How to compare two JS array and create new JS array using changes

Question

I have two JS arrays. They are arr1 and arr2. I want to create another array called arr3. Now it wants to compare arr1 and arr2 and insert all elements which arr2 contains and arr1 not contain.

Example:

arr1:

  0:
    des: "cont1"
    note: "cont1"
    pro_code: "XXY"
  1:
    des: "cont2"
    note: "cont2"
    pro_code: "NNB"
  2:
    des: "cont4"
    note: "cont4"
    pro_code: "QQA"
  3:
    des: "cont5"
    note: "cont5"
    pro_code: "GFD"

arr2:

  0:
    des: "cont1"
    note: "cont1"
    pro_code: "XXY"
  1:
    des: "cont2"
    note: "cont2"
    pro_code: "NNB"
  2:
    des: "cont3"
    note: "cont3"
    pro_code: "QAS"

In arr2 contains pro_code: QAS. But it is not in arr1. So it should be contained in arr3.

In arr1 contains arr1[4] pro_code: GFD. it should not be contained in arr3. Becase aar1 can contain additional elements.

Here arr2 must contain elements that are in arr1. There can not additional elements in arr2.

Expected output:

const arr3 = [{
  des: "cont3",
  note: "cont3",
  pro_code: "QAS"
}

My tried code, not workes. Please help me to solve this.

@NinaScholz I updated the question with Expected output: I am using this in vue computed: there some() is not working. — user_v12
– user_v12, Commented Oct 19, 2020 at 8:14

Nina Scholz · Accepted Answer · 2020-10-19 08:25:08Z

2

You could take a Set for all pro_code values and filter the second array by checking if the value is not in the set.

const
    array1 = [{ des: "cont1", note: "cont1", pro_code: "XXY" }, { des: "cont2", note: "cont2", pro_code: "NNB" }, { des: "cont4", note: "cont4", pro_code: "QQA" }, { des: "cont5", note: "cont5", pro_code: "GFD" }],
    array2 = [{ des: "cont1", note: "cont1", pro_code: "XXY" }, { des: "cont2", note: "cont2", pro_code: "NNB" }, { des: "cont3", note: "cont3", pro_code: "QAS" }],
    pro_codes = new Set(array1.map(({ pro_code }) => pro_code)),
    result = array2.filter(({ pro_code }) => !pro_codes.has(pro_code));

console.log(result);

answered Oct 19, 2020 at 8:25

Nina Scholz

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Comments

Gabriele Petrioli · Accepted Answer · 2020-10-19 09:31:25Z

1

In short, your requirement is to filter the arr2 and keep elements that are not in arr1.

const arr3 = arr2.filter(function(arr2item) {
  // only keep this item if it is not in arr1
  return !arr1.some(function(arr1item) {
    return arr1item.pro_code === arr2item.pro_code;
  })
});

const arr1 = [{des: "cont1", note: "cont1", pro_code: "XXY"}, {des: "cont2", note: "cont2", pro_code: "NNB"}, {des: "cont4", note: "cont4", pro_code: "QQA"}, {des: "cont5", note: "cont5", pro_code: "GFD"}];
const arr2 = [{des: "cont1", note: "cont1", pro_code: "XXY"}, {des: "cont2", note: "cont2", pro_code: "NNB"}, {des: "cont3", note: "cont3", pro_code: "QAS"}]

const arr3 = arr2.filter(function(arr2item) {
  return !arr1.some(function(arr1item) {
    return arr1item.pro_code === arr2item.pro_code;
  })
});

console.log(arr3);

edited Oct 19, 2020 at 9:31

answered Oct 19, 2020 at 7:41

Gabriele Petrioli

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1 Comment

Gabriele Petrioli Over a year ago

@connexo i am not comparing the objects, but the pro_code string property.

Sapna Dorbi · Accepted Answer · 2020-10-19 09:36:55Z

1

Here's another optimized solution

const arr1 = [{
  des: "cont1",
  note: "cont1",
  pro_code: "XXY"
}, {
  des: "cont2",
  note: "cont2",
  pro_code: "NNB"
}, {
  des: "cont4",
  note: "cont4",
  pro_code: "QQA"
}, {
  des: "cont5",
  note: "cont5",
  pro_code: "GFD"
}];

const arr2 = [{
  des: "cont1",
  note: "cont1",
  pro_code: "XXY"
}, {
  des: "cont2",
  note: "cont2",
  pro_code: "NNB"
}, {
  des: "cont3",
  note: "cont3",
  pro_code: "QAS"
}];

const results = arr2.filter(({ pro_code: id1 }) => !arr1.some(({ pro_code: id2 }) => id2 === id1));

console.log(results);

edited Oct 19, 2020 at 9:36

answered Oct 19, 2020 at 8:42

Sapna Dorbi

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