How can I check if the array of objects has duplicate property values and get the last value that is repeated?

Question

I want to browse an array to check if the elements are repeated and then recover the last element .

const tabs = [
  { name: 'john', details:"john1" },
  { name: 'michael' , details:"michael1" },
  { name: 'beris' , details:"beris1" },
  { name: 'john'  , details:"john2"},
  { name: 'john'  , details:"john3"},
  { name: 'beris' , details:"beris2" },
   { name: 'beris' , details:"beris3" },
]

I would like to have as result

const tabs = [
   { name: 'john'  , details:"john3"},
    { name: 'beris' , details:"beris3" },
]

What have you tried so far? Hint: a map/dictionary would be helpful. — Domenico De Felice
– Domenico De Felice, Commented Jun 17, 2020 at 9:19
Why isn't michael in the result? because it's not repeated? — Kaddath
– Kaddath, Commented Jun 17, 2020 at 9:19

Sagar Kulkarni · Accepted Answer · 2020-06-17 09:41:22Z

1

const tabs = [
  { name: 'john', details:"john1" },
  { name: 'michael' , details:"michael1" },
  { name: 'beris' , details:"beris1" },
  { name: 'john'  , details:"john2"},
  { name: 'john'  , details:"john3"},
  { name: 'beris' , details:"beris2" },
  { name: 'beris' , details:"beris3" },
];

// logic to create newTabs array having only latest duplicates

let names=[];
let newTabs=[];

tabs.forEach((element)=>{
	if(names.includes(element.name)){
	newTabs = newTabs.filter(function( obj ){
		return obj.name !== element.name;
		});
	newTabs.push(element);
  }else{
	names.push(element.name);
  }
});

// output newTabs array that has only latest duplicate objects!
console.log(newTabs);

edited Jun 17, 2020 at 9:41

answered Jun 17, 2020 at 9:32

Sagar Kulkarni

5356 silver badges16 bronze badges

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Comments

Nick Parsons · Accepted Answer · 2020-06-17 09:38:54Z

You can group objects with the same name into an array using a Map, where the key of the array is the name of the objects it stores. Then you can use Array.from() to convert the values from the map into an array, which you can then filter to only contain arrays where the occurrence is greater than one. For each array, you can then .map() it to get the last element:

const tabs = [{ name: 'john', details:"john1" }, { name: 'michael' , details:"michael1" }, { name: 'beris' , details:"beris1" }, { name: 'john'  , details:"john2"}, { name: 'john'  , details:"john3"}, { name: 'beris' , details:"beris2" }, { name: 'beris' , details:"beris3" },];

const res = Array.from(tabs.reduce((m, o) => 
  m.set(o.name, (m.get(o.name)|| []).concat(o))
, new Map).values()).filter(({length}) => length > 1).map(vals => vals.pop());
console.log(res);

Moritz Roessler · Accepted Answer · 2020-06-17 09:42:36Z

1

You can group your items by name and take the last element of the group if it's longer than 2.

const lkp = tabs.reduce((acc, cur) => {
  acc[cur.name] = acc[cur.name] || [];
  acc[cur.name].push(cur); 
  return acc;
}, {});
const filtered = Object.values(lkp).reduce((acc, cur) => {
    if (cur.length > 1) acc.push(cur.pop());
    return acc;
}, [])
console.log (filtered)

<script>
const tabs = [
  { name: 'john', details:"john1" },
  { name: 'michael' , details:"michael1" },
  { name: 'beris' , details:"beris1" },
  { name: 'john'  , details:"john2"},
  { name: 'john'  , details:"john3"},
  { name: 'beris' , details:"beris2" },
   { name: 'beris' , details:"beris3" },
]
</script>

answered Jun 17, 2020 at 9:42

Moritz Roessler

8,67329 silver badges57 bronze badges

1 Comment

spiritualProgrammer Over a year ago

it also works for the method but I have not yet read the reduce () method, thank you too.

Tanmoy Majumdar · Accepted Answer · 2020-06-17 09:29:56Z

0

you can try the following

var map = {}
tabs.forEach(item => map[item.name] = item)
const result = Object.values(map)

edited Jun 17, 2020 at 9:29

answered Jun 17, 2020 at 9:23

Tanmoy Majumdar

4583 silver badges7 bronze badges

1 Comment

spiritualProgrammer Over a year ago

it returns an element that does not repeat (michael).

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