How to find symmetrical difference using JavaScript?-CodePudding

I've been working on an algorithm that will find the symmetric difference of two arrays (i.e. only items that are in 1 of the 2 arrays, but not both). I've come up with the following so far:

function diffArray(arr1, arr2) {
  let newArr1 = arr1.slice();
  let newArr2 = arr2.slice();
  if (newArr1.length > newArr2.length) {
    return newArr1.filter((item) => !(item in newArr2));
  } else {
    return newArr2.filter((item) => !(item in newArr1));
  }
};

const result = diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
console.log(result);

But when testing with diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]) the output is [4, 5] instead of just [4]. The issue seems to happen with whatever the last value of either array is, but I can't seem to figure out what about my code is causing the issue. Thanks in advance for any help.

CodePudding user response：

The in operator checks if the value on the left matches a property name on the right.

In you want to check if one of the property values is in an array, use the includes method.

function diffArray(arr1, arr2) {
  let newArr1 = arr1.slice();
  let newArr2 = arr2.slice();
  if (newArr1.length > newArr2.length) {
    return newArr1.filter((item) => !newArr2.includes(item));
  } else {
    return newArr2.filter((item) => !newArr1.includes(item));
  }
};

const result = diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
console.log(result);

CodePudding user response：

Updating the answer for another condition

const result = diffArray([1, 2, 3, 4], [1, 2, 3, 5]);

function diffArray(arr1, arr2){
  const diff1 = arr1.filter(item => !arr2.includes(item)); // if item not available, the value will be filtered
  const diff2 = arr2.filter(item => !arr1.includes(item));// if item not available, the value will be filtered
  return diff1.concat(diff2);
}
// const result = diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
const result = diffArray([1, 2, 3, 4], [1, 2, 3, 5]); // This condition will fail on the above code
console.log(result);