Filtering multi-dimensional array based on another array elements

Question

I have two arrays in the code below - which matches is the main one and the other played which serves it is purpose for elements to filter out in the main array:

var matches = [[1,4],[3,1],[5,2],[3,4],[4,5],[2,1]];
var played = [2,5];

I need to filter out elements in matches based on played array, which means if there is any 2 or 5 in, then remove it altogether. Also the played array can be any length, min is 1.

Expected output should be

[[1,4],[3,1],[3,4]];

So I have tried this piece of code, but it doesn't yield the result I want.

var result = matches.map(x => x.filter(e => played.indexOf(e) < 0))

So anyway to achieve this?

Nina Scholz · Accepted Answer · 2020-03-30 20:35:36Z

1

You could check with some and exclude the unwanted arrays.

var matches = [[1, 4], [3, 1], [5, 2], [3, 4], [4, 5], [2, 1]],
    played = [2, 5],
    result = matches.filter(a => !a.some(v => played.includes(v)));

console.log(result);

answered Mar 30, 2020 at 20:35

Nina Scholz

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CertainPerformance · Accepted Answer · 2020-03-30 20:33:50Z

1

While filtering, check that .every one of the subarray elements are not included in [2, 5]:

var matches = [[1,4],[3,1],[5,2],[3,4],[4,5],[2,1]];
var played = [2,5];

const result = matches.filter(
  subarr => played.every(
    num => !subarr.includes(num)
  )
);
console.log(result);

answered Mar 30, 2020 at 20:33

CertainPerformance

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TheMaster · Accepted Answer · 2020-03-30 20:42:12Z

0

Another way would be to create a Set of played to avoid iterating it again and again:

var matches = [[1,4],[3,1],[5,2],[3,4],[4,5],[2,1]];
var played = new Set([2,5]);
var out = matches.filter(a => !a.some(num => played.has(num)));
console.info(out)

answered Mar 30, 2020 at 20:42

TheMaster

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