How to remove duplicates in subarrays?

Question

It is not a duplicate! The link above doesn't show how to delete all other subarrays values by index of the third subarray (see expected array).

This code finds duplicates:

    var array = [];
    
    array[0] = [0, "1:7001", "1:7002", "1:7003", "1:7004", "1:7005"];
    array[1] = [1, "1:8001", "1:8002", "1:8003", "1:8004", "1:8005"];
    array[2] = ["asw", 32884789084, 32919691801, 32919691801, 32919691801, 32857764298];
    
    var dup = array[2].filter((v, i, a) => a.indexOf(v) !== i);
    
    console.log(dup);

But I don't know how to delete elements in subarrays. Expected array should look like:

[0, "1:7001", "1:7002", "1:7005"]; 
[1, "1:8001", "1:8002", "1:8005"];
["asw", 32884789084, 32919691801, 32857764298];

@brk there are duplicates in the 3rd subarray. check properly. — Code Maniac, Dec 15 '18 at 12:53
You want to delete duplicate value from third array only, m i right? — Chandan Kumar, Dec 15 '18 at 12:55

Code Maniac · Answer 1 · 2018-12-18T08:21:25.980

2

You can do it like this with help of map and Set

So with map here i am iterating each element of array variable. and than making a set out of each subarray (which means it will have only unique items). and before returning i am using spread operator to de-structure.

var array = [];
    
    array[0] = [0, "1:7001", "1:7002", "1:7003", "1:7004", "1:7005"];
    array[1] = [1, "1:8001", "1:8002", "1:8003", "1:8004", "1:8005"];
    array[2] = ["asw", 32884789084, 32919691801, 32919691801, 32919691801, 32857764298];
    
    
    var dup = array.map(e=>[...new Set(e)])
    
    console.log(dup);

If you want to remove elements from other subarray based on 3rd subarray you can do it like this.

var array = [];
     
array[0] = [0, "1:7001", "1:7002", "1:7003", "1:7004", "1:7005"];
array[1] = [1, "1:8001", "1:8002", "1:8003", "1:8004", "1:8005"];
array[2] = ["asw", 32884789084, 32919691801, 32919691801, 32919691801, 32857764298];
        
let op = array.map((e,ind,a)=>{
  return e.filter((el,i)=>{
    return  i == 0 ? true : a[2][i] !== a[2][i-1]
  })
})
        
        console.log(op);

edited Dec 18 '18 at 08:21

answered Dec 15 '18 at 12:53

Code Maniac

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Could you explain a bit how it works? By the way it's not removing elements from first and second subarrays. – F. Vosnim Dec 15 '18 at 12:54
@F.Vosnim check the updated one and i don't see any duplicates in the first and second one. – Code Maniac Dec 15 '18 at 12:59
Delete all other subarrays values by index of the third subarray (see expected array) – F. Vosnim Dec 15 '18 at 13:00
ES6 Style :: const arr = ["asw", 32884789084, 32919691801, 32919691801, 32919691801, 32857764298]; let s = new Set(arr) let it = s.values() console.log(Array.from(it)) – Chandan Kumar Dec 15 '18 at 13:01
@Chandan Kumar, thank you. But it's not helping to delete other subarrays values. – F. Vosnim Dec 15 '18 at 13:08
You can put this logic inside for long and change arr value – Chandan Kumar Dec 15 '18 at 13:11
@F.Vosnim check the updated answer. – Code Maniac Dec 15 '18 at 13:12
@Code Maniac, thank you very much! But it's keeping last element, not the first – F. Vosnim Dec 15 '18 at 13:14
@F.Vosnim oh yeah. now it is keeping :) – Code Maniac Dec 15 '18 at 13:43
@Code Maniac, I found that if `array[2] = ["a","b","c","c","a","b","e"]` it removes only `c` and does not remove second `"a","b"`. – F. Vosnim Dec 16 '18 at 16:11

How to remove duplicates in subarrays?

1 Answers1