Remove duplicates with max performance

Question

Has someone an idea how the function have to look like if I want the following results:

In:

{
  3: [ "1-2", "4-5" ],
  4: [ "1-2", "6-9" ]
}

Out:

{
  3: [ "4-5" ],
  4: [ "1-2", "6-9" ]
}

I want to remove all duplicate values and leave the last occurrence in the object. If there are no elements left in a property, it should be removed. I tried the suggestions from https://stackoverflow.com/a/9229821/3852382 but it does not solve my problem.

"last" has no meaning in an object: object properties are unordered. — trincot, Aug 12 '17 at 14:53
@trincot I mean the last occurrence. I know object properties don't have an order. — iPaat, Aug 12 '17 at 14:55
OK, so you would be fine if the output was `{ 3: ["1-2", "4-5"], 4: ["6-9"] }` then? — trincot, Aug 12 '17 at 14:57
@D-reaper: I would do some iterations with the algorithm with big objects. — iPaat, Aug 12 '17 at 14:59
@trincot If the input was `{ 4: ["1-2", "6-9"], 3: ["1-2", "4-5"] }`. After the object is cleaned, I want to iterate over it to do some replacements within a string. If it would not an object would it be easier then? — iPaat, Aug 12 '17 at 15:01
Your object is not "organised" in any sense of the word so "If the input was `{ 4: ["1-2", "6-9"], 3: ["1-2", "4-5"] }`" makes no sense. @trincot's question still stands. — Andy, Aug 12 '17 at 15:08
Okay. If I have an object and iterate over it, something comes first and something comes last. I don't know how JavaScript decides which comes first and which comes last. But I would have the one, that JavaScript decides to be the last, the occurrence which remains. Or is that what u want to say, that if I iterate two times over an object, it is not clear which comes first and which comes last? In this case, what should I do to "organize" it? — iPaat, Aug 12 '17 at 15:12
It may differ from one JS engine to another. You shouldn't base your code on that because you may get the wrong result. — Andy, Aug 12 '17 at 15:19
@Andy I got a solution which works and solve my problem. But I have to think about what u said. So what do you suggest to do? — iPaat, Aug 12 '17 at 15:21
you can always manipulate the object by adding to an array, sorting it like you expect it to behave, and then perform operation on the orignal object using this sorted array as a reference — marvel308, Aug 12 '17 at 15:24
@marvel308 Would you please add an example to your answer as "best practice"? Let's say that the higher property number is the one which should remain? — iPaat, Aug 12 '17 at 15:32
check the first answer in https://stackoverflow.com/questions/5467129/sort-javascript-object-by-key, I'll add it in a while — marvel308, Aug 12 '17 at 15:40
In mainstream JS engines, "integer" properties mainly get listed in a sorted fashion regardless of in what order they are added. — Redu, Aug 12 '17 at 22:31

score 0 · Accepted Answer · answered Aug 12 '17 at 15:15

0

I think you can do this in the following way

let obj = {
  3: [ "1-2", "4-5" ],
  4: [ "1-2", "6-9" ]
};

let reverseMap = {};
for(property in obj){
 for(let element of obj[property]){
  if(reverseMap[element] == undefined || reverseMap[element] == null){
   ;
  }
  else {
   obj[reverseMap[element]].splice(obj[reverseMap[element]].indexOf(element), 1);
  }
  reverseMap[element] = property;
 }
}

console.log(obj);

answered Aug 12 '17 at 15:15

marvel308

10,288
1
21
32

That is exactly what I mean! Thanks a lot! – iPaat Aug 12 '17 at 15:19

score 0 · Answer 2 · answered Aug 12 '17 at 22:13

You may do as follows;

var obj = {3: [ "1-2", "4-5" ], 4: [ "1-2", "6-9" ]},
    res = Object.keys(obj)
                .reduceRight((m,k) => obj[k].reduce((t,e) => t[e] ? t 
                                                                  : (t[e] = true, (t.r[k] = t.r[k] ? t.r[k].concat(e)
                                                                                                   : [e]), t), m), {r:{}}).r;
console.log(res);

Remove duplicates with max performance

2 Answers2