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?
CodePudding user response:
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);
CodePudding user response:
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