Given the following two array objects:
let origArr = [ { value: 2, label: 'Dog' },
{ value: 4, label: 'Cat' },
{ value: 16, label: 'Horse' }
]
let newArr = [
{ value: 2, label: 'Dog' },
{ value: 3, label: 'Bird' },
{ value: 0, label: 'Fish' }
]
what would be the best way to return a new array of just the value of the difference between the origArr and the newArr, that is, where origArr element(s) are not in newArr ?
In the above example, I am after a new array called diffArr = [4,16]
Please note: that all value values within both origArr and the newArr are unique
Not sure if there is an ES2016 means?
You can use filter and some to do the job.
let origArr = [ { value: 2, label: 'Dog' },
{ value: 4, label: 'Cat' },
{ value: 16, label: 'Horse' }
]
let newArr = [
{ value: 2, label: 'Dog' },
{ value: 3, label: 'Bird' },
{ value: 0, label: 'Fish' }
]
const res = origArr.filter(x => !newArr.some(y => y.value === x.value));
console.log(res); // assuming you want the entire filtered objects
const justValues = res.map(x => x.value); // if you want just the filtered values
console.log(justValues)some - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some
filter - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
You can get all unique value from newArr using Set and array#map and then filter value which doesn't exist in the set and get all values.
const origArr = [ { value: 2, label: 'Dog' }, { value: 4, label: 'Cat' }, { value: 16, label: 'Horse' } ],
newArr = [ { value: 2, label: 'Dog' }, { value: 3, label: 'Bird' }, { value: 0, label: 'Fish' } ],
unique = new Set(newArr.map(o => o.value)),
difference = origArr
.filter(({value}) => !unique.has(value))
.map(o => o.value);
console.log(difference);