JS reducer: return combined array of objects

Question

I have such two arrays of objects:

const first = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: false}, {text: 'sometext3', applicable: true}];

const second = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: true}];

As a result I want to get such array:

const result = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: true}, {text: 'sometext3', applicable: true}];

so => just add to second array all non-existing items from first array, filtered by 'text' key.

is it possible to do via reducers? or maybe any better way?

Rakesh Makluri · Accepted Answer · 2019-01-28 12:22:04Z

1

Just iterate over first array and check whether each item present in second array or not, if not present then push in to second array.

first.forEach((item) => {
    const index = second.findIndex((st) => st.text === item.text);
    if(index < 0) {
        second.push(item);
    }
})

answered Jan 28, 2019 at 12:22

Rakesh Makluri

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3 Comments

byCoder Over a year ago

it's a dirty way :)

Rakesh Makluri Over a year ago

Map we can't use, as it'll return new array. Even if we use reduce on 'first' array and pass 'second' array as accumulator, you have to check each element is present in second array or not.

Rakesh Makluri Over a year ago

How about, first iterating over 'second' and storing text's in map, then iterate over 'first' array and check each item is present in map or not and then push. Which will avoid finding of elements every time. We need to traverse each array once. So complexity also less.

ellipsis · Accepted Answer · 2019-01-28 11:39:43Z

0

You can use maps

const first = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: false}, {text: 'sometext3', applicable: true}];

const second = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: true}];

var a=second.map((e)=>e.text);
first.map((e)=>{a.includes(e.text)?false:second.push(e)})
console.log(second);

answered Jan 28, 2019 at 11:39

ellipsis

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Comments

lakshman.pasala · Accepted Answer · 2019-01-28 11:45:35Z

0

You can use a filter to remove duplicates using a key.

function concatAndRemoveDuplicates(arr1, arr2) {
  return arr1.concat(
    arr2.filter((o) => !arr2.map((x) => x.text).includes(o.text)),
  );
}

answered Jan 28, 2019 at 11:45

lakshman.pasala

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1 Comment

byCoder Over a year ago

wrong, it should pick up all items like in my result, your solution will set second item with false applicable

Gabriele Petrioli · Accepted Answer · 2019-01-28 14:54:51Z

0

You could do

const first = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: false}, {text: 'sometext3', applicable: true}];

const second = [{text: 'sometext1', applicable: true}, {text: 'sometext2', applicable: true}];

const results = [...second, ...first.filter(({text:firstText})=>!second.find(({text})=>text===firstText))]

console.log(results);

but it does not clone the objects.

answered Jan 28, 2019 at 14:54

Gabriele Petrioli

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

As per your tags, I see that you have the possibility to use ES6 so you could do it with Set for example to have unique values in your array and then filter it:

const grouped = [...second, ...first];
const result = 
   [ ...new Set(grouped.map(({ text}) => text))]
   .map(text => grouped.find(obj => obj.text === text))

Tests

Here is a link with tests:

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jan 28, 2019 at 11:56

user1230234

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3 Comments

byCoder Over a year ago

not sure that it's working for 100% cases, sometimes it's taking wrong object, so applicable value is ovewrriten from first - what is wrong

byCoder Over a year ago

probably due to sorting order

user1230234 Over a year ago

@brabertaser19 Could you confirm it with examples? I will update the answer with tests. Please feel free to add your tests and point the issue

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