I have arrays like;
var john = { name: "John Smith", age: 23 };
var mary = { name: "Mary Key", age: 18 };
var bob = { name: "Bob-small", age: 6 };
var people = [john, mary, bob];
var john2 = { name: "John Smith", age: 23 };
var people2 = [john2];
What I would like to do is subtract people2 from people and get result;
[mary, bob];
How can I achieve this? TIA
The difference of two sets, A and B, is defined as the set of all those elements of A which are not in B. If we implement it naively, computing the difference of two sets of sizes m and n respectively would take O(m * n) time. Not very efficient:
const john1 = { name: "John Smith", age: 23 };
const john2 = { name: "John Smith", age: 23 };
const mary = { name: "Mary Key", age: 18 };
const bob = { name: "Bob-small", age: 6 };
const people1 = [john1, mary, bob];
const people2 = [john2];
const eqPerson = (p, q) => p.name === q.name && p.age === q.age;
const result = people1.filter(p => people2.every(q => !eqPerson(p, q)));
console.log(result); // [mary, bob]Fortunately, there's a faster way to compute the set difference for large sets using hashing.
const john1 = { name: "John Smith", age: 23 };
const john2 = { name: "John Smith", age: 23 };
const mary = { name: "Mary Key", age: 18 };
const bob = { name: "Bob-small", age: 6 };
const people1 = [john1, mary, bob];
const people2 = [john2];
const hashPerson = ({ name, age }) => `${name} ${age}`;
const hashSet = new Set(people2.map(hashPerson));
const result = people1.filter(p => !hashSet.has(hashPerson(p)));
console.log(result); // [mary, bob]The advantage is that creating a hash set takes O(n) time and calculating the difference takes O(m) time. Hence in total it only takes O(m + n) time instead of O(m * n) time. In addition, you can reuse the hash set in the future.
Here is an easy solution:
var diff = people.filter(function(item) {
return !people2.some(function(test){
return test.name === item.name && test.age === item.age;
});
});
Make sure the function passed to people2.some correctly checks that the two objects are equal, since == would fail, as you have references to different objects with identical properties.
So, here goes the final running code. Copy , paste it and check for the output.It is first of all separating the strings (comma-separation). Then it is comparing them individually . Check this link: http://jsfiddle.net/DXRZ4/
<html>
<head>
<script>
var a = new Array(); var i=1; var people = new Array();
function fun() {
var str = "john,john,bob",
l = str.split(",");
for(i=0; i<3; i++) {
a[i] = l[i];
// this will show the comma separated strings
document.writeln(l[i]);
}
for(i=0; i<3; i++) {
t=a[i];
j=a[i+1];
if((a[i]==a[i+1]) || (a[i]==a[i+2])) {
// it will store the position of string which are same
p=i;
}
}
for(i=0; i<3; i++) {
if(p!=i){
document.writeln("Subtracted Strings are:-");
document.writeln(l[i]);
}
}
}
</script>
<body>
<input type="button" name="b1" onClick="fun()">
</body>
</html>
