I have an array like this:
arr = [{id: 1, name: 'John' }, {id: 2, name: 'Sam' }, {id: 3, name: 'Bob' }]
I need to check if any of arr objects have name Sam. What is the most elegant way? I can only think of cycling with each.
I need to check if any of arr objects have name Sam
Enumerable#any? is a good way to go.
arr = [ {id: 1, name: 'John' }, {id: 2, name: 'Sam' }, {id: 3, name: 'Bob' }]
arr.any? {|h| h[:name] == "Sam"}
# => true
Now if you also want to see which Array object has the value Sam in it,you can use Enumerable#find for the same:
arr.find {|h| h[:name] == "Sam"}
# => {:id=>2, :name=>"Sam"}
You can also choose select or count methods
Enumberable#select
> arr = [{id: 1, name: 'John' }, {id: 2, name: 'Sam' }, {id: 3, name: 'Bob' }]
> arr.select { | h | h[:name] == 'Sam' }
# => [{:id=>2, :name=>"Sam"}]
Enumberable#count
> arr.count { | h | h[:name] == 'Sam' }
# => 1
You can use Enumberable#find_all to return all object that match the constrain
arr = [{:id=>1,:first_name=>'sam'},{:id=>2,:first_name=>'sam'},{:id=>3,:first_name=>'samanderson'},{:id=>4,:first_name=>'samuel'}]
arr.find_all{|obj| obj.first_name == 'sam'}
# => [{:id=>1,:first_name=>'sam'},{:id=>2,:first_name=>'sam'}]
Related
I have two Arrays of Hashes which simulate two tables in a database, with one key in the first hash referencing a separately-named key in the second hash, example below:
cars = [ { id: 1, color: 'red', owner_id: 1 }, { id: 2, color: 'black', owner_id: 1 } ]
owners = [ { id: 1, name: 'Alice' }, { id: 2, name: 'Bob' } ]
I'd like to try to accomplish a "join" on these two hashes, resulting in a new Array of Hashes, so that the keys and values from owners will be merged into any of the cars hashes where the cars' :owner_id matches an owner's :id. So in the above example, the result would look like this:
[ { id: 1, color: 'red', owner_id: 1, name: 'Alice' }, { id: 2, color: 'black', owner_id: 1, name: 'Alice' } ]
Anyone have any thoughts on how I could achieve this? Thank you!
[EDIT] Updated to clarify that I would like the results to be placed in a new Array of Hashes, rather than mutating either of the original Arrays.
def join(referers, referees, on_referer, on_referee)
referers.map do |record|
referees.find do |referee_record|
record[on_referer] == referee_record[on_referee]
end.merge(record)
end
end
cars = [ { id: 1, color: 'red', owner_id: 1 }, { id: 2, color: 'black', owner_id: 1 } ]
owners = [ { id: 1, name: 'Alice' }, { id: 2, name: 'Bob' } ]
join(cars, owners, :owner_id, :id)
# => [{:id=>1, :name=>"Alice", :color=>"red", :owner_id=>1},
# {:id=>2, :name=>"Alice", :color=>"black", :owner_id=>1}]
Edit: I just noticed that it is the key :owner_id in cars that is to be matched with the :id in owners. I assumed the key :id in cars was to be matched. I will leave my answer as is, considering that the modification is trivial and that it may be easier to follow if the match is to be on the same key names.
Assuming that:
you want to modify (mutate) cars; and
for each element h of owners there is an element g of cars for which h[:id] == g[:id],
it's just
owners.each { |h| cars.find { |g| g[:id] == h[:id] }.update(h) }
cars #=> [{:id=>1, :color=>"red", :owner_id=>1, :name=>"Alice"},
# {:id=>2, :color=>"black", :owner_id=>1, :name=>"Bob"}]
On the other hand, if:
you do not wish to mutate cars or
for a given element h of owners there may be no element g of cars for which h[:id]==g[:id] or
you just want to improve efficiency,
you could first create a hash for cars or owners whose keys are values of :id.
Suppose:
owners = [ { id: 3, name: 'Alice' }, { id: 2, name: 'Bob' } ]
We could create a hash for owners:
owners_by_id = owners.each_with_object({}) { |g,h| h.update(g[:id]=>g) }
#=> {3=>{:id=>3, :name=>"Alice"}, 2=>{:id=>2, :name=>"Bob"}}
and then write:
cars.map do |h|
g = {}.merge(h)
id = g[:id]
g.update(owners_by_id[id]) if owners_by_id.key?(id)
g
end
#=> [{:id=>1, :color=>"red", :owner_id=>1},
# {:id=>2, :color=>"black", :owner_id=>1, :name=>"Bob"}]
Assuming that the hashes at the same position in the arrays correspond:
[cars, owners].transpose.map{|h1, h2| h1.merge(h2)}
Otherwise, your example is bad.
I am using ActiveRecord. It has a handy method called group_by. When I use it with my activerecord objects, i get the below hash:
{["junior"]=>[#<Lead id: 1, created_at: "2015-02-13 02:34:39", updated_at: "2015-02-13 02:35:27", case_enabled: true>, #<Lead id: 2, created_at: "2015-02-13 20:48:19", updated_at: "2015-02-13 20:48:19", case_enabled: nil>, ["senior"]=>[#<Lead id: 3, created_at: "2015-02-13 20:48:19", updated_at: "2015-02-13 20:48:19", case_enabled: nil>, #<Lead id: 4, created_at: "2015-02-13 20:49:16", updated_at: "2015-02-13 20:49:16", case_enabled: nil>]}
However, I want a hash with subhashes that contain the collection as an ActiveRecord::Relation and column data. So this is what I come up with:
i = 0
r = group.reduce({}) do |acc, (k,v)|
h = {}
active_record_relation = where(id: v.map(&:id))
h["#{k.first}_collection"] = active_record_relation
h["#{k.first}_columns"] = Classification.where(code: k.first).first.default_fields
acc[i] = h
i += 1
acc
end
And it gives me the results I want:
{0=>{"junior_collection"=>#<ActiveRecord::Relation [# ... ]>, "junior_columns"=>[ ... ]}, 1=>{"senior_collection"=>#<ActiveRecord::Relation [# ... ]>, "senior_columns"=>[ ... ]}}
The fact that I had to add the i variable makes me feel like this is not the ruby way to do this. But I looked at the docs and I didn't find a way to add an index to reduce, since I am already passing a hash into reduce. Is there another way?
Your way is probably good enough but you can avoid separately tracking the index by doing .each.with_index.reduce(...) { |acc, ((k,v),i)| ... }, like so:
h = {'a' => 'b', 'c' => 'd', 'e' => 'f'}
h.each.with_index.reduce('OK') do |acc, ((k, v), i)|
puts "acc=#{acc}, k=#{k}, v=#{v}, i=#{i}"
acc
end
# acc=OK, k=a, v=b, i=0
# acc=OK, k=c, v=d, i=1
# acc=OK, k=e, v=f, i=2
# => "OK"
Not sure if it's more Rubyish than your way =\
I have an array of ids order say
order = [5,2,8,6]
and another array of hash
[{id: 2,name: name2},{id: 5,name: name5}, {id: 6,name: name6}, {id: 8,name: name8}]
I want it sorted as
[{id: 5,name: name5},{id: 2,name: name2}, {id: 8,name: name8}, {id: 6,name: name6}]
What could be best way to implement this? I can implement this with iterating both and pushing it to new array but looking for better solution.
Try this
arr = [
{:id=>2, :name=>"name2"}, {:id=>5, :name=>"name5"},
{:id=>6, :name=>"name6"}, {:id=>8, :name=>"name8"}
]
order = [5,2,8,6]
arr.sort_by { |a| order.index(a[:id]) }
# => [{:id=>5, :name=>"name5"}, {:id=>2, :name=>"name2"},
#{:id=>8, :name=>"name8"}, {:id=>6, :name=>"name6"}]
Enumerable#in_order_of (Rails 7+)
Starting from Rails 7, there is a new method Enumerable#in_order_of.
A quote right from the official Rails docs:
in_order_of(key, series)
Returns a new Array where the order has been set to that provided in the series, based on the key of the objects in the original enumerable.
[ Person.find(5), Person.find(3), Person.find(1) ].in_order_of(:id, [ 1, 5, 3 ])
=> [ Person.find(1), Person.find(5), Person.find(3) ]
If the series include keys that have no corresponding element in the Enumerable, these are ignored. If the Enumerable has additional elements that aren't named in the series, these are not included in the result.
It is not perfect in a case of hashes, but you can consider something like:
require 'ostruct'
items = [{ id: 2, name: 'name2' }, { id: 5, name: 'name5' }, { id: 6, name: 'name6' }, { id: 8, name: 'name8' }]
items.map(&OpenStruct.method(:new)).in_order_of(:id, [5,2,8,6]).map(&:to_h)
# => [{:id=>5, :name=>"name5"}, {:id=>2, :name=>"name2"}, {:id=>8, :name=>"name8"}, {:id=>6, :name=>"name6"}]
Sources:
Official docs - Enumerable#in_order_of.
PR - Enumerable#in_order_of #41333.
Rails 7 adds Enumerable#in_order_of.
I have an Array of Hashes with the same keys, storing people's data.
I want to remove the hashes that have the same values for the keys :name and :surname. The rest of the values can differ, so calling uniq! on array won't work.
Is there a simple solution for this?
You can pass a block to uniq or uniq!, the value returned by the block is used to compare two entries for equality:
irb> people = [{name: 'foo', surname: 'bar', age: 10},
{name: 'foo', surname: 'bar' age: 11}]
irb> people.uniq { |p| [p[:name], p[:surname]] }
=> [{:name=>"foo", :surname=>"bar", :age=>10}]
arr=[{name: 'john', surname: 'smith', phone:123456789},
{name: 'thomas', surname: 'hardy', phone: 671234992},
{name: 'john', surname: 'smith', phone: 666777888}]
# [{:name=>"john", :surname=>"smith", :phone=>123456789},
# {:name=>"thomas", :surname=>"hardy", :phone=>671234992},
# {:name=>"john", :surname=>"smith", :phone=>666777888}]
arr.uniq {|h| [h[:name], h[:surname]]}
# [{:name=>"john", :surname=>"smith", :phone=>123456789},
# {:name=>"thomas", :surname=>"hardy", :phone=>671234992}]
unique_people = {}
person_array.each do |person|
unique_people["#{person[:name]} #{person[:surname]}"] = person
end
array_of_unique_people = unique_people.values
This should do the trick.
a.delete_if do |h|
a.select{|i| i[:name] == h[:name] and i[:surname] == h[:surname] }.count > 1
end
I have an Array of objects:
[
#<User id: 1, name: "Kostas">,
#<User id: 2, name: "Moufa">,
...
]
And I want to convert this into an Hash with the id as the keys and the objects as the values. Right now I do it like so but I know there is a better way:
users = User.all.reduce({}) do |hash, user|
hash[user.id] = user
hash
end
The expected output:
{
1 => #<User id: 1, name: "Kostas">,
2 => #<User id: 2, name: "Moufa">,
...
}
users_by_id = User.all.map { |user| [user.id, user] }.to_h
If you are using Rails, ActiveSupport provides Enumerable#index_by:
users_by_id = User.all.index_by(&:id)
You'll get a slightly better code by using each_with_object instead of reduce.
users = User.all.each_with_object({}) do |user, hash|
hash[user.id] = user
end
You can simply do (using the Ruby 3 syntax of _1)
users_by_id = User.all.to_h { [_1.id, _1] }