How to reduce an array while merging one of it's field as well

Question

I have this,

var o = [{
  "id": 1, // its actually a string in real life
  "course": "name1",
  // more properties
}, 
{
  "id": 1, // its actually a string in real life
  "course": "name2",
  // more properties
}];

I want this,

var r = [{
  "id": 1, // its actually a string in real life
  "course": ["name1", "name2"],
}];

I am trying this,

var flattened = [];
for (var i = 0; i < a.length; ++i) {
  var current = a[i];
  if(flattened.)
}

but I am stuck, I am not sure what to do next, array will have more then 2 records but this was just an example.

THERE are more fields but I removed them for simplicity, I won't be using them in final array.

@NickParsons i think it's better to call it a string in this scenario, sorry for confusion — Mathematics, Mar 12 '19 at 11:59
Possible duplicate of [Group array items using object](https://stackoverflow.com/questions/31688459/group-array-items-using-object) and [Merge objects in array based on property](https://stackoverflow.com/questions/43281272) — adiga, Mar 12 '19 at 12:09

Nina Scholz · Answer 1 · 2019-03-12T12:10:36.890

You could reduce the array and find the object.

var array = [{ id: 1, course: "name1" }, { id: 1, course: "name2" }],
    flat = array.reduce((r, { id, course }) => {
        var temp = r.find(o => id === o.id);
        if (!temp) {
            r.push(temp = { id, course: [] });
        }
        temp.course.push(course);
        return r;
    }, []);

console.log(flat);

The same by taking a Map.

var array = [{ id: 1, course: "name1" }, { id: 1, course: "name2" }],
    flat = Array.from(
        array.reduce((m, { id, course }) => m.set(id, [...(m.get(id) || []) , course]), new Map),
        ([id, course]) => ({ id, course })
    );

console.log(flat);

score 1 · Accepted Answer · answered Mar 12 '19 at 12:07

This way you will get the data flattened in the shape you want

const o = [
  {
    id: 1,
    course: "name1"
  },
  {
    id: 1,
    course: "name2"
  },
  {
    id: 2,
    course: "name2"
  }
];

const r = o.reduce((acc, current) => {
  const index = acc.findIndex(x => x.id === current.id);
  if (index !== -1) {
    acc[index].course.push(current.course);
  } else {
    acc.push({id:current.id, course: [current.course]});
  }
  return acc
}, []);

console.log(r);

Nick Parsons · Answer 3 · 2019-03-12T13:13:23.233

You could use .reduce to create an object of keys, and then use that object to set keys to be of the id. This way you can add to the same course array by targetting the id of the object. Lastly, you can get the values of the object to get your result.

See example below:

var o = [{
  "id": 1,
  "course": "name1",
  "foo": 1
}, 
{
  "id": 1,
  "course": "name2",
  "bar": 2
}];

var res = Object.values(o.reduce((acc, {id, course, ...rest}) => {
  if(id in acc) 
    acc[id] = {...acc[id], course: [...acc[id].course, course], ...rest};
  else acc[id] = {id, course: [course], ...rest};
  return acc;
}, {}));

console.log(res);

jo_va · Answer 4 · 2019-03-12T12:35:44.443

You can do this with reduce and Object.entries. This example works for any number of properties:

const o = [
  { id: 1, course: 'name1', time: 'morning', topic: 'math' },
  { id: 1, course: 'name2', time: 'afternoon' },
  { id: 2, course: 'name3', time: 'evening' }
];

const result = o.reduce((out, { id, ...rest }) => {
  out[id] = out[id] || {};
  const mergedProps = Object.entries(rest).reduce((acc, [k, v]) => {
    return { ...acc, [k]: [...(out[id][k] || []), v] };
  }, out[id]);
  out[id] = { id, ...mergedProps };
  return out;
}, {});

console.log(result);

If you only care about the id and course fields, you can simplify to this:

const o = [
  { id: 1, course: 'name1', time: 'morning', topic: 'math' },
  { id: 1, course: 'name2', time: 'afternoon' },
  { id: 2, course: 'name3', time: 'evening' }
];

const result = o.reduce((out, { id, course }) =>
  ({ ...out, [id]: { id, course: [...((out[id] || {}).course || []), course] } })
, {});

console.log(result);

score 0 · Answer 5 · edited Mar 12 '19 at 12:59

0

You can use reduce to accumulate the results. Search in the current result (accumulator a) for an object (el) with the same id, if found, append course to existing object and return the same accumulator, otherwise put into the accumulator with course as an array.

var res = o.reduce((a, {id,course}) => {
    var found = a.find(el => el.id == id);
    return found ? found.course.push(course) && a : [...a, {id, course: [course]}];
}, []);

edited Mar 12 '19 at 12:59

Vadim

8,701
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answered Mar 12 '19 at 12:14

ttulka

10,309
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Please provide an explanation or some description as to why this would be the solution. – halfpastfour.am Mar 12 '19 at 12:50

AlexOwl · Answer 6 · 2019-03-12T12:24:56.080

function merge(array, key = 'id') {
  const obj = {}
  
  for(const item of array) {
    const existing = obj[item[key]]
    if(existing) {
      for(const [name, value] of Object.entries(item)) {
        if(name === key) continue;
        
        if(existing[name]) {
          existing[name] = [ ...(existing[name].$custom ? existing[name] : [existing[name]]), value ]
          existing[name].$custom = true;
        } else {
          existing[name] = value;          
        }
      }
    } else {
      obj[item[key]] = { ...item }
    }
  }
  
  return Object.values(obj)
}

var o = [
{
  "id": 1,
  "single": "test"
},
{
  "id": 1,
  "course": "name1",
  "multifield": "test"
}, 
{
  "id": 1,
  "course": "name2"
},
{
  "id": 1,
  "newfield": "test"
}, {
  "id": 2,
  "anotherid": "test",
  "array": [1,3,4]
}, {
  "id": 2,
  "array": "text"
}];

console.log(merge(o))

My solution benefits are 1. You can specify custom id key 2. Not merged properties are not arrays — AlexOwl, Mar 12 '19 at 12:18

How to reduce an array while merging one of it's field as well

6 Answers6