Exploring the intricacies of nested arrays

Question

Exploring the intricacies of nested arrays

I am searching for a solution to compare arrays within 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 instance, I would like to determine how many arrays in small are present in large. A function that takes the above arrays as arguments and returns 2, as [2, 2] and [3, 0] from small exist in large.

What is your approach to accomplish this task?

javascript arrays

Answer 1

Answer №1

To simplify the comparison process, convert one of the arrays into a set of hashes and then filter the second array based on this 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

To simplify the comparison process, convert one of the arrays into a set of hashes and then filter the second array based on this 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 3

Answer №2

If you are looking for a solution, you may consider the following code snippet:

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);
}

Hope this helps you with your task!

Answer 4

If you are looking for a solution, you may consider the following code snippet:

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);
}

Hope this helps you with your task!

Answer 5

Answer №3

Utilize the every and some methods to compare the arrays with each other.

If you desire an array containing the subarrays that match, employ the filter method:

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

This code snippet filters out the subarray from small that has a matching subarray in large with the same length and elements/items.

Live Example:

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);

If you just need a count, then utilize the reduce method instead of filter for counting the matched items (based on 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);

Live Example:

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 consist solely of numbers, you could simplify the code by using Array#toString instead of comparing each element and the length:

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

This approach converts both arrays into strings and compares the two strings directly. It can be implemented with reduce as well.

Answer 6

Utilize the every and some methods to compare the arrays with each other.

If you desire an array containing the subarrays that match, employ the filter method:

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

This code snippet filters out the subarray from small that has a matching subarray in large with the same length and elements/items.

Live Example:

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);

If you just need a count, then utilize the reduce method instead of filter for counting the matched items (based on 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);

Live Example:

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 consist solely of numbers, you could simplify the code by using Array#toString instead of comparing each element and the length:

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

This approach converts both arrays into strings and compares the two strings directly. It can be implemented with reduce as well.

Answer 7

Answer №4

Implement a nested map() function for comparing two arrays and finding matching elements using stringification.

let arrayA = [[1, 3], [2, 2], [2, 3], [3, 0]];
let arrayB = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
var matches = [];
arrayA.map(function(itemA){
    arrayB.map(function(itemB){
        if(JSON.stringify(itemA) == JSON.stringify(itemB)){
            matches.push(itemA);
        }
    });
});
console.log(matches);

Answer 8

Implement a nested map() function for comparing two arrays and finding matching elements using stringification.

let arrayA = [[1, 3], [2, 2], [2, 3], [3, 0]];
let arrayB = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
var matches = [];
arrayA.map(function(itemA){
    arrayB.map(function(itemB){
        if(JSON.stringify(itemA) == JSON.stringify(itemB)){
            matches.push(itemA);
        }
    });
});
console.log(matches);

Answer 9

Answer №5

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

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


const largeArrayToString = largeArray.map(item => item.toString())
const matchingItemsFound = smallArray.filter(item => largeArrayToString.includes(item.toString()))

console.log(`Number of matches: ${matchingItemsFound.length}`)

Answer 10

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

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


const largeArrayToString = largeArray.map(item => item.toString())
const matchingItemsFound = smallArray.filter(item => largeArrayToString.includes(item.toString()))

console.log(`Number of matches: ${matchingItemsFound.length}`)

Exploring the intricacies of nested arrays

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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