比较数组中的数组

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}`)