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]);

Answer:

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.

Comments

[Jon Randy 🎖️]

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

[Adam Crockett 🌀]

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