How to merge values within array of objects

Question

I have an array with the following values:

let orders = [
{n: 0, orderId: null, lat: 49.396315, lng: 15.609607, type: "start"}, 
{n: 2.5, orderId: 26889, lat: 48.98951, lng: 14.45937, type: "odes"}, 
{n: 9, orderId: 26969, lat: 50.0823536, lng: 12.3791268, type: "odes"}, 
{n: 9, orderId: 26995, lat: 50.0823536, lng: 12.3791268, type: "odes"}, 
{n: 31, orderId: null, lat: 50.7343366, lng: 15.0501002, type: "end"}, 
{n: 31, orderId: 26969, lat: 50.7343366, lng: 15.0501002, type: "prij"}, 
{n: 31, orderId: 26995, lat: 50.7343366, lng: 15.0501002, type: "prij"},
];

and I would like to have following result:

[
{n: 0, orderId: null, lat: 49.396315, lng: 15.609607, type: "start"}, 
{n: 2.5, orderId: 26889, lat: 48.98951, lng: 14.45937, type: "odes"}, 
{n: 9, orderId: [26969,26995], lat: 50.0823536, lng: 12.3791268, type: ["odes", "odes"]}, 
{n: 31, orderId: [null,26969,26995], lat: 50.7343366, lng: 15.0501002, type: ["end", "prij", "prij"]}, 
]

Basically, I want to merge values if n, lat and long are the same, any ideas on how to elegantly do this?

Thanks in advance

You can just use a reducer and add to accumulator with the condition you have just mentioned. — Berk Kurkcuoglu
– Berk Kurkcuoglu, Commented Oct 27, 2020 at 15:22

Mister Jojo · Accepted Answer · 2020-10-28 02:56:13Z

Array reduce can do that:

let orders = 
    [ { n:  0,   orderId:  null, lat: 49.396315,  lng: 15.609607,  type: 'start' } 
    , { n:  2.5, orderId: 26889, lat: 48.98951,   lng: 14.45937,   type: 'odes'  } 
    , { n:  9,   orderId: 26969, lat: 50.0823536, lng: 12.3791268, type: 'odes'  } 
    , { n:  9,   orderId: 26995, lat: 50.0823536, lng: 12.3791268, type: 'odes'  } 
    , { n: 31,   orderId:  null, lat: 50.7343366, lng: 15.0501002, type: 'end'   } 
    , { n: 31,   orderId: 26969, lat: 50.7343366, lng: 15.0501002, type: 'prij'  } 
    , { n: 31,   orderId: 26995, lat: 50.7343366, lng: 15.0501002, type: 'prij'  } 
    ] 
    
let res = orders.reduce((a,c,i,t)=>
  {
  if ( i                                // is equivalent to  test (i> 0)
    && c.n   === t[i-1].n
    && c.lat === t[i-1].lat
    && c.lgn === t[i-1].lgn )
    {
    let el = a[a.length-1]              // last accumulator element
    if (!Array.isArray( el.orderId ))  
      {                                 // change elements to array type
      el.orderId = [ el.orderId ]
      el.type    = [ el.type ]
      }
    el.orderId.push(c.orderId)         // add new values on
    el.type.push(c.type)
    }
  else
    a.push({...c})   // this one is the first with the keys combinations
  return a
  }
  ,[])  // answer initialisation with an empty array
 
console.log( res )

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If you don't want to create a new array, but just "cleaning" your original one, do:

let orders = 
    [ { n:  0,   orderId:  null, lat: 49.396315,  lng: 15.609607,  type: 'start' } 
    , { n:  2.5, orderId: 26889, lat: 48.98951,   lng: 14.45937,   type: 'odes'  } 
    , { n:  9,   orderId: 26969, lat: 50.0823536, lng: 12.3791268, type: 'odes'  } 
    , { n:  9,   orderId: 26995, lat: 50.0823536, lng: 12.3791268, type: 'odes'  } 
    , { n: 31,   orderId:  null, lat: 50.7343366, lng: 15.0501002, type: 'end'   } 
    , { n: 31,   orderId: 26969, lat: 50.7343366, lng: 15.0501002, type: 'prij'  } 
    , { n: 31,   orderId: 26995, lat: 50.7343366, lng: 15.0501002, type: 'prij'  } 
    ] 

for (let i=orders.length; (--i>0); )  // for i = orders.length-1 to 1 
  {
  let el_c = orders[i]      // pointers on current 
    , el_p = orders[i-1]   // and previous array elements
    ;
  if ( el_c.n   === el_p.n
    && el_c.lat === el_p.lat
    && el_c.lng === el_p.lng )    // if same keys
    {
    el_p.orderId = [ el_p.orderId, el_c.orderId ].flat() 
    el_p.type    = [ el_p.type,    el_c.type    ].flat()
    orders.splice(i,1)    // remove duplicate
  } } 
  
console.log( orders )

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Thanks a lot, really elegant solution! Just in the first code section, you haven't declared the el ;)
@kybernaut.cz thank you! I was also happy to have had this inspiration for the second solution :)

Terry · Accepted Answer · 2020-10-27 15:27:48Z

You can take advantage of Array.prototype.reduce, but some tweaking of the internal logic is required since the grouped properties can be string/number or an array thereof.

In the reduce callback, you can attempt to search for an existing entry by the grouping criteria, which you say it should be n, lat and lng. If an entry is not found, you simply push the current item into your accumulator.

If an entry is found, then you need to perform some logic to convert the pre-existing string/number value into an array, and then append a new value to it.

See proof-of-concept code below:

let orders = [
  {n: 0, orderId: null, lat: 49.396315, lng: 15.609607, type: "start"}, 
  {n: 2.5, orderId: 26889, lat: 48.98951, lng: 14.45937, type: "odes"}, 
  {n: 9, orderId: 26969, lat: 50.0823536, lng: 12.3791268, type: "odes"}, 
  {n: 9, orderId: 26995, lat: 50.0823536, lng: 12.3791268, type: "odes"}, 
  {n: 31, orderId: null, lat: 50.7343366, lng: 15.0501002, type: "end"}, 
  {n: 31, orderId: 26969, lat: 50.7343366, lng: 15.0501002, type: "prij"}, 
  {n: 31, orderId: 26995, lat: 50.7343366, lng: 15.0501002, type: "prij"},
];

// Receives a maybe array and pushes a new entry into it
function pushNewEntry(target, entry) {
  if (!Array.isArray(target)) {
    return [target, entry];
  } else {
    target.push(entry);
    return target;
  }
}

const groupedOrders = orders.reduce((acc, cur) => {
  const { n, orderId, lat, lng, type } = cur;

  // Find existing entry based on matching `n`, `lat`, and `lng`
  const existingEntry = acc.find(x => {
    return x.n === n && x.lat === lat && x.lng === lng;
  });

  if (!existingEntry) {
    acc.push(cur);
  } else {
    existingEntry.orderId = pushNewEntry(existingEntry.orderId, orderId);
    existingEntry.type = pushNewEntry(existingEntry.type, type);
  }
  
  return acc;
}, []);

console.log(groupedOrders);

Thanks for your help, I've picked the shortest answer as the most elegant, but your solution works nicely as well. Thanks for opening my eyes to the solution!

farvilain · Accepted Answer · 2020-10-27 15:31:47Z

1

A simple naive solution consist of using a combined key, like ${n}-${lat}-${lng}. I added '-' separator as it seems to be absent of n/lat/lng.

const results = {};

orders.forEach( elem => {
    const key = `${elem.n}-${elem.lat}-${elem.lng}`;

    //Already present, just add type and orderId
    //Values can be duplicate, but you don't talk about that
    if ( result[key] ) {
        results[key].type.push( elem.type );
        results[key].orderId.push( elem.orderId );
    } else {
        results[key] = {
            n: elem.n,
            orderId: [elem.orderId],
            lat: elem.lat,
            lng: elem.lng,
            type: [elem.type]
        };
    }
});

Using the .reduce method of Array could be better, you should give a look.

answered Oct 27, 2020 at 15:31

farvilain

2,5622 gold badges15 silver badges24 bronze badges

1 Comment

Karolína Vyskočilová Over a year ago

Thanks for your help, I've picked the shortest answer as the most elegant, but your solution works nicely as well. Thanks for opening my eyes to the solution!

Shashank Garg · Accepted Answer · 2020-10-27 15:39:56Z

1

let orders = [
  { n: 0, orderId: null, lat: 49.396315, lng: 15.609607, type: "start" },
  { n: 2.5, orderId: 26889, lat: 48.98951, lng: 14.45937, type: "odes" },
  { n: 9, orderId: 26969, lat: 50.0823536, lng: 12.3791268, type: "odes" },
  { n: 9, orderId: 26995, lat: 50.0823536, lng: 12.3791268, type: "odes" },
  { n: 31, orderId: null, lat: 50.7343366, lng: 15.0501002, type: "end" },
  { n: 31, orderId: 26969, lat: 50.7343366, lng: 15.0501002, type: "prij" },
  { n: 31, orderId: 26995, lat: 50.7343366, lng: 15.0501002, type: "prij" }
];

let map = new Map();

orders.forEach((record) => {
  let key = `n: ${record.n}, lat: ${record.lat}, lng: ${record.lng}`;

  if (map.has(key)) {
    let data = map.get(key);
    data.push(record);
    map.set(key, data);
  } else {
    map.set(key, [record]);
  }
});

const nOrders = [];

map.forEach((value, key) => {
  nOrders.push({
    n: value[0].n,
    lat: value[0].lat,
    lng: value[0].lng,
    orderId:
      value.length === 1
        ? value[0].orderId
        : value.map((entry) => entry.orderId),
    type: value.length === 1 ? value[0].type : value.map((entry) => entry.type)
  });
});

console.log(nOrders);

answered Oct 27, 2020 at 15:39

Shashank Garg

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1 Comment

Karolína Vyskočilová Over a year ago

Thanks for your help, I've picked the shortest answer as the most elegant, but your solution works nicely as well. Thanks for opening my eyes to the solution!

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