JavaScript Do Arrays se croix
let intersection = arrA.filter(x => arrB.includes(x));
Eager Echidna
let intersection = arrA.filter(x => arrB.includes(x));const arr1 = [{ id: 1 }, { id: 2 }]
const arr2 = [{ id: 1 }, { id: 3 }]
const intersection = arr1.filter(item1 => arr2.some(item2 => item1.id === item2.id))
// intersection => [{ id: 1 }]// Generic helper function that can be used for the three operations:
const operation = (list1, list2, isUnion = false) =>
list1.filter(
(set => a => isUnion === set.has(a.userId))(new Set(list2.map(b => b.userId)))
);
// Following functions are to be used:
const inBoth = (list1, list2) => operation(list1, list2, true),
inFirstOnly = operation,
inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);
// Sample data
const list1 = [
{ userId: 1234, userName: 'XYZ' },
{ userId: 1235, userName: 'ABC' },
{ userId: 1236, userName: 'IJKL' },
{ userId: 1237, userName: 'WXYZ' },
{ userId: 1238, userName: 'LMNO' }
];
const list2 = [
{ userId: 1235, userName: 'ABC' },
{ userId: 1236, userName: 'IJKL' },
{ userId: 1252, userName: 'AAAA' }
];
console.log('inBoth:', inBoth(list1, list2));
console.log('inFirstOnly:', inFirstOnly(list1, list2));
console.log('inSecondOnly:', inSecondOnly(list1, list2));const intersection = (a, b) => {
b = new Set(b); // recycling variable
return [...new Set(a)].filter(e => b.has(e));
};
console.log(intersection([1, 2, 3, 1, 1], [1, 2, 4])); // Array [ 1, 2 ]