You could define three functions inBoth, inFirstOnly, and inSecondOnly which all take two lists as arguments, and return a list as can be understood from the function name. The main logic could be put in a common function operation that all three rely on.
Here are a few implementations for that operation to choose from, for which you can find a snippet further down:
- Plain old JavaScript
forloops - Arrow functions using
filterandsomearray methods - Optimised lookup with a
Set
Plain old for loops
// Generic helper function that can be used for the three operations:
function operation(list1, list2, isUnion) {
var result = [];
for (var i = 0; i < list1.length; i++) {
var item1 = list1[i],
found = false;
for (var j = 0; j < list2.length && !found; j++) {
found = item1.userId === list2[j].userId;
}
if (found === !!isUnion) { // isUnion is coerced to boolean
result.push(item1);
}
}
return result;
}
// Following functions are to be used:
function inBoth(list1, list2) {
return operation(list1, list2, true);
}
function inFirstOnly(list1, list2) {
return operation(list1, list2);
}
function inSecondOnly(list1, list2) {
return inFirstOnly(list2, list1);
}
// Sample data
var list1 = [
{ userId: 1234, userName: 'XYZ' },
{ userId: 1235, userName: 'ABC' },
{ userId: 1236, userName: 'IJKL' },
{ userId: 1237, userName: 'WXYZ' },
{ userId: 1238, userName: 'LMNO' }
];
var 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)); Arrow functions using filter and some array methods
This uses some ES5 and ES6 features:
// Generic helper function that can be used for the three operations:
const operation = (list1, list2, isUnion = false) =>
list1.filter( a => isUnion === list2.some( b => a.userId === 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));Optimising lookup
The above solutions have a O(n²) time complexity because of the nested loop — some represents a loop as well. So for large arrays you’d better create a (temporary) hash on user-id. This can be done on-the-fly by providing a Set (ES6) as argument to a function that will generate the filter callback function. That function can then perform the look-up in constant time with has:
// 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));