Iterate over multi-dimensional array and return aggregated sum based on first index

Question

I've really been wrecking my brain over this one so any assistance is greatly appreciated. I currently have an array of objects and each of these objects has two properties, a timestamp (ms since epoch) and an arbitrary for that timestamp. Like so:

[
    [1518739200000, 1], 
    [1518739200000, 1],
    [1518739200000, 12],
    [1518739200000, 16],
    [1518739200000, 16],
    [1518825600000, 16],
    [1518825600000, 20],
    [1518825600000, 20],
    [1518825600000, 8],
]

What I'm trying to do is to condense the multidimensional array into another multidimensional array which has only the unique values of the first index, the timestamp, and an aggregate count as the second value. In example, I'm looking for this from the given set of data included above.

[
    [1518739200000, 46],
    [1518825600000, 64],
]

I've tried using .reduce in a few ways but haven't had any success yet for this particular kind manipulation.

Possible duplicate of Javascript : How group and sum values from multidimensional array — PM 77-1
– PM 77-1, Commented May 8, 2018 at 19:00
I don't see how the example works... the sum for the shortest timestamp isn't 32 as far as I can see... — Joe Iddon
– Joe Iddon, Commented May 8, 2018 at 19:05
The duplicate is a bit dated. Probably easier to reduce and Map. — HMR
– HMR, Commented May 8, 2018 at 19:11

Ele · Accepted Answer · 2018-05-08 19:05:43Z

1

An alternative using the function reduce

let array = [[1518739200000, 1], [1518739200000, 1],[1518739200000, 12],[1518739200000, 16],[1518739200000, 16],[1518825600000, 16],[1518825600000, 20],[1518825600000, 20],[1518825600000, 8]];
let result = Object.values(array.reduce((a, [key, value]) => {
  (a[key] || (a[key] = [key, 0]))[1] += value;
  return a;
}, {}));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered May 8, 2018 at 19:05

Ele

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3 Comments

Joe Iddon Over a year ago

Very cool stuff, I'd love to be able to right js as succinct as this. Would you mind explaining how (x = [1,2])[1] += 1 works, but (x = 1) += 1 doesn't. Thanks! +1

Steve Frost Over a year ago

Interesting, can you explain why you use Object.values & the empty object at the end. I assume thats the initial value though I see we get back a multidimensional array (as requested) so what is the empty object there to do? Thanks!

Joe Iddon Over a year ago

The idea is to generate an object with the key as the common times and the value as the running total. On each iteration, either the number is added to the total of that key if it already exists or it creates a new key for that time with the running total as 0 and then immediately adds the current number. The whole reduce call would return an object with 2d arrays as values, but if passed into Object.values, you just get an array of the values.

HMR · Accepted Answer · 2018-05-08 19:09:15Z

0

You could reduce the array to a Map and use the Map to create an array of arrays using Array.from.

var data = [
  [1518739200000, 1], 
  [1518739200000, 1],
  [1518739200000, 12],
  [1518739200000, 16],
  [1518739200000, 16],
  [1518825600000, 16],
  [1518825600000, 20],
  [1518825600000, 20],
  [1518825600000, 8],
];

console.log(
  Array.from(
    data.reduce(
      (result,[current,num])=>
        result.set(
          current,
          (result.get(current)||0)+num
        ),
      new Map()
    )
  )
)

answered May 8, 2018 at 19:09

HMR

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Comments

ibrahim tanyalcin · Accepted Answer · 2018-05-08 19:29:36Z

Use reduce

var x = [
        [1518739200000, 1], 
        [1518739200000, 1],
        [1518739200000, 12],
        [1518739200000, 16],
        [1518739200000, 16],
        [1518825600000, 16],
        [1518825600000, 20],
        [1518825600000, 20],
        [1518825600000, 8],
    ]

x.reduce(function(ac,d,i,a){if(!ac.some(function(dd,ii){return dd[0] === d[0] && dd[1] === d[1]})){ac.push(d)
}return ac},[]);

The above gets unique values for BOTH indexes, for the first one only:

x.reduce(function(ac,d,i,a){if(!ac.some(function(dd,ii){return dd[0] === d[0]})){ac.push(d)}return ac},[])

But then, the question is, this gets the first encountered entry, do you have any preference for the second?

UPDATE:Oww, I didn't see the aggreate, this is for it:

x.reduce(function(ac,d,i,a){
    var index;
    if(
        !ac.some(function(dd,ii){
            return dd[0] === d[0] ? (index=ii,true) : false
        })
    ){
        ac.push(d.slice())
    } else {
        ac[index][1] += d[1];
    }
    return ac
},[])

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