I have an array of hashes similar to this:
[
{"student": "a","scores": [{"subject": "math","quantity": 10},{"subject": "english", "quantity": 5}]},
{"student": "b", "scores": [{"subject": "math","quantity": 1 }, {"subject": "english","quantity": 2 } ]},
{"student": "a", "scores": [ { "subject": "math", "quantity": 2},{"subject": "science", "quantity": 5 } ] }
]
Is there a simpler way of getting the output similar to this except looping through the array and finding a duplicate and then combining them?
[
{"student": "a","scores": [{"subject": "math","quantity": 12},{"subject": "english", "quantity": 5},{"subject": "science", "quantity": 5 } ]},
{"student": "b", "scores": [{"subject": "math","quantity": 1 }, {"subject": "english","quantity": 2 } ]}
]
Rules for merging duplicate objects:
Is there a simpler way of getting the output similar to this except looping through the array and finding a duplicate and then combining them?
Not that I know of. IF you explain where this data is comeing form the answer may be different but just based on the Array
of Hash
objects I think you will haev to iterate and combine.
While it is not elegant you could use a solution like this
arr = [
{"student"=> "a","scores"=> [{"subject"=> "math","quantity"=> 10},{"subject"=> "english", "quantity"=> 5}]},
{"student"=> "b", "scores"=> [{"subject"=> "math","quantity"=> 1 }, {"subject"=> "english","quantity"=> 2 } ]},
{"student"=> "a", "scores"=> [ { "subject"=> "math", "quantity"=> 2},{"subject"=> "science", "quantity"=> 5 } ] }
]
#Group the array by student
arr.group_by{|student| student["student"]}.map do |student_name,student_values|
{"student" => student_name,
#combine all the scores and group by subject
"scores" => student_values.map{|student| student["scores"]}.flatten.group_by{|score| score["subject"]}.map do |subject,subject_values|
{"subject" => subject,
#combine all the quantities into an array and reduce using `+`
"quantity" => subject_values.map{|h| h["quantity"]}.reduce(:+)
}
end
}
end
#=> [
{"student"=>"a", "scores"=>[
{"subject"=>"math", "quantity"=>12},
{"subject"=>"english", "quantity"=>5},
{"subject"=>"science", "quantity"=>5}]},
{"student"=>"b", "scores"=>[
{"subject"=>"math", "quantity"=>1},
{"subject"=>"english", "quantity"=>2}]}
]
I know that you specified your expected result but I wanted to point out that making the output simpler makes the code simpler.
arr.map(&:dup).group_by{|a| a.delete("student")}.each_with_object({}) do |(student, scores),record|
record[student] = scores.map(&:values).flatten.map(&:values).each_with_object(Hash.new(0)) do |(subject,score),obj|
obj[subject] += score
obj
end
record
end
#=>{"a"=>{"math"=>12, "english"=>5, "science"=>5}, "b"=>{"math"=>1, "english"=>2}}
With this structure getting the students is as easy as calling .keys
and the scores would be equally as simple. I am thinking something like
above_result.each do |student,scores|
puts student
scores.each do |subject,score|
puts " #{subject.capitalize}: #{score}"
end
end
end
The console out put would be
a
Math: 12
English: 5
Science: 5
b
Math: 1
English: 2
There are two common ways of aggregating values in such instances. The first is to employ the method Enumerable#group_by , as @engineersmnky has done in his answer. The second is to build a hash using the form of the method Hash#update (aka merge!
) that uses a block to resolve the values of keys which are present in both of the hashes being merged. My solution uses the latter approach, not because I prefer it to the group_by
, but just to show you a different way it can be done. (Had engineersmnky used update
, I would have gone with group_by
.)
Your problem is complicated somewhat by the particular data structure you are using. I found that the solution could be simplfied and made easier to follow by first converting the data to a different structure, update the scores, then convert the result back to your data structure. You may want to consider changing the data structure (if that's an option for you). I've addressed that issue in the "Discussion" section.
Code
def combine_scores(arr)
reconstruct(update_scores(simplify(arr)))
end
def simplify(arr)
arr.map do |h|
hash = Hash[h[:scores].map { |g| g.values }]
hash.default = 0
{ h[:student]=> hash }
end
end
def update_scores(arr)
arr.each_with_object({}) do |g,h|
h.update(g) do |_, h_scores, g_scores|
g_scores.each { |subject,score| h_scores[subject] += score }
h_scores
end
end
end
def reconstruct(h)
h.map { |k,v| { student: k, scores: v.map { |subject, score|
{ subject: subject, score: score } } } }
end
Example
arr = [
{ student: "a", scores: [{ subject: "math", quantity: 10 },
{ subject: "english", quantity: 5 }] },
{ student: "b", scores: [{ subject: "math", quantity: 1 },
{ subject: "english", quantity: 2 } ] },
{ student: "a", scores: [{ subject: "math", quantity: 2 },
{ subject: "science", quantity: 5 } ] }]
combine_scores(arr)
#=> [{ :student=>"a",
# :scores=>[{ :subject=>"math", :score=>12 },
# { :subject=>"english", :score=> 5 },
# { :subject=>"science", :score=> 5 }] },
# { :student=>"b",
# :scores=>[{ :subject=>"math", :score=> 1 },
# { :subject=>"english", :score=> 2 }] }]
Explanation
First consider the two intermediate calculations:
a = simplify(arr)
#=> [{ "a"=>{ "math"=>10, "english"=>5 } },
# { "b"=>{ "math"=> 1, "english"=>2 } },
# { "a"=>{ "math"=> 2, "science"=>5 } }]
h = update_scores(a)
#=> {"a"=>{"math"=>12, "english"=>5, "science"=>5}
# "b"=>{"math"=> 1, "english"=>2}}
Then
reconstruct(h)
returns the result shown above.
+ simplify
arr.map do |h|
hash = Hash[h[:scores].map { |g| g.values }]
hash.default = 0
{ h[:student]=> hash }
end
This maps each hash into a simpler one. For example, the first element of arr
:
h = { student: "a", scores: [{ subject: "math", quantity: 10 },
{ subject: "english", quantity: 5 }] }
is mapped to:
{ "a"=>Hash[[{ subject: "math", quantity: 10 },
{ subject: "english", quantity: 5 }].map { |g| g.values }] }
#=> { "a"=>Hash[[["math", 10], ["english", 5]]] }
#=> { "a"=>{"math"=>10, "english"=>5}}
Setting the default value of each hash to zero simplifies the update step, which follows.
+ update_scores
For the array of hashes a
that is returned by simplify
, we compute:
a.each_with_object({}) do |g,h|
h.update(g) do |_, h_scores, g_scores|
g_scores.each { |subject,score| h_scores[subject] += score }
h_scores
end
end
Each element of a
(a hash) is merged into an initially-empty hash, h
. As update
(same as merge!
) is used for the merge, h
is modified. If both hashes share the same key (eg, "math"), the values are summed; else subject=>score
is added to h
.
Notice that if h_scores
does not have the key subject
, then:
h_scores[subject] += score
#=> h_scores[subject] = h_scores[subject] + score
#=> h_scores[subject] = 0 + score (because the default value is zero)
#=> h_scores[subject] = score
That is, the key-value pair from g_scores
is merely added to h_scores
.
I've replaced the block variable representing the subject with a placeholder _
, to reduce the chance of errors and to inform the reader that it is not used in the block.
+ reconstruct
The final step is to convert the hash returned by update_scores
back to the original data structure, which is straightforward.
Discussion
If you change the data structure, and it meets your requirements, you may wish to consider changing it to that produced by combine_scores
:
h = { "a"=>{ math: 10, english: 5 }, "b"=>{ math: 1, english: 2 } }
Then to update the scores with:
g = { "a"=>{ math: 2, science: 5 }, "b"=>{ english: 3 }, "c"=>{ science: 4 } }
you would merely to the following:
h.merge(g) { |_,oh,nh| oh.merge(nh) { |_,ohv,nhv| ohv+nhv } }
#=> { "a"=>{ :math=>12, :english=>5, :science=>5 },
# "b"=>{ :math=> 1, :english=>5 },
# "c"=>{ :science=>4 } }
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.