1 
let obj = [
    { id : 1 },
    { id : 10 },
    { brand : 12 },
    { id : 15 },
    { id : 18 },
    { image_link : 'some link' },
    { price : 10 },
    { brand : 111 },
    { image_link : 'some link 2' }
];

I have this array of object. I want filter this object so that I can get an object without duplicate keys.

I am trying this:

let uniqueIds = [];

let unique = obj.filter( (element ) => {
    
    let key = Object.keys(element)[0];                
    let isDuplicate = uniqueIds.includes(element.key);

    if (!isDuplicate) {
        uniqueIds.push(element.key);
        return true;
    }
    return false;   
});

console.log( unique )

But everytime it's showing me :

[ { id: 1 } ]

My expected output:

[
    { id : 18 },
    { price : 10 },
    { brand : 111 },
    { image_link : 'some link 2' }
];
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2
  • what is the expected output? Commented Oct 29, 2022 at 6:28
  • brnad is not brand Commented Oct 29, 2022 at 6:32

3 Answers 3

Reset to default
3

You can filter your array based on whether the object's key is the last occurrence of that key (checked using lastIndexOf) in the array:

let obj = [
    { id : 1 },
    { id : 10 },
    { brand : 12 },
    { id : 15 },
    { id : 18 },
    { image_link : 'some link' },
    { price : 10 },
    { brand : 111 },
    { image_link : 'some link 2' }
];

const keys = obj.map(o => Object.keys(o)[0])

const result = obj.filter((o, i) => keys.lastIndexOf(Object.keys(o)[0]) == i)

console.log(result)

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7 Comments

I have added my exptected output on my question. Can you check it please?
@Shibbir please see my edit, that matches your required output
@Shibbir good to hear! :)
one more help. Can I remove the keys which value is empty?
@Shibbir sorry, I misinterpreted what you wanted. I thought for that sample data you would want { id : 18 } in the output instead of { id : '' }, not to remove id completely. In that case you should add the filter on result instead i.e. const result = obj.filter((o, i) => keys.lastIndexOf(Object.keys(o)[0]) == i).filter(o => Object.values(o)[0])
|
1

The reason is that you need to use key instead of element.key

Note: it seems you have a typo brnad is not brand

let obj = [
    { id : 1 },
    { id : 10 },
    { brand : 12 },
    { id : 15 },
    { id : 18 },
    { image_link : 'some link' },
    { price : 10 },
    { brand : 111 },
    { image_link : 'some link 2' }
];

let uniqueIds = [];
let unique = obj.filter( (element ) => {
    
    let key = Object.keys(element)[0];                
    let isDuplicate = uniqueIds.includes(key);

    if (!isDuplicate) {
       uniqueIds.push(key);
       return true;
    }
    return false;   
});

console.log( unique )

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4 Comments

I have added my exptected output on my question. Can you check it please?
@Shibbir I change brnad to brand, you can check it now again
Can you please check my exptected output on my questions?
@Shibbir so you want to the latter key to override previous,in this case forEach might be a better choice or as Nick's answer
0

You have used element.key in two places it should be just key.

And since you need the final duplicate element you need to do a reverse of the input array and again do another reverse to unique array.

let obj = [
  { id : 1 },
  { id : 10 },
  { brand : 12 },
  { id : 15 },
  { id : 18 },
  { image_link : 'some link' },
  { price : 10 },
  { brand : 111 },
  { image_link : 'some link 2' }
];

let uniqueIds = [];

// reverse the input array before start filtering
obj.reverse()

let unique = obj.filter( (element ) => {
    
    let key = Object.keys(element)[0];                
    let isDuplicate = uniqueIds.includes(key);

    if (!isDuplicate) {
        uniqueIds.push(key);
        return true;
    }
    return false;   
});
// to make the output order according to the input array
unique.reverse()
// to make the input array to the initial order 
obj.reverse()

console.log(unique)

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1 Comment

id should be 18

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