys & hash2.keys)))]
=> {3=>"hello", 4=>"world"}

Which can be reduced a bit to:

keys = (hash1.keys & hash2.keys)
Hash[keys.zip(hash2.values_at(*keys))]

The trick is in Array's & method. The documentation says:

Set Intersection — Returns a new array containing elements common to the two arrays, excluding any duplicates. The order is preserved from the original array.

---

Here are some benchmarks to show what is the most efficient way to do this:

require 'benchmark'

HASH1 = {1 => "a" , 2 => "b" , 3 => "c" , 4 => "d"} 
HASH2 = {3 => "hello", 4 => "world" , 5 => "welcome"} 

def tinman
  keys = (HASH1.keys & HASH2.keys)
  Hash[keys.zip(HASH2.values_at(*keys))]
end

def santhosh
  HASH2.select {|key, value| HASH1.has_key? key }
end
def santhosh_2
  HASH2.select {|key, value| HASH1[key] }
end

def priti
  HASH2.select{|k,v| HASH1.assoc(k) }
end

def koraktor
  HASH1.keep_if { |k, v| HASH2.key? k }
end
def koraktor2
  HASH2.keep_if { |k, v| HASH1.key? k }
end

N = 1_000_000
puts RUBY_VERSION
puts "N= #{N}"

puts [:tinman, :santhosh, :santhosh_2, :priti, :koraktor, :koraktor2].map{ |s| "#{s.to_s} = #{send(s)}" }
Benchmark.bm(11) do |x|
  x.report('tinman') { N.times { tinman() }}
  x.report('santhosh_2') { N.times { santhosh_2() }}
  x.report('santhosh') { N.times { santhosh() }}
  x.report('priti') { N.times { priti() }}
  x.report('koraktor') { N.times { koraktor() }}
  x.report('koraktor2') { N.times { koraktor2() }}
end

Ruby 1.9.3-p448:

1.9.3
N= 1000000
tinman = {3=>"hello", 4=>"world"}
santhosh = {3=>"hello", 4=>"world"}
santhosh_2 = {3=>"hello", 4=>"world"}
priti = {3=>"hello", 4=>"world"}
koraktor = {3=>"c", 4=>"d"}
koraktor2 = {3=>"hello", 4=>"world"}
                  user     system      total        real
tinman        2.430000   0.000000   2.430000 (  2.430030)
santhosh_2    1.000000   0.020000   1.020000 (  1.003635)
santhosh      1.090000   0.010000   1.100000 (  1.104067)
priti         1.350000   0.000000   1.350000 (  1.352476)
koraktor      0.490000   0.000000   0.490000 (  0.484686)
koraktor2     0.480000   0.000000   0.480000 (  0.483327)

Running under Ruby 2.0.0-p247:

2.0.0
N= 1000000
tinman = {3=>"hello", 4=>"world"}
santhosh = {3=>"hello", 4=>"world"}
santhosh_2 = {3=>"hello", 4=>"world"}
priti = {3=>"hello", 4=>"world"}
koraktor = {3=>"c", 4=>"d"}
koraktor2 = {3=>"hello", 4=>"world"}
                  user     system      total        real
tinman        1.890000   0.000000   1.890000 (  1.882352)
santhosh_2    0.710000   0.010000   0.720000 (  0.735830)
santhosh      0.790000   0.020000   0.810000 (  0.807413)
priti         1.030000   0.010000   1.040000 (  1.030018)
koraktor      0.390000   0.000000   0.390000 (  0.389431)
koraktor2     0.390000   0.000000   0.390000 (  0.389072)

Koraktor's original code doesn't work, but he turned it around nicely with his second code pass, and walks away with the best speed. I added the santhosh_2 method to see what effect removing key? would have. It sped the routine up a little, but not enough to catch up to Koraktor's.

---

Just for documentation purposes, I tweaked Koraktor's second code to remove the key? method also, and shaved more time from it. Here's the added method and the new output:

def koraktor3
  HASH2.keep_if { |k, v| HASH1[k] }
end

1.9.3
N= 1000000
tinman = {3=>"hello", 4=>"world"}
santhosh = {3=>"hello", 4=>"world"}
santhosh_2 = {3=>"hello", 4=>"world"}
priti = {3=>"hello", 4=>"world"}
koraktor = {3=>"c", 4=>"d"}
koraktor2 = {3=>"hello", 4=>"world"}
koraktor3 = {3=>"hello", 4=>"world"}
                  user     system      total        real
tinman        2.380000   0.000000   2.380000 (  2.382392)
santhosh_2    0.970000   0.020000   0.990000 (  0.976672)
santhosh      1.070000   0.010000   1.080000 (  1.078397)
priti         1.320000   0.000000   1.320000 (  1.318652)
koraktor      0.480000   0.000000   0.480000 (  0.488613)
koraktor2     0.490000   0.000000   0.490000 (  0.490099)
koraktor3     0.390000   0.000000   0.390000 (  0.389386)

2.0.0
N= 1000000
tinman = {3=>"hello", 4=>"world"}
santhosh = {3=>"hello", 4=>"world"}
santhosh_2 = {3=>"hello", 4=>"world"}
priti = {3=>"hello", 4=>"world"}
koraktor = {3=>"c", 4=>"d"}
koraktor2 = {3=>"hello", 4=>"world"}
koraktor3 = {3=>"hello", 4=>"world"}
                  user     system      total        real
tinman        1.840000   0.000000   1.840000 (  1.832491)
santhosh_2    0.720000   0.010000   0.730000 (  0.737737)
santhosh      0.780000   0.020000   0.800000 (  0.801619)
priti         1.040000   0.010000   1.050000 (  1.044588)
koraktor      0.390000   0.000000   0.390000 (  0.387265)
koraktor2     0.390000   0.000000   0.390000 (  0.388648)
koraktor3     0.320000   0.000000   0.320000 (  0.327859)

Answer 4

hash3 = hash1.keep_if { |k, v| hash2.key? k }

This won't have the same effect as the code in the question, instead it will return:

hash3 #=> { 3 => "c", 4 => "d" }

The order of the hashes is important here. The values will always be taken from the hash that #keep_if is send to.

hash3 = hash2.keep_if { |k, v| hash1.key? k }
#=> {3 => "hello", 4 => "world"}