I would like to combine arrrays #a, #b, and #c into a single array with multiple data elements, for example OpenStruct:
#a = ["my", "foo", "bar"]
#b = ["yan", "can", "cook"]
#c = ["in", "your", "dreams"]
the output would be like:
[
{ :a => "my", :b => "yan", :c => "in" },
{ :a => "foo", :b => "can", :c => "your" },
{ :a => bar, :b => "cook", :c => "dreams" }
]
What is the fastest way to do this? Should I consider another class?
Here is one solution, I am not quite convinced it is the neatest though:
#a.zip(#b, #c).map {|t| {:a => t[0], :b => t[1], :c => t[2]}}
Functional approach:
[#a, #b, #c].transpose.map { |xs| Hash[[:a, :b, :c].zip(xs)] }
#=> [{:a=>"my", :b=>"yan", :c=>"in"}, {:a=>"foo", :b=>"can", :c=>"your"}, {:a=>"bar", :b=>"cook", :c=>"dreams"}]
The best way to know the fastest way is do a benchmark. Based on previous answers:
require 'benchmark'
#a = ["my", "foo", "bar"]
#b = ["yan", "can", "cook"]
#c = ["in", "your", "dreams"]
$n = 500_000
Benchmark.bmbm do |x|
x.report("Boris Strandjev") do $n.times do
#a.zip(#b, #c).map {|t| {:a => t[0], :b => t[1], :c => t[2]}}
end end
x.report("tokland") do $n.times do
[#a, #b, #c].transpose.map { |xs| Hash[[:a, :b, :c].zip(xs)] }
end end
x.report("mu is too short") do $n.times do
(0 ... [#a, #b, #c].max_by(&:length).length).map { |i| { :a => #a[i], :b => #b[i], :c => #c[i] } }
end end
x.report("KL-7") do $n.times do
#a.each_with_index.map { |a, i| { :a => a, :b => #b[i], :c => #c[i] } }
end end
end
Output:
Rehearsal ---------------------------------------------------
Boris Strandjev 4.540000 0.015000 4.555000 ( 4.571261)
tokland 7.145000 0.000000 7.145000 ( 7.268415)
mu is too short 5.304000 0.047000 5.351000 ( 5.560318)
KL-7 4.914000 0.000000 4.914000 ( 5.030287)
----------------------------------------- total: 21.965000sec
user system total real
Boris Strandjev 4.462000 0.016000 4.478000 ( 4.553260)
tokland 7.129000 0.031000 7.160000 ( 7.309418)
mu is too short 5.366000 0.031000 5.397000 ( 5.447312)
KL-7 4.898000 0.016000 4.914000 ( 4.997286)
If there is no need to worry about arrays having different length I think that's the fastest way (as it iterate over array only once):
#a.each_with_index.map { |a, i| { :a => a, :b => #b[i], :c => #c[i] } }
Something like this should work:
(0 ... [#a, #b, #c].max_by(&:length).length).map { |i| { :a => #a[i], :b => #b[i], :c => #c[i] } }
That doesn't assume they're all the same length but you'll get nil values if they're not.
Related
I would like to generate a proc that, when called, returns successive values for any Enumerable. In other words:
a_proc = generate_proc([:a, :b, :c])
a_proc.call # => :a
a_proc.call # => :b
etc.
Ideally, I'd want to translate it before returning, too, so it might get used like:
a_proc = generate_proc([:a, :b, :c ]) { |e| "Element: #{e.inspect}" }
a_proc.call # => "Element: :a"
a_proc.call # => "Element: :b"
a_proc.call # => "Element: :c"
This isn't exactly what you are asking for, but I think it may meet your needs:
an_enum = [:a, :b, :c].each
an_enum.next # => :a
an_enum.next # => :b
And:
an_enum = [:a, :b, :c].map { |e| "Element: #{e.inspect}" }.each
# or, if you want to defer calling the block...
an_enum = [:a, :b, :c].lazy.map { |e| "Element: #{e.inspect}" }
an_enum.next # => "Element: :a"
an_enum.next # => "Element: :b"
an_enum.next # => "Element: :c"
For more information, check out Enumerator and Enumerator::Lazy.
UPDATE FROM FUTURE, SMARTER, HANDSOMER ME:
To answer your original question, you can turn any object's method into a bound, callable object with Object#method. We can even do it to Enumerator#next!
an_enum = [:a, :b, :c].each
a_proc = an_enum.method(:next)
a_proc.call # => :a
a_proc.call # => :b
a_proc.call # => :c
def generate_proc x, &postproc
myobj = Fiber.new { x.each { |v| Fiber.yield v }}
postproc ? -> () { postproc.call myobj.resume } :
-> () { myobj.resume }
end
And this is exactly what you're asking for:
def generate_proc(list, &transform)
enum = list.to_enum
if transform
proc { transform[enum.next] }
else
proc { enum.next }
end
end
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 }] }]
Have a hash:
h = {:a => "val1", :b => "val2", :c => "val3"}
I can refer to the hash value:
h[:a], h[:c]
but I would like to refer by numeric index:
h[0] => val1
h[2] => val3
Is it possible?
h.values will give you an array requested.
> h.values
# ⇒ [
# [0] "val1",
# [1] "val2",
# [2] "val3"
# ]
UPD while the answer with h[h.keys[0]] was marked as correct, I’m a little bit curious with benchmarks:
h = {:a => "val1", :b => "val2", :c => "val3"}
Benchmark.bm do |x|
x.report { 1_000_000.times { h[h.keys[0]] = 'ghgh'} }
x.report { 1_000_000.times { h.values[0] = 'ghgh'} }
end
#
# user system total real
# 0.920000 0.000000 0.920000 ( 0.922456)
# 0.820000 0.000000 0.820000 ( 0.824592)
Looks like we’re spitting on 10% of productivity.
h = {:a => "val1", :b => "val2", :c => "val3"}
keys = h.keys
h[keys[0]] # "val1"
h[keys[2]] # "val3"
So you need both array indexing and hash indexing ?
If you need only the first one, use an array.
Otherwise, you can do the following :
h.values[0]
h.values[1]
I have a hash:
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
What is the best way to extract a sub-hash like this?
h1.extract_subhash(:b, :d, :e, :f) # => {:b => :B, :d => :D}
h1 #=> {:a => :A, :c => :C}
ActiveSupport, at least since 2.3.8, provides four convenient methods: #slice, #except and their destructive counterparts: #slice! and #except!. They were mentioned in other answers, but to sum them in one place:
x = {a: 1, b: 2, c: 3, d: 4}
# => {:a=>1, :b=>2, :c=>3, :d=>4}
x.slice(:a, :b)
# => {:a=>1, :b=>2}
x
# => {:a=>1, :b=>2, :c=>3, :d=>4}
x.except(:a, :b)
# => {:c=>3, :d=>4}
x
# => {:a=>1, :b=>2, :c=>3, :d=>4}
Note the return values of the bang methods. They will not only tailor existing hash but also return removed (not kept) entries. The Hash#except! suits best the example given in the question:
x = {a: 1, b: 2, c: 3, d: 4}
# => {:a=>1, :b=>2, :c=>3, :d=>4}
x.except!(:c, :d)
# => {:a=>1, :b=>2}
x
# => {:a=>1, :b=>2}
ActiveSupport does not require whole Rails, is pretty lightweight. In fact, a lot of non-rails gems depend on it, so most probably you already have it in Gemfile.lock. No need to extend Hash class on your own.
If you specifically want the method to return the extracted elements but h1 to remain the same:
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h2 = h1.select {|key, value| [:b, :d, :e, :f].include?(key) } # => {:b=>:B, :d=>:D}
h1 = Hash[h1.to_a - h2.to_a] # => {:a=>:A, :c=>:C}
And if you want to patch that into the Hash class:
class Hash
def extract_subhash(*extract)
h2 = self.select{|key, value| extract.include?(key) }
self.delete_if {|key, value| extract.include?(key) }
h2
end
end
If you just want to remove the specified elements from the hash, that is much easier using delete_if.
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h1.delete_if {|key, value| [:b, :d, :e, :f].include?(key) } # => {:a=>:A, :c=>:C}
h1 # => {:a=>:A, :c=>:C}
Ruby 2.5 added Hash#slice:
h = { a: 100, b: 200, c: 300 }
h.slice(:a) #=> {:a=>100}
h.slice(:b, :c, :d) #=> {:b=>200, :c=>300}
If you use rails, Hash#slice is the way to go.
{:a => :A, :b => :B, :c => :C, :d => :D}.slice(:a, :c)
# => {:a => :A, :c => :C}
If you don't use rails, Hash#values_at will return the values in the same order as you asked them so you can do this:
def slice(hash, *keys)
Hash[ [keys, hash.values_at(*keys)].transpose]
end
def except(hash, *keys)
desired_keys = hash.keys - keys
Hash[ [desired_keys, hash.values_at(*desired_keys)].transpose]
end
ex:
slice({foo: 'bar', 'bar' => 'foo', 2 => 'two'}, 'bar', 2)
# => {'bar' => 'foo', 2 => 'two'}
except({foo: 'bar', 'bar' => 'foo', 2 => 'two'}, 'bar', 2)
# => {:foo => 'bar'}
Explanation:
Out of {:a => 1, :b => 2, :c => 3} we want {:a => 1, :b => 2}
hash = {:a => 1, :b => 2, :c => 3}
keys = [:a, :b]
values = hash.values_at(*keys) #=> [1, 2]
transposed_matrix =[keys, values].transpose #=> [[:a, 1], [:b, 2]]
Hash[transposed_matrix] #=> {:a => 1, :b => 2}
If you feels like monkey patching is the way to go, following is what you want:
module MyExtension
module Hash
def slice(*keys)
::Hash[[keys, self.values_at(*keys)].transpose]
end
def except(*keys)
desired_keys = self.keys - keys
::Hash[[desired_keys, self.values_at(*desired_keys)].transpose]
end
end
end
Hash.include MyExtension::Hash
You can use slice!(*keys) which is available in the core extensions of ActiveSupport
initial_hash = {:a => 1, :b => 2, :c => 3, :d => 4}
extracted_slice = initial_hash.slice!(:a, :c)
initial_hash would now be
{:b => 2, :d =>4}
extracted_slide would now be
{:a => 1, :c =>3}
You can look at slice.rb in ActiveSupport 3.1.3
module HashExtensions
def subhash(*keys)
keys = keys.select { |k| key?(k) }
Hash[keys.zip(values_at(*keys))]
end
end
Hash.send(:include, HashExtensions)
{:a => :A, :b => :B, :c => :C, :d => :D}.subhash(:a) # => {:a => :A}
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
keys = [:b, :d, :e, :f]
h2 = (h1.keys & keys).each_with_object({}) { |k,h| h.update(k=>h1.delete(k)) }
#=> {:b => :B, :d => :D}
h1
#=> {:a => :A, :c => :C}
if you use rails, it may be convenient to use Hash.except
h = {a:1, b:2}
h1 = h.except(:a) # {b:2}
Both delete_if and keep_if are part of Ruby core. Here you can achieve what you would like to without patching the Hash type.
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h2 = h1.clone
p h1.keep_if { |key| [:b, :d, :e, :f].include?(key) } # => {:b => :B, :d => :D}
p h2.delete_if { |key, value| [:b, :d, :e, :f].include?(key) } #=> {:a => :A, :c => :C}
For futher info, check the links below from the documentation:
delete_if
keep_if
As others have mentioned, Ruby 2.5 added the Hash#slice method.
Rails 5.2.0beta1 also added it's own version of Hash#slice to shim the functionality for users of the framework that are using an earlier version of Ruby.
https://github.com/rails/rails/commit/01ae39660243bc5f0a986e20f9c9bff312b1b5f8
If looking to implement your own for whatever reason, it's a nice one liner as well:
def slice(*keys)
keys.each_with_object(Hash.new) { |k, hash| hash[k] = self[k] if has_key?(k) }
end unless method_defined?(:slice)
if you want to extract from data base record also it is better to use slice
hash = { a: 1, b: 2, c: 3, d: 4 }
hash.slice!(:a, :b) # => {:c=>3, :d=>4}
hash # => {:a=>1, :b=>2}
https://api.rubyonrails.org/classes/Hash.html#method-i-slice-21
class Hash
def extract(*keys)
key_index = Hash[keys.map{ |k| [k, true] }] # depends on the size of keys
partition{ |k, v| key_index.has_key?(k) }.map{ |group| Hash[group] }
end
end
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h2, h1 = h1.extract(:b, :d, :e, :f)
Here is a quick performance comparison of the suggested methods, #select seems to be the fastest
k = 1_000_000
Benchmark.bmbm do |x|
x.report('select') { k.times { {a: 1, b: 2, c: 3}.select { |k, _v| [:a, :b].include?(k) } } }
x.report('hash transpose') { k.times { Hash[ [[:a, :b], {a: 1, b: 2, c: 3}.fetch_values(:a, :b)].transpose ] } }
x.report('slice') { k.times { {a: 1, b: 2, c: 3}.slice(:a, :b) } }
end
Rehearsal --------------------------------------------------
select 1.640000 0.010000 1.650000 ( 1.651426)
hash transpose 1.720000 0.010000 1.730000 ( 1.729950)
slice 1.740000 0.010000 1.750000 ( 1.748204)
----------------------------------------- total: 5.130000sec
user system total real
select 1.670000 0.010000 1.680000 ( 1.683415)
hash transpose 1.680000 0.010000 1.690000 ( 1.688110)
slice 1.800000 0.010000 1.810000 ( 1.816215)
The refinement will look like this:
module CoreExtensions
module Extractable
refine Hash do
def extract(*keys)
select { |k, _v| keys.include?(k) }
end
end
end
end
And to use it:
using ::CoreExtensions::Extractable
{ a: 1, b: 2, c: 3 }.extract(:a, :b)
This code injects the functionality you're asking for into the Hash class:
class Hash
def extract_subhash! *keys
to_keep = self.keys.to_a - keys
to_delete = Hash[self.select{|k,v| !to_keep.include? k}]
self.delete_if {|k,v| !to_keep.include? k}
to_delete
end
end
and produces the results you provided:
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
p h1.extract_subhash!(:b, :d, :e, :f) # => {b => :B, :d => :D}
p h1 #=> {:a => :A, :c => :C}
Note: this method actually returns the extracted keys/values.
Here's a functional solution that can be useful if you're not running on Ruby 2.5 and in the case that you don't wan't to pollute your Hash class by adding a new method:
slice_hash = -> keys, hash { hash.select { |k, _v| keys.include?(k) } }.curry
Then you can apply it even on nested hashes:
my_hash = [{name: "Joe", age: 34}, {name: "Amy", age: 55}]
my_hash.map(&slice_hash.([:name]))
# => [{:name=>"Joe"}, {:name=>"Amy"}]
Just an addition to slice method, if the subhash keys which you want to separate from original hash is going to be dynamic you can do like,
slice(*dynamic_keys) # dynamic_keys should be an array type
We can do it by looping on keys only we want to extract and just checking the key is exist and then extract it.
class Hash
def extract(*keys)
extracted_hash = {}
keys.each{|key| extracted_hash[key] = self.delete(key) if self.has_key?(key)}
extracted_hash
end
end
h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h2 = h1.extract(:b, :d, :e, :f)
We have the following datastructures:
{:a => ["val1", "val2"], :b => ["valb1", "valb2"], ...}
And I want to turn that into
[{:a => "val1", :b => "valb1"}, {:a => "val2", :b => "valb2"}, ...]
And then back into the first form. Anybody with a nice looking implementation?
This solution works with arbitrary numbers of values (val1, val2...valN):
{:a => ["val1", "val2"], :b => ["valb1", "valb2"]}.inject([]){|a, (k,vs)|
vs.each_with_index{|v,i| (a[i] ||= {})[k] = v}
a
}
# => [{:a=>"val1", :b=>"valb1"}, {:a=>"val2", :b=>"valb2"}]
[{:a=>"val1", :b=>"valb1"}, {:a=>"val2", :b=>"valb2"}].inject({}){|a, h|
h.each_pair{|k,v| (a[k] ||= []) << v}
a
}
# => {:a=>["val1", "val2"], :b=>["valb1", "valb2"]}
Using a functional approach (see Enumerable):
hs = h.values.transpose.map { |vs| h.keys.zip(vs).to_h }
#=> [{:a=>"val1", :b=>"valb1"}, {:a=>"val2", :b=>"valb2"}]
And back:
h_again = hs.first.keys.zip(hs.map(&:values).transpose).to_h
#=> {:a=>["val1", "val2"], :b=>["valb1", "valb2"]}
Let's look closely what the data structure we are trying to convert between:
#Format A
[
["val1", "val2"], :a
["valb1", "valb2"], :b
["valc1", "valc2"] :c
]
#Format B
[ :a :b :c
["val1", "valb1", "valc1"],
["val2", "valb2", "valc3"]
]
It is not diffculty to find Format B is the transpose of Format A in essential , then we can come up with this solution:
h={:a => ["vala1", "vala2"], :b => ["valb1", "valb2"], :c => ["valc1", "valc2"]}
sorted_keys = h.keys.sort_by {|a,b| a.to_s <=> b.to_s}
puts sorted_keys.inject([]) {|s,e| s << h[e]}.transpose.inject([]) {|r, a| r << Hash[*sorted_keys.zip(a).flatten]}.inspect
#[{:b=>"valb1", :c=>"valc1", :a=>"vala1"}, {:b=>"valb2", :c=>"valc2", :a=>"vala2"}]
m = {}
a,b = Array(h).transpose
b.transpose.map { |y| [a, y].transpose.inject(m) { |m,x| m.merge Hash[*x] }}
My attempt, perhaps slightly more compact.
h = { :a => ["val1", "val2"], :b => ["valb1", "valb2"] }
h.values.transpose.map { |s| Hash[h.keys.zip(s)] }
Should work in Ruby 1.9.3 or later.
Explanation:
First, 'combine' the corresponding values into 'rows'
h.values.transpose
# => [["val1", "valb1"], ["val2", "valb2"]]
Each iteration in the map block will produce one of these:
h.keys.zip(s)
# => [[:a, "val1"], [:b, "valb1"]]
and Hash[] will turn them into hashes:
Hash[h.keys.zip(s)]
# => {:a=>"val1", :b=>"valb1"} (for each iteration)
This will work assuming all the arrays in the original hash are the same size:
hash_array = hash.first[1].map { {} }
hash.each do |key,arr|
hash_array.zip(arr).each {|inner_hash, val| inner_hash[key] = val}
end
You could use inject to build an array of hashes.
hash = { :a => ["val1", "val2"], :b => ["valb1", "valb2"] }
array = hash.inject([]) do |pairs, pair|
pairs << { pair[0] => pair[1] }
pairs
end
array.inspect # => "[{:a=>["val1", "val2"]}, {:b=>["valb1", "valb2"]}]"
Ruby documentation has a few more examples of working with inject.