Is there a Javascript function for checking if an array of arrays contains a specific array?

Question

I'm making a Javascript application where I need to check an array of arrays (a nested array) for specific arrays containing two objects each. The nested array looks somewhat like this: [[obj1, obj2], [obj2, obj3], [obj3, obj1]]

The problem I have is that even though this part of the application doesn't raise any errors, it does not work as intended. I tried to approach it like this: array.includes([obj1, obj2]) , but this returns false, even though the array does contain [obj1, obj2] .

So why isn't this working the way it's intended, and what's a better approach for this situation?

Edit: I want to check with strict comparison of the objects included in the array, no matter the reference of the array containing them. How can I do that?

Answer 1

includes compares with === , apparently your search array is only equivalent to one of the entries, not the same actual array.

You can use some with the callback checking whether every entry in the array being visited matches your search array.

const flag = arrayOfArrays.some(
    array => array.length === search.length && array.every(
        (entry, index) => entry === search[index]
   )
);

Note that this assumes the inner entries are comparable with === . If not, substitute the appropriate way of comparing them.

Live Example:

 function check(arrayOfArrays, search) { return arrayOfArrays.some( array => array.length === search.length && array.every( (entry, index) => entry === search[index] ) ); } console.log( "yes:", check( [[1, 2], [3, 4], [5, 6], [7, 8]], [5, 6] ) ); console.log( "no:", check( [[1, 2], [3, 4], [5, 6], [7, 8]], [6, 5] // Note the different order ) ); 

If you want to find the array in the array of arrays, you can use find instead of some . If you want to find its index in the array of arrays, you can use findIndex .

Answer 2

In JS each array(object) is unique, even if it has the same content as another one:

 console.log( {} == {} ); // false console.log( [] == [] ); // false console.log( [1, 2] == [1, 2] ); // false 

So, you have to loop and compare each element one by one:

 let arr = [ [0, 0, 0], [1, 1, 1], [2, 2, 2], ] console.log( arrInArr( arr, [3, 3, 3] ) ); // false arr[3] = [3, 3, 3]; // new elem added console.log( arrInArr( arr, [3, 3, 3] ) ); // true function arrInArr(mama, child){ for( let i = 0; i < mama.length; i++ ){ // don't even starting to compare if length are not equal if( mama[i].length != child.length ){ continue } let match = 0; // To count each exact match for( let j = 0; j < child.length; j++ ){ if( mama[i][j] === child[j] ){ // We made sure, that they have equal length, and can use index 'j' // to compare each element of arrays. match++; } else { break; } } if( match === child.length ){ // If each element exactly matched return true; } } return false; } 

Answer 3

If you use a package like lodash, it has a method for comparing if two objects are equal by content rather than by the much stricter === comparison: https://lodash.com/docs/4.17.15#isEqual

Using _.isEqual , you could then do something like this:

array.findIndex(item => _.isEqual(item, testObject)) >= 0

Lodash might even have a method to directly search an array with a deep-equals comparison too, but the above would get the job done.