Javascript - Remove duplicate ID from array of objects

Question

Take the following two arrays:

const array1 = [
  {
    props: {
      type : 'text',
      id : 'item1',
      name : 'item1',
      value : '@item1@',
    },
  },
  {
    props: {
      type: 'hidden',
      id: 'item2',
      name: 'item2',
      value: '@item2@',
    },
  }
];

const array2 = [
  {
    props: {
      type: 'hidden',
      id: 'item1',
      name: 'item1',
      value: '@item1@',
    },
  }
];

What I'm trying to do is concatenate them into a single array, and remove any duplicates based on the id property. However the caveat here is that the object that does NOT have a type of hidden must have presidence.

So I should basically be left with the contents of array1, as the duplicate item from array2 has the type value of hidden, like so:

// Result
[
  {
    props: {
      type : 'text', // Note the type here is "text"
      id : 'item1',
      name : 'item1',
      value : '@item1@',
    },
  },
  {
    props: {
      type: 'hidden',
      id: 'item2',
      name: 'item2',
      value: '@item2@',
    },
  }
];

I can easily concatenate them using:

const array = array1.concat(array2);

My idea was to then use a Filter but I'm having a bit of a brain melt. Here is what I have come up with so far:

const concat = (array1, array2) => {
  const array = array1.concat(array2);
  const ids = [];

  // Create array of ID's
  for (const i in array1) {
    ids.push(array1[i].props.id);
  }

  return array.filter((obj) => {
    if (obj.props.type !== 'hidden' && ids.includes(obj.props.id)) {
      return true;
    }

    return false;
  });
};

Would using a Reduce be a better approach here?

Here is a JSFiddle of what I have so far: https://jsfiddle.net/64uprbhn/

Do you want to discard all type: "hidden" or always keep them? I'm a bit confused because I don't think you mistyped "must have presidence", sounds like you meant to say "must have precedence". But I want to double check. — VLAZ
– VLAZ, Commented Oct 8, 2019 at 13:07
@VLAZ Let's say 2 objects contain the same ID (item1)....one of the object's had type:hidden and the other is type:xxx. The one that has type:hidden should be the one to be removed as the duplicate — Lodder
– Lodder, Commented Oct 8, 2019 at 13:15

georg · Accepted Answer · 2019-10-08 13:47:30Z

5

You can

Create a Map of id => object from the first array
Go over the second array and either
- add the item if not already in the map
- check if the current item is not type: "hidden" and overwrite the other one
- otherwise discard the item
Create a new array from the Map

const array1 = [
  {
    props: {
      type : 'text',
      id : 'item1',
      name : 'item1',
      value : '@item1@',
    },
  },
  {
    props: {
      type: 'hidden',
      id: 'item2',
      name: 'item2',
      value: '@item2@',
    },
  },
  {
    props: {
      type: 'hidden',
      id: 'item3',
      name: 'item3',
      value: '@item3@',
    },
  }
];

const array2 = [
  {
    props: {
      type: 'hidden',
      id: 'item1',
      name: 'item1',
      value: '@item1@',
    },
  },
  {
    props: {
      type: 'text',
      id: 'item3',
      name: 'item3',
      value: '@item3@',
    },
  }
];

//1. create a map from id => object pairs
const map = new Map(array1.map(obj => [obj.props.id, obj]));

//2. go over the second array
array2.forEach(obj => {
  if (!map.has(obj.props.id) //not a duplicate
    || obj.props.type !== "hidden") { //is not hidden
    //insert
    map.set(obj.props.id, obj);
  }
});

//3. convert back into array
const merged = [...map.values()];

console.log(merged);

For the record, you can basically do the same (or pretty similar) thing using .filter but you'll have to do a O(n) lookup for each item. A Map ensures much faster lookups.

edited Oct 8, 2019 at 13:47

georg

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answered Oct 8, 2019 at 13:18

VLAZ

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1 Comment

Lodder Over a year ago

Legend! Thank you so much. I didn't even think of using a Map

VLAZ · Accepted Answer · 2019-10-08 13:30:59Z

0

const array1 = [
  {
    props: {
      type : 'text',
      id : 'item1',
      name : 'item1',
      value : '@item1@',
    },
  },
  {
    props: {
      type: 'hidden',
      id: 'item2',
      name: 'item2',
      value: '@item2@',
    },
  }
];

const array2 = [
  {
    props: {
      type: 'hidden',
      id: 'item1',
      name: 'item1',
      value: '@item1@',
    },
  }
];

function getUnique(arr, comp) {

  const unique = arr
    .map(e => e[comp])

    // store the keys of the unique objects
    .map((e, i, final) => final.indexOf(e) === i && i)

    // eliminate the dead keys & store unique objects
    .filter(e => arr[e]).map(e => arr[e]);

  return unique;
}

const arr = array1.concat(array2);

console.log(getUnique(arr, 'id'));

edited Oct 8, 2019 at 13:30

VLAZ

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answered Oct 8, 2019 at 13:09

Anton Skybun

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1 Comment

Lodder Over a year ago

Thank you, but this doesn't preserve items that do not have a duplicate

maninak · Accepted Answer · 2022-05-12 13:52:30Z

0

Here's what worked for me

const someItems = [{ id: 1 }, { id: 2 }, { id: 1 }]

function getUniqueItems(items) {
  const uniqueIds = new Set(items.map((item) => item.id))

  const itemsWithUniqueIds = [...uniqueIds].map(id => items.find(item => item.id === id)).filter(Boolean)

  return itemsWithUniqueIds
}

console.log(getUniqueItems(someItems))

If you are using TypeScript, TS will complain about the .filter(Boolean). In that case just replace Boolean with (item: any | undefined): item is any => Boolean(item). Of course, you can then go ahead and also replace any with whichever type makes sense in your case.

answered May 12, 2022 at 13:52

maninak

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