reduce array of array into a flat array of object

Question

I'm stuck at transforming a data structure:

let d = [
  { no: 1, score: 7000 },
  { no: 2, score: 10000 },
  [
    { no: 1, score: 8500 },
    { no: 2, score: 6500 }
  ]
];

    
d = d.reduce((accum, o) => {
   
}, [])

How can I produce this?

[{name: 'no 1', score: [7000, 8500]}, {name: 'no 2', score: [10000, 6500]}]

Answer 1

Here is one way to do it with simple reduce,

const result = d.flat().reduce((acc: {name: string, score: number[]}[], curr) => {
  const { no, score } = curr;
  let item = acc.find(a => a.name === `no ${no}`);
  if (!item) {
    item = { name: `no ${no}`, score: []};
    acc.push(item);
  }

  item.score.push(score);
  return acc;
    
}, []);

console.log(result)

Answer 2

In your case, you not only need to flatten the list, but also group by no property.

For flatting, you can use Array.prototype.flat(). It's quite a new feature, so if you don't use polyfills, you probably can't use it. So you can check for alternative implementations.

For grouping, you can reduce to the object where the key is no property. Note, that if multiple no properties exist, you need to save an array of all score values.

Example:

const d = [{ no: 1, score: 7000 },
    { no: 2, score: 10000 },
   [ { no: 1, score: 8500 },
    { no: 2, score: 6500 }]]


const grouped = d.flat().reduce((prev, cur) => {
    if (cur.no in prev) {
        prev[cur.no].score.push(cur.score)
    } else {
        prev[cur.no] = {
            name: 'no ' + cur.no,
            score: [cur.score]
        }
    }
    return prev
}, {})

console.log(Object.values(grouped))

In the example we use modifications. It is possible to do it without modifications - return a new copy during each reduction iteration. However, depending on array size, there can be performance issues. Also, it's safe to do modifications, because we create a new object in this case.

Answer 3

let d = [{ no: 1, score: 7000 },
    { no: 2, score: 10000 },
   [ { no: 1, score: 8500 },
    { no: 2, score: 6500 }]]
    
    const result=d.flat().reduce((acc,curr)=>{
    if(acc[curr.no]){
      acc[curr.no].score.push(curr.score)
    } else {
      const keys=Object.keys(curr)
      acc[curr.no]={ name: keys[0]+ ' '+curr.no, score:[curr.score]}
    }
    return acc;
    },{})
console.log(Object.values(result))

Answer 4

Array.prototype.flat() call before the actual grouping, then use reduce function create the result.

let d = [{
    no: 1,
    score: 7000
},
{
    no: 2,
    score: 10000
},
[{
        no: 1,
        score: 8500
    },
    {
        no: 2,
        score: 6500
    }
]
]

const result = d.flat().reduce((result, element) => {
const key = element.no;
if (!result[key]) {
    result[key] = {
        name: `no ${key}`,
        score: []
    }
}

result[key].score.push(element.score);
return result;
}, {})
console.log(Object.values(result))

Answer 5

Here's a sleek functional solution for typescript, since you included the tag-

const arr = [
    { no: 1, score: 7000 },
    { no: 2, score: 10000 },
    [
        { no: 1, score: 8500 },
        { no: 2, score: 6500 },
    ],
];

const result = Object.entries(
    arr.flat().reduce((accum: { [key: number]: number[] }, el: { no: number; score: number }) => {
        accum[el.no] = (accum[el.no] ?? []).concat(el.score);
        return accum;
    }, {})
).map(([num, scores]) => ({ no: Number(num), scores: scores }));

console.log(result);

Result-

[
  { no: 1, scores: [ 7000, 8500 ] },
  { no: 2, scores: [ 10000, 6500 ] }
]

This flattens the inner arrays first using Array.prototype.flat. Then it uses reduce to construct an object that has the no values as keys and score values as an array of values.

In the end, the reduce results in { 1: [7000, 8500], 2: [10000, 6500] } - turn that into entries using Object.entries to get [['1', [7000, 8500]], ['2', [10000, 6500]]]

Finally, map over the entries to turn the ['1', [7000, 8500]] format into { no: 1, scores: [ 7000, 8500 ] } format and you're done!

Answer 6

You could take a dynamic approach which groups by the given key and takes all other properties for a new array.

const
    groupBy = key => (r, value) => {
        if (Array.isArray(value)) return value.reduce(group, r);
        const { [key]: _, ...o } = value;
        Object.entries(o).forEach(([k, v]) => ((r[_] ??= { [key]: _ })[k] ??= []).push(v));
        return r;
    }
    data = [{ no: 1, score: 7000 }, { no: 2, score: 10000 }, [{ no: 1, score: 8500 }, { no: 2, score: 6500 }]],
    group = groupBy('no'),
    result = Object.values(data.reduce(group, {}));

console.log(result);

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Answer 7

You can use Array.prototype.flat()

or if you have any array like,

let d = [
  { no: 1, score: 7000 },
  { no: 2, score: 10000 },
  [
    { no: 1, score: 8500 },
    { no: 2, score: 6500 }
  ]
];

and then use d.flat()

Answer 8

Try this:

let d = [
    { no: 1, score: 7000 },
    { no: 2, score: 10000 },
    [
        { no: 1, score: 8500 },
        { no: 2, score: 6500 }
    ]
]


const reducer = (arr, start = []) => arr.reduce((acc, next) => {
    if (Array.isArray(next)) return reducer(next, acc);
    for (const value of acc) {
        if (value.name === `no ${next.no}`) {
            value.score = [...value.score, next.score];
            return acc;
        }
    }
    
    return [...acc, {
        name: `no ${next.no}`,
        score: [next.score]
    }];
}, start);

console.log(reducer(d));