如何使用 .some() 检查数组中是否存在元素？

Question

I am trying to see if a particular pair of numbers exist in an array. I know that .some() works effectively when it's a simple array, but I can't seem to get it to work in nested arrays.

Thank you in advance.

 const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]]; // checks whether an element exists const exists = (i) => i == [3, 1] || [1, 3]; console.log(array.some(exists)); // expected output: true

Answer 1

You need to check every value because of different arrays, you have different object references, and you misused the logical OR operator. This does not include a comparison with another value.

 const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]]; const exists = ([a, b]) => a === 3 && b === 1 || a === 1 && b === 3; console.log(array.some(exists));

Answer 2

Here is a general solution.

const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const expected = [[1, 3], [3, 1]];
const exists = (i) => expected.some((j) => {
    return i.every((k, n) => j[n] === k);
});

console.log(array.some(exists));

This could be wrapped in a function to make it reusable.

const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const expected = [[1, 3], [3, 1]];

const containsAny = (array, expected) => {
  return array.some((i) => expected.some((j) => {
    return i.every((k, n) => j[n] === k);
  }))
};


console.log(containsAny(array, expected));

Answer 3

You could make a custom function out of it which makes it reusable. This is a simple example. By flattening the array you can check if values are the same.

Got my inspiration on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some

 const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]]; const checkArrayFor = (needle, haystack) => { const flattenNeedle = needle.join(':'); const flattenArray = haystack.map(item => item.join(':')); return flattenArray.includes(flattenNeedle); }; console.log("Check for [1, 0]", checkArrayFor([1, 0], array)); console.log("Check for [1, 1]", checkArrayFor([1, 1], array));