Sum array values to hash

Question

I have an array like

[
    [358, 202102, 5],
    [358, 202112, 5],
    [358, 202102, 10],
    [311, 202103, 5],
    [311, 202101, 1],
    [311, 202101, 1],
    [311, 202115, 8],
    [311, 202101, 1],
    [311, 202101, 1]
]

I need a hash like this:

{
    358 => { 
        202102 => 15,
        202112 => 5
    },
    311 => {
        202103 => 5,
        202101 => 4,
        202115 => 8
    }
}

The order does not matter. Only the third values from each row of the array must be added up.

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– Jörg W Mittag, Commented Sep 9, 2021 at 14:30
idownvotedbecau.se/nocode idownvotedbecau.se/noattempt idownvotedbecau.se/nomcve — Jörg W Mittag
– Jörg W Mittag, Commented Sep 9, 2021 at 14:30
A small suggestion: if you attach a variable to all example inputs (e.g., a = [[358, 202102, 5],...]) readers can refer to those variables in their answers and comments without having to define them. — Cary Swoveland
– Cary Swoveland, Commented Sep 9, 2021 at 18:21

marc_s · Accepted Answer · 2024-09-02 19:28:00Z

4

You can solve this using Enumerable#each_with_object as follows

a = [
    [358, 202102, 5],
    [358, 202112, 5],
    [358, 202102, 10],
    [311, 202103, 5],
    [311, 202101, 1],
    [311, 202101, 1],
    [311, 202115, 8],
    [311, 202101, 1],
    [311, 202101, 1]
]

a.each_with_object(Hash.new {|h,k| h[k] = Hash.new(0)}) do |(k1,k2,val), obj| 
  obj[k1][k2] += val 
end 
# => {358=>{202102=>15, 202112=>5}, 311=>{202103=>5, 202101=>4, 202115=>8}}

What this does is iterates over the Array (a) with an accumulator Object (obj).

This object is a Hash with a default proc where every time a new key is added its value is assigned as a new Hash with a default value of 0.

In the block arguments we deconstruct the Array into 3 arguments (k1,k2,val) so on first pass it would look something like this:

k1 = 358
k2 = 202102
val = 5 
obj[k1] 
#=> {358=> {}} 
obj[k1][k2] 
#=> {358 => {202102 => 0 }}
obj[k1][k2] += val
obj
#=> {358 => {202102 => 5 }}

edited Sep 2, 2024 at 19:28

marc_s

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answered Sep 9, 2021 at 14:03

engineersmnky

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Comments

Cary Swoveland · Accepted Answer · 2021-09-09 23:02:35Z

3

Another way is to use the methods Hash#update (aka merge!) and Hash#merge that employ blocks to determine the values of keys that are present in both hashes being merged. See the docs for definitions of the three block variables. I've used an underscore for the first block variables (the common key of the hashes being merged) to signify that it is not used in the block calculation.

a = [
    [358, 202102, 5],
    [358, 202112, 5],
    [358, 202102, 10],
    [311, 202103, 5],
    [311, 202101, 1],
    [311, 202101, 1],
    [311, 202115, 8],
    [311, 202101, 1],
    [311, 202101, 1]
]

a.each_with_object({}) do |(x,y,n),h|
  h.update(x=>{y=>n}) do |_,old1,new1|
    old1.merge(new1) { |_,old2,new2| old2+new2 }
  end
end
  #=> {358=>{202102=>15, 202112=>5}, 311=>{202103=>5, 202101=>4, 202115=>8}}

edited Sep 9, 2021 at 23:02

answered Sep 9, 2021 at 19:19

Cary Swoveland

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3 Comments

engineersmnky Over a year ago

Only comment purely from a readability standpoint is that you distinguished o and n in the merge context (oo and nn respectively); however n already has an outer context as well as in |(x,y,n),h| so a further distinction could be valuable.

engineersmnky Over a year ago

Sorry I was disingenuous I have a secondary comment h does not need to be defined using the block syntax for Hash; literal {} would be sufficient. You are not actually utilizing the default proc because first pass creates a nested Hash via update.

Cary Swoveland Over a year ago

@engineersmnky, both good points; I did an edit. Thanks.

nwnoll · Accepted Answer · 2021-09-09 14:06:58Z

2

At first I create a hash of hashes with the default value 0. After that I just add the values of the hash of hashes up based on the last value in the array.

aoa =
[
    [358, 202102, 5],
    [358, 202112, 5],
    [358, 202102, 10],
    [311, 202103, 5],
    [311, 202101, 1],
    [311, 202101, 1],
    [311, 202115, 8],
    [311, 202101, 1],
    [311, 202101, 1]
]


hash = Hash.new { |h,k| h[k] = Hash.new(0) }
aoa.each { |ary|  hash[ary[0]][ary[1]] += ary[2]}

pp hash

output:

{358=>{202102=>15, 202112=>5}, 311=>{202103=>5, 202101=>4, 202115=>8}}

answered Sep 9, 2021 at 14:06

nwnoll

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