Cumulative sum of specific keys with array output using reduce

Question

Say I have the following array:

let arr = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}]

I would like to compute the cumulative sum of each key, but I would also like the output to be an array of the same length with the cumulative values at each step. The final result should be:

[{a: 1, b: 2}, {a: 3, b: 6}, {a: 11, b: 5}]

My issue is that I am not able to obtain the array as desired. I only get the final object with this:

let result = arr.reduce((accumulator, element) => {
  if(accumulator.length === 0) {
    accumulator = element
  } else {
    for(let i in element){
      accumulator[i] = accumulator[i] + element[i]
    }
  }
  return accumulator
}, [])

console.log(result); // {a: 11, b: 5}

user3297291 · Accepted Answer · 2021-12-07 08:30:09Z

What you're after sounds like the scan() higher-order function (borrowing the idea from ramda.js), which allows you to return an accumulated result for each element within your array. The scan method is similar to how the .reduce() method behaves, except that it returns the accumulator for each element. You can build the scan() function yourself like so:

let arr = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}];

const scan = ([x, ...xs], fn) => xs.reduce((acc, elem) => {
  return [...acc, fn(acc.at(-1), elem)];
}, xs.length ? [x] : []);

const res = scan(arr, (x, y) => ({a: x.a+y.a, b: x.b+y.b}));
console.log(res);

You might consider further improvements such as providing an initial value to the scan method (similar to how reduce accepts one). Also, if you need better browser support the .at() method currently has limited browser support, so you may instead consider creating your own at() function:

const at = (arr, idx) => idx >= 0 ? arr[idx] : arr[arr.length + idx];

DecPK · Accepted Answer · 2021-12-07 05:36:17Z

2

You can easily achieve the result using reduce as

let arr = [
  { a: 1, b: 2 },
  { a: 2, b: 4 },
  { a: 8, b: -1 },
];

const result = arr.reduce((acc, curr, i) => {
  if (i === 0) acc.push(curr);
  else {
    const last = acc[i - 1];
    const newObj = {};
    Object.keys(curr).forEach((k) => (newObj[k] = curr[k] + last[k]));
    acc.push(newObj);
  }
  return acc;
}, []);

console.log(result);

answered Dec 7, 2021 at 5:36

DecPK

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Comments

ProDec · Accepted Answer · 2021-12-07 05:19:56Z

Something like this:

const arr = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}]

const result = arr.reduce((accumulator, element, index) => {
  if(accumulator.length === 0) {
    accumulator.push(element)
  } else {
    const sum = {};
    for(let i in element) {
      sum[i] = element[i] + (accumulator[index - 1][i] || 0)
    }
    accumulator.push(sum)
  }
  return accumulator
}, [])

console.log(result);

Another option is keep sum result using a Map, it helps if keys in elements of the array are not always same.

const arr = [{a: 1, b: 2}, {a: 2}, {a: 8, b: -1}];
const map = new Map();
const result = arr.map((element) => {
  const sum = {};
  for (let i in element) {
    sum[i]= element[i] + (map.get(i) || 0);
    map.set(i, sum[i]);
  }
  return sum;
});

console.log(result);

Kinglish · Accepted Answer · 2021-12-07 05:45:43Z

1

Here is a bit more concise reduce, probably not as readable as a consequence...

array.reduce((y,x,i) => ( i===0 ? y : [...y, {a: x.a + y[i-1].a, b: x.b + y[i-1].b}]),[array[0]])

let array = [{a: 1, b: 2}, {a: 2, b: 4}, {a: 8, b: -1}]
let culm = array.reduce((y,x,i) => ( i===0 ? y : [...y, {a: x.a + y[i-1].a, b: x.b + y[i-1].b}]),[array[0]])
console.log(culm)

edited Dec 7, 2021 at 5:45

answered Dec 7, 2021 at 5:31

Kinglish

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customcommander · Accepted Answer · 2021-12-07 14:03:04Z

0

Given:

const xs =
  [ {a: 1, b:  2}
  , {a: 2, b:  4}
  , {a: 8, b: -1}];

Define a function sum such as:

const sum = ([head, ...tail]) =>
  tail.reduce((x, y) =>
    ({a: (x.a+y.a), b: (x.b+y.b)}), head);

sum(xs);
//=> {a: 11, b: 5}

Then apply that function in a map on larger slices of xs:

xs.map((_, i, arr) => sum(arr.slice(0, i+1)));
//=> [ {a:  1, b: 2}
//=> , {a:  3, b: 6}
//=> , {a: 11, b: 5}]

answered Dec 7, 2021 at 14:03

customcommander

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