Filter Array Not in Another Array

Need to filter one array based on another array. Is there a util function in knock out ? else i need to go with javascript

First :

var obj1 = [{ "visible": "true", "id": 1
}, { "visible": "true", "id": 2
}, { "visible": "true", "id": 3
}, { "Name": "Test3", "id": 4
}];

Second :

var obj2 = [ 2,3]

Now i need to filter obj1 based on obj2 and return items from obj1 that are not in obj2 omittng 2,3 in the above data (Comparison on object 1 Id)

output:

[{ "visible": "true", "id": 1
}, { "Name": "Test3", "id": 4
}];

3 Answers

You can simply run through obj1 using filter and use indexOf on obj2 to see if it exists. indexOf returns -1 if the value isn't in the array, and filter includes the item when the callback returns true.

var arr = obj1.filter(function(item){ return obj2.indexOf(item.id) === -1;
});

With newer ES syntax and APIs, it becomes simpler:

const arr = obj1.filter(i => !obj2.includes(i.id))

Even if the answer provided by @Joseph are good, it can lead to performance issues if you have a very large array. The complexity is O(n^2), but it can be improved to O(n). Below I put the code I'd use to perform this task:

//create a map for storing object id
const map = new Map();
//add existing id from the array to the map
for (item of obj2) { map.set(item.id, true);
}
//create a new array that contains all items that aren't included in the //map
//the map lookUp is O(1) complexity
const arr = obj1.filter(item => !map[item.id]);

In this way we avoid nested iteration and have O(n) complexity

To create your output array, create a function that will iterate through obj1 and populate a new array based on whether the id of every obj in the iteration exists in obj2.

var obj1 = [{ "visible": "true", "id": 1
}, { "visible": "true", "id": 2
}, { "visible": "true", "id": 3
}, { "Name": "Test3", "id": 4
}];
var obj2 = [2,3]
var select = function(arr) { var newArr = []; obj1.forEach(function(obj) { if obj2.indexOf(obj.id) !== -1 { newArr.push(obj) }; }; return newArr;
};