Re-framing array of objects from another array in javascript

Question

Want to create one array from another array of objects as below using javascript. arr is the input array and it should output as outArr

Input Array:

var arr = [
    {name: 'A', flag: 0},
    {email: 'B', flag: 0},
    {address: 'C', flag: 0},
    {name: 'A1', flag: 1},
    {email: 'B1', flag: 1},
    {address: 'C1', flag: 1}
]

Output Array:

var outArr = [
   {
     name: 'A',
     email: 'B',
     address: 'C'
   },
   {
     name: 'A1',
     email: 'B1',
     address: 'C1'
   }
]

What did you try and what problems are you facing?

dee-see
– dee-see

2019-01-20 17:44:08 +00:00
Commented Jan 20, 2019 at 17:44 — dee-see
– dee-see, Commented Jan 20, 2019 at 17:44
Are you asking how you would group by flag?

Jerinaw
– Jerinaw

2019-01-20 17:46:10 +00:00
Commented Jan 20, 2019 at 17:46 — Jerinaw
– Jerinaw, Commented Jan 20, 2019 at 17:46

Nina Scholz · Accepted Answer · 2019-01-20 17:51:42Z

2

You could get the rest of the object and add it to an object of the result array with the flag as index.

var array = [{ name: 'A', flag: 0 }, { email: 'B', flag: 0 }, { address: 'C', flag: 0 }, { name: 'A1', flag: 1 }, { email: 'B1', flag: 1 }, { address: 'C1', flag: 1 }],
    result = array.reduce((r, { flag, ...o }) => {
        Object.assign(r[flag] = r[flag] || {}, o);
        return r;
    }, []);

console.log(result);

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

answered Jan 20, 2019 at 17:51

Nina Scholz

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Comments

kemicofa · Accepted Answer · 2019-01-20 17:47:15Z

1

Using Array#reduce, destructuring, and spread syntax, Map and Map#values.

const data = [
    {name: 'A', flag: 0},
    {email: 'B', flag: 0},
    {address: 'C', flag: 0},
    {name: 'A1', flag: 1},
    {email: 'B1', flag: 1},
    {address: 'C1', flag: 1}
  ];
  
const res = data.reduce((a,{flag, ...rest})=>{
  return a.set(flag, {...a.get(flag), ...rest});
}, new Map()).values();

console.log([...res]);

answered Jan 20, 2019 at 17:47

kemicofa

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Comments

Code Maniac · Accepted Answer · 2019-01-20 17:50:15Z

0

Using normal javascript reduce you can do it like this.

var arr = [
    {name: 'A', flag: 0},
    {email: 'B', flag: 0},
    {address: 'C', flag: 0},
    {name: 'A1', flag: 1},
    {email: 'B1', flag: 1},
    {address: 'C1', flag: 1}
  ]
  
  
let output = Object.values( arr.reduce((op,cur)=>{
  let key = Object.keys(cur)[0]
  if(op[cur.flag]){
    op[cur.flag][key] = cur[key]
  } else {
    op[cur.flag] = {key:cur[key]}
  }
  return op;
},{}) )

console.log(output)

answered Jan 20, 2019 at 17:50

Code Maniac

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Comments

Ori Drori · Accepted Answer · 2019-01-20 17:54:34Z

Group the items by flag, then _.merge() the items, are remove flag using _.omit().

const arr = [{"name":"A","flag":0},{"email":"B","flag":0},{"address":"C","flag":0},{"name":"A1","flag":1},{"email":"B1","flag":1},{"address":"C1","flag":1}]

const fn = _.flow(
  arr => _.groupBy(arr, 'flag'),
  groups => _.map(groups, g => _.omit(_.merge({}, ...g), 'flag')),
)

const result = fn(arr)

console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

Or use lodash/fp to get a terser functional solution:

const arr = [{"name":"A","flag":0},{"email":"B","flag":0},{"address":"C","flag":0},{"name":"A1","flag":1},{"email":"B1","flag":1},{"address":"C1","flag":1}]

const fn = _.flow(
  _.groupBy('flag'),
  _.map(_.flow(
    _.mergeAll,
    _.omit('flag')
  )),
)

const result = fn(arr)

console.log(result)

<script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'></script>

Karolis Stakėnas · Accepted Answer · 2019-01-20 17:54:42Z

0

var arr = [
    {name: 'A', flag: 0},
    {email: 'B', flag: 0},
    {address: 'C', flag: 0},
    {name: 'A1', flag: 1},
    {email: 'B1', flag: 1},
    {address: 'C1', flag: 1}
  ]

var outputarray = []

for (var i = 0; i < arr.length; i++) {

  if ((i % 3 === 0)) {

    outputarray.push({
    name: arr[i].name,
    email: arr[i + 1].email,
    address: arr[i + 2].address
   })

  }

}

Heres a simple answer with a for loop.

answered Jan 20, 2019 at 17:54

Karolis Stakėnas

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