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Randy Rivera

Posted on • Updated on

Comparing Two Different Arrays

• Second we will simply compare two arrays and return a new array with any items only found in one of the two given arrays, but not both. Remember the keyword "not both".
• Problem Below:
``````function arrays(arr1, arr2) {

}

arrays([1, 2, 4, 5], [1, 2, 3, 4, 5]);
``````

``````function arrays(arr1, arr2) {
let merge = arr1.concat(arr2);

return merge.filter(function(num) { // <--- num are all the numbers in merge. [1, 2, 4, 5, 1, 2, 3, 4, 5]
if (arr1.indexOf(num) === -1 || arr2.indexOf(num) === -1) {
return num;
}
})

}

console.log(arrays([1, 2, 4, 5], [1, 2, 3, 4, 5])); // will display [3]
``````
• We're just checking the two arrays and return a new array that contains only the items that are not in either of the original arrays. In this case 3.
• What we did was merge the list to make it easy to compare and used filter to get the new array in which you will need to create a callback function.

Jon Randy ποΈ • Edited

Your filter function only need return true or false, and as @adam_cyclones points out, you should really only check the unique values in the merged array (no point checking the same numbers more than once):

``````function diffArray(arr1, arr2) {
let merge = [...new Set([...arr1, ...arr2])]

return merge.filter(function(num) {
return arr1.indexOf(num) === -1 || arr2.indexOf(num) === -1
})

}
``````

or, shorter:

``````const diffArray = (arr1, arr2) => [...new Set([...arr1, ...arr2])].filter(
num => arr1.indexOf(num) === -1 || arr2.indexOf(num) === -1
)
``````

DevFranPR

Nice one

I saw something useful when creating an exlusion list. Use a Set on the second array to reduce complexity in code and speed things up. Perfect for es6 > targets

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