从另一个对象数组中删除对象数组

Question

Assume we have the following arrays of objects to be compared based on property id :

a = [{'id':'1', 'name':'a1'}, {'id':'2', 'name':'a2'}, {'id':'3', 'name':'a3'}]

and

b = [[{'id':'2', 'name':'a2'}, ]

How can I subtract b from a? So that we have c = a - b which should be equal to [ {'id':'1', 'name':'a1'}, {'id':'3', 'name':'a3'}] .

I have tried using this:

var c= a.filter(function(item) {
                    return !b.includes(item.id);
                });

but still not working.

Answer 1

How about this solution? It assumes that 'b' is also an array so for each element of 'a' you check if there is a matching object in 'b'. If there is a matching object then return a false in the filter function so that that element is discarded.

 var a = [{ 'id': '1', 'name': 'a1' }, { 'id': '2', 'name': 'a2' }, { 'id': '3', 'name': 'a3' }] var b = [{ 'id': '2', 'name': 'a2' }] var c = a.filter(function(objFromA) { return !b.find(function(objFromB) { return objFromA.id === objFromB.id }) }) console.log(c)

Answer 2

Here is a nice one line answer :)

Basically, you can filter, as you were trying to do already. Then you can also filter b for each a element and if the length of the filtered b is zero, then you return true because that means the a element is unique to a.

 var a = [{ 'id': '1', 'name': 'a1' }, { 'id': '2', 'name': 'a2' }, { 'id': '3', 'name': 'a3' }]; var b = [{ 'id': '2', 'name': 'a2' }]; c = a.filter( x => !b.filter( y => y.id === x.id).length); console.log(c);

Answer 3

First, you build just a map of the ids you want to delete. Then, you filter your first array with it, like that:

 var a = [{ 'id': '1', 'name': 'a1' }, { 'id': '2', 'name': 'a2' }, { 'id': '3', 'name': 'a3' }]; var b = [{ 'id': '2', 'name': 'a2' }]; var idsToDelete = b.map(function(elt) {return elt.id;}); var result = a.filter(function(elt) {return idsToDelete.indexOf(elt.id) === -1;}); console.log(result)

Answer 4

Easy with new ES6 Syntax

Second and Third way are more performant i guess....

a.filter(i => !b.filter(y => y.id === i.id).length); // One Way
a.filter(i => !b.find(f => f.id === i.id)); // Second Way
a.filter(i => b.findIndex(f => f.id === i.id)) // Third Way