Comparing arrays in arrays

Question

I'm looking for a way to make comparisons between arrays in arrays.

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];

let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];

For example, I would like to be able to find out how many of the arrays in small are found in large. Some function that, given the arrays above as arguments, would return 2, since [2, 2] and [3, 0] from small are found in large.

How would you go about doing that?

Answer 1

You can convert one of the arrays into a Set of hashes, and than filter the 2nd array using the set:

const small = [[1, 3], [2, 2], [2, 3], [3, 0]];

const large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];

const containsCount = (arr1, arr2, hashFn) => {
  const arr1Hash = new Set(arr1.map(hashFn));
  
  return arr2.filter(s => arr1Hash.has(hashFn(s))).length;
}

const result = containsCount(small, large, ([a, b]) => `${a}-${b}`); 

console.log(result);

Answer 2

You can try something like:

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];

let z = zeta(small, large);
console.log(z);

function zeta(a, b) {
  let join = m => m.join();
  let x = a.map(join);
  let y = b.map(join);
  
  return x.reduce((n, m) => (y.indexOf(m)>0) ? ++n : n, 0);
}

I hope this helps!

Answer 3

Use every and some to compare the arrays with each other.

If you want to get an array containing the subarrays that match, use filter:

let result = small.filter(arr => 
  large.some(otherArr =>
    otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
  )
);

Which filters the subarray from small that some subarray from large has the same length and the same elements/items.

Demo:

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];

let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];

let result = small.filter(arr => 
  large.some(otherArr =>
    otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
  )
);

console.log(result);

And if you want just a count, then use reduce instead of filter to count the mathched items (this makes use of the fact that the numeric value of true is 1 and that of false is 0):

let count = small.reduce((counter, arr) => 
  counter + large.some(otherArr =>
    otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
  )
, 0);

Demo:

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];

let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];

let count = small.reduce((counter, arr) =>
  counter + large.some(otherArr =>
    otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
  )
, 0);

console.log(count);

Note: If the subarrays contain only numbers, the code could be simplified to use Array#toString instead of every and length comparaison:

let result = small.filter(arr => large.some(otherArr => "" + otherArr === "" + arr));

Which casts both arrays into strings and compares the two strings instead. This can be used with the reduce as well.

Answer 4

Create two nesting map() function oute for small and inner for large then use JSON.stringify()

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
var same=[];
    small.map(function(element){
        large.map(function(element2){
            if(JSON.stringify(element)==JSON.stringify(element2)){
        same.push(element);
  }
 });
});
console.log(same);

Answer 5

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];

let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];


let largeArrayStringForm = large.map(item => item.toString())
let matchingItems = small.filter(item => largeArrayStringForm.includes(item.toString()))

console.log(`matchCount: ${matchingItems.length}`)