Get repeating objects from object array

Question

I have this initial array and want to extract the repeating flights based on from.flightNo and to.flightNo

const myArray = [
        { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },
        { from: { flightNo: 'A12' }, to: { flightNo: 'B45' }, code: 22 },
        { from: { flightNo: 'A12' }, to: { flightNo: 'B52' }, code: 21 },
        { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },
        { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },
        { from: { flightNo: 'A15' }, to: { flightNo: 'B69' }, code: 24 },
        { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },
      ];

  result: 
 [
    { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },
    { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },
    { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },
    { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },
  ]

I wrote this but I can only get the first repeating one and doesn't look very pretty.

  const duplicates = myArray
    .map((item) => {
      let count = 0;
      for (let i = 0; i < myArray.length; i++) {
        if (count > 1) {
          count = 0;
          return item;
        }
        if (
          myArray[i]?.from?.flightNo == item?.from?.flightNo &&
          myArray[i]?.to?.flightNo == item?.to?.flightNo
        )
          count++;
      }
    })
    .filter((notUndefined) => notUndefined !== undefined);

Any suggestions?

Som Shekhar Mukherjee · Accepted Answer · 2021-04-08 19:33:13Z

2

Create a Map out of the array whose key is from.flightNo + to.flightNo and value is an array of all the objects that match this from.flightNo + to.flightNo
Keep only the values that have length greater than 1.

const myArray = [
  { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },
  { from: { flightNo: 'A12' }, to: { flightNo: 'B45' }, code: 22 },
  { from: { flightNo: 'A12' }, to: { flightNo: 'B52' }, code: 21 },
  { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },
  { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },
  { from: { flightNo: 'A15' }, to: { flightNo: 'B69' }, code: 24 },
  { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },
];

console.log(
  Array.from(
    myArray.reduce(
      (m, o) =>
        m.set(
          o.from.flightNo + o.to.flightNo,
          (m.get(o.from.flightNo + o.to.flightNo) || []).concat(o)
        ),
      new Map()
    )
  ).reduce((a, [k, v]) => (v.length > 1 ? a.concat(v) : a), [])
);

Based on the discussions below in the comment I've updated my code to the following:

const myArray = [
  { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },
  { from: { flightNo: 'A12' }, to: { flightNo: 'B45' }, code: 22 },
  { from: { flightNo: 'A12' }, to: { flightNo: 'B52' }, code: 21 },
  { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },
  { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },
  { from: { flightNo: 'A15' }, to: { flightNo: 'B69' }, code: 24 },
  { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },
];

const result = Array.from(
  myArray.reduce(
    (m, o) => (
      m.has(o.from.flightNo + "-" + o.to.flightNo)
        ? m.get(o.from.flightNo + "-" + o.to.flightNo).push(o)
        : m.set(o.from.flightNo + "-" + o.to.flightNo, [o]),
      m
    ),
    new Map()
  )
)
  .reduce((a, [k, v]) => (v.length > 1 && a.push(v), a), [])
  .flat();

console.log(result);

edited Apr 8, 2021 at 19:33

answered Apr 8, 2021 at 15:51

Som Shekhar Mukherjee

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12 Comments

trincot Over a year ago

Nice. Same comment about + as I made for Ravikumar's answer. For the rest it is very functional programming style (+1). Downside is that arrays are copied over and over again, and this can give a worst case time complexity of O(n²) -- when all flights have the same pair of codes.

Som Shekhar Mukherjee Over a year ago

@trincot I didn't get how is it an O(n2) solution. myArray should be reduced to a Map in O(n) time then Array.from() that Map should also take O(n) time and the final reduce should also take O(n) time. And all these happen independently so there's no nesting, so why do you say its O(n2)? Please help!

Som Shekhar Mukherjee Over a year ago

Oh you mean to say when I spread the array, that is where the problem lies, right @trincot?

trincot Over a year ago

That is really engine dependent. I just ran that second snippet there on FF, and concat turns out to be faster. Most important though is that they both have a O(n+m) time complexity (m for the second array, if used to join two arrays)

trincot Over a year ago

Yes, with push you resolve the complexity issue. However, the comment about adding the two codes (that I made to another answer) still holds, and (partly?) explains why your code runs faster, since in my code I took precautions against such code mixup.

|

trincot · Accepted Answer · 2021-04-08 15:29:46Z

2

You can create a map, keyed by the flight number pairs (and use JSON.stringify to stringify that pair), with an empty array for each key. Then iterate the data to populate those arrays.

Finally extract those sub arrays that have more than one element, and flatten the result:

const myArray = [{ from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },{ from: { flightNo: 'A12' }, to: { flightNo: 'B45' }, code: 22 },{ from: { flightNo: 'A12' }, to: { flightNo: 'B52' }, code: 21 },{ from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },{ from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },{ from: { flightNo: 'A15' }, to: { flightNo: 'B69' }, code: 24 },{ from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },];

let map = new Map(myArray.map(o => [JSON.stringify([o.from.flightNo, o.to.flightNo]), []]));
for (let o of myArray) map.get(JSON.stringify([o.from.flightNo, o.to.flightNo])).push(o);
let result = [...map.values()].filter(arr => arr.length > 1).flat();

console.log(result);

answered Apr 8, 2021 at 15:29

trincot

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3 Comments

Jamiec Over a year ago

I dont think its just the flight numbers - the code feeds into the duplicate too

trincot Over a year ago

I verified the output, @Jamiec, and it is identical to what the OP expects. Did I miss something?

Som Shekhar Mukherjee Over a year ago

@trincot If you could spare some time and review my solution, it would be really helpful.

Ravikumar · Accepted Answer · 2021-04-09 07:18:50Z

1

You can achieve this by grouping all the elements by unique key then filter out unnecessary elements by group length

const myArray = [
        { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },
        { from: { flightNo: 'A12' }, to: { flightNo: 'B45' }, code: 22 },
        { from: { flightNo: 'A12' }, to: { flightNo: 'B52' }, code: 21 },
        { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },
        { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },
        { from: { flightNo: 'A15' }, to: { flightNo: 'B69' }, code: 24 },
        { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },
      ];
      
      
      function getDuplicates(myArray) {
      //grouping by unique key
       return Object.values(myArray.reduce((acc, r) => {
            const key = r.from.flightNo+r.to.flightNo;
            if (!acc[key]) {
              acc[key] = [];
            }
            acc[key].push(r);
            return acc;
      }, {}))
      // merging all duplicated items
      .reduce((acc, list) => list.length > 1 ? acc.concat(list) : acc, []);
      }
      
      console.log(getDuplicates(myArray));

edited Apr 9, 2021 at 7:18

answered Apr 8, 2021 at 15:39

Ravikumar

2,20516 silver badges18 bronze badges

4 Comments

trincot Over a year ago

Be careful here: if one flight is from "A1B" to "2C3" and another is from "A1" to "B2C3", then they will be considered the same.

Ravikumar Over a year ago

I don't think, as @arled clearly mentioned that he need duplicated flights exactly matching from and to with same flightNo with the same order. Agreed, need to be careful (rewrite the matching logic) if what you said is the criteria.

Som Shekhar Mukherjee Over a year ago

@Ravikumar You can simply add some special character between the strings like "A1B" + "&" + "2C3".

Ravikumar Over a year ago

@SomShekharMukherjee why do you want that? do you see any additional gain out of it?

Wimanicesir · Accepted Answer · 2021-04-08 15:28:23Z

0

You could loop your flights and make a new code with the from and to flight numbers. Then check this string if it's already added. If not, add to result.

const myArray = [
        { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 23 },
        { from: { flightNo: 'A12' }, to: { flightNo: 'B45' }, code: 22 },
        { from: { flightNo: 'A12' }, to: { flightNo: 'B52' }, code: 21 },
        { from: { flightNo: 'A11' }, to: { flightNo: 'B11' }, code: 20 },
        { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 25 },
        { from: { flightNo: 'A15' }, to: { flightNo: 'B69' }, code: 24 },
        { from: { flightNo: 'A14' }, to: { flightNo: 'B44' }, code: 26 },
];

var codes = [];
var result = [];

myArray.forEach(function (flight) {
  var newCode = flight.from.flightNo + flight.to.flightNo;
  if (codes.indexOf(newCode) === -1) {
    codes.push(newCode);
   }
  else {
   result.push(flight)
   }


});

console.log(result)

answered Apr 8, 2021 at 15:28

Wimanicesir

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1 Comment

trincot Over a year ago

I didn't downvote, but the OP expects 4 results, not 2. Secondly, indexOf represents a scan over an array, which makes the time complexity O(n²), while O(n) is possible. Still, I don't think that is worth a downvote, so +1

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