Filtering one array of objects with another array of objects

Question

There are a few similar questions out there, but not quite one that answers what I need to do. I have 2 arrays, both filled with objects and nothing else.

arr1 = [
   {name:'one'},
   {name:'two'},
   {name:'three'},
]

arr2 = [
   {name:'four'},
   {name:'two'},
   {name:'six'},
]

What I want to do is return arr2 such that it doesn't have anything that matches in arr1. For example:

filteredArr = [
   {name:'four'},
   {name:'six'},
]

A requirement is I want to use array.prototype.filter to achieve this. My problem is, whenever I attempt to solve this, it gives me duplicates in my returned array. Here's what I've come up with:

Repl.it Link

let state = {
    one: [
        {name:'one'},
        {name:'two'},
        {name:'three'}
    ],
    two: [
        {name:'one'},
        {name:'four'},
        {name:'five'},
        {name:'three'}
    ]
}

let {one, two} = state

let newStuff = []

two.filter(s => {
    one.filter(t => {
        s.name !== t.name ? newStuff.push(s) : null
    })
})

console.log(newStuff)

I know the logic of what it's doing and why I'm returning duplicates, but I don't know how to write it so that it simply returns state.two with all of the objects from state.one filtered out.

Hassan Imam · Accepted Answer · 2018-01-12 14:31:20Z

2

Use array#reduce with array#some. Iterate arr1 and check if the name value exists in the other array if it exists ignore it, otherwise add to your result array.

var arr1 = [ {name:'one'}, {name:'two'}, {name:'three'}], 
    arr2 = [ {name:'four'}, {name:'two'}, {name:'six'}],
    result = arr2.reduce((r,{name}) => !arr1.some(o => o.name === name) ? (r.push({name}), r) : r, []);
console.log(result);

answered Jan 12, 2018 at 14:31

Hassan Imam

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