Optimising ES6 intersection of Arrays function

Question

I have the following Arrays:

this.originalSelectedRows = [1,2,3,4,5];
this.selectedRows = [3,4,5,6,7,8];

I want to know the following:

How many numbers are in this.originalSelectedRows, but not in this.selectedRows

Which in this case would be 2:

1 and 2

and how many numbers are in this.selectedRows, but not this.originalSelectedRows

Which in this case would be 3:

6, 7 and 8

I have the following, which works fine:

    let isOnListCount = 0;
    let isNotOnListCount = 0

    this.selectedRows.map((row) => {
      const isSelected = this.originalSelectedRows.filter(x => x === row);
        if(isSelected.length > 0)
          isOnListCount++;
        else
          isNotOnListCount++;
    });

But I want to know if this can be achieved in a neater fashion usng some of ES6's new features, or original JS

You are looking for the **difference** between the two arrays. Google for that. — , Oct 13 '17 at 13:09
Refer this https://stackoverflow.com/questions/40998880/get-the-different-of-arrays-in-es6 — user-developer, Oct 13 '17 at 13:13
This question has an answer with a lot of ideas, including using ES6 generators. https://stackoverflow.com/questions/40998880/get-the-different-of-arrays-in-es6/41001435#41001435 — , Oct 13 '17 at 13:18
If the arrays are always sorted, replace the `.includes` in the below answers with a binary search. — 4castle, Oct 13 '17 at 13:19
why does your code works fine, if you get as result 3 and 3? — Nina Scholz, Oct 13 '17 at 13:45

Nina Scholz · Answer 1 · 2017-10-13T13:15:24.803

2

You could take the length as counter and reduce the value with common items.

var array1 = [1, 2, 3, 4, 5],
    array2 = [3, 4, 5, 6, 7, 8],
    count1 = array1.length,
    count2 = array2.length;
    
array1.forEach(function (a) {
    if (array2.includes(a)) {
        count1--;
        count2--;
    }
});
console.log(count1, count2);

edited Oct 13 '17 at 13:15

answered Oct 13 '17 at 13:12

Nina Scholz

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Readability is more important than short fancy code. You always have a way of making the most simple logic completely incomprehensible to new programmers. – Cerbrus Oct 13 '17 at 13:15
Specifically `&& (count1--, count2--)` is something new users will not recognize. `&&` as shortcut `if` and the `( , )` aren't common structures that they will recognize. – Cerbrus Oct 13 '17 at 13:26
@NinaScholz I know how much you wanted to use `reduce`, and write more obscure code. Here you go: `const common = arr.reduce((result, elt) => result + array2.includes(elt));`. Note the clever way we add a boolean. Now the answers are simply `array1.length - common` and `array2.length - common`. – Oct 13 '17 at 17:09
@torazaburo, you know (i hope so), that i am able to write that in reduce style, but it is not necessary in this case (for saving a single line of code). – Nina Scholz Oct 13 '17 at 21:56

score 1 · Answer 2 · answered Oct 13 '17 at 13:09

1

Sets will do a great job here :)

let set1 = new Set(this.originalSelectedRows)
let set2 = new Set(this.selectedRows)
let difference = new Set([...set1].filter(n => !set2.has(n)))

They are also relatively fast as well, but not as fast as a specially crafted solution :P

I am assuming that all of your items are unique though :)

answered Oct 13 '17 at 13:09

user2662833

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I love your smilies--do they mean there is some kind of joke in there that I missed? :-) – Oct 13 '17 at 13:12
I really do need to ease up on the smiles! Dang! – user2662833 Oct 13 '17 at 13:18

score 1 · Answer 3 · answered Oct 13 '17 at 13:13

I'd convert one array into a set and than filter the other array for items which aren't in the set.

const
  valuesA = [1,2,3,4,5],
  valuesB = [3,4,5,6,7,8];
  
function getUniqueValues(sourceArray, testArray) {
  const 
    set = new Set(testArray);
    
  return sourceArray.filter(value => !set.has(value));
}

console.log(`Values in A and not in B: ${getUniqueValues(valuesA, valuesB)}`);
console.log(`Values in B and not in A: ${getUniqueValues(valuesB, valuesA)}`);

Suren Srapyan · Accepted Answer · 2017-10-13T13:10:56.740

0

You can just use Array#includes function to check if the element exists in the second array or not

const originalSelectedRows = [1,2,3,4,5];
const selectedRows = [3,4,5,6,7,8];

let isOnListCount = 0;
let isNotOnListCount = 0;

selectedRows.forEach(row => 
 originalSelectedRows.includes(row) ? isOnListCount++ : isNotOnListCount++
);

console.log(isOnListCount);
console.log(isNotOnListCount);

edited Oct 13 '17 at 13:10

answered Oct 13 '17 at 13:05

Suren Srapyan

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@Alex You've accepted an answer which gives an answer different from that you yourself said was correct, which was a `2` for the first case. – Oct 13 '17 at 16:45
You are right. I am going to remove this answer. I didn't notice that my result is wrong. Remove the acceptance, to let me remove this – Suren Srapyan Oct 13 '17 at 16:51
Why don't you just fix the answer, instead of deleting it? – Cody Gray - on strike Oct 14 '17 at 13:00

Optimising ES6 intersection of Arrays function

4 Answers4