Difference of two javascript arrays of objects

Question

I have two arrays like this:

owners: [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "[email protected]"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "[email protected]"
  }
]

users: [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "[email protected]"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "[email protected]"
  },
  {
    userID: "88c4d7ac",
    username: "User3.Name3",
    firstName: "User3",
    lastName: "Name3",
    email: "[email protected]"
  }
]

I would like to get an array of users which contains only the elements which are not in the owners array.

I tried different approaches. Finally, I ended up with the solution:

const usersItems = users.map(user => {
    // Check whether the user is already an owner
    if (owners.findIndex(owner => owner.userID === user.userID) === -1) {
        return  owner
    } else {
        return null;
    }
});

console.log(usersItems);

// Filter out all items which are null
const newUsersItems = usersItems.filter(user => {
    if (user) return user;
});

console.log(usersItems);

To me, it doesn't' look like a clean solution. Is there a cleaner and easier way to do this? As a result, I would like to have:

newUsers: [
  {
    userID: "88c4d7ac",
    username: "User3.Name3",
    firstName: "User3",
    lastName: "Name3",
    email: "[email protected]"
  }
]

schu34 · Accepted Answer · 2019-02-26 17:36:01Z

5

You could get rid of your map and just use the filter, (that's exactly what filter is for) somthing like

const filtered = users.filter(user => {
    // Check whether the user is already an owner
   return owners.findIndex(owner => owner.userID === user.userID) === -1

});

would probably work

answered Feb 26, 2019 at 17:36

schu34

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1 Comment

schu34 Over a year ago

Worth noting that if performance on large datasets matters to you Shidersz answer below is worth looking at

user3210641 · Accepted Answer · 2019-02-26 17:41:17Z

You can use combination of filter and some functions like this:

const owners = [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "[email protected]"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "[email protected]"
  }
];

const users = [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "[email protected]"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "[email protected]"
  },
  {
    userID: "88c4d7ac",
    username: "User3.Name3",
    firstName: "User3",
    lastName: "Name3",
    email: "[email protected]"
  }
];

const result = users.filter(user => !owners.some(owner => owner.userID === user.userID));

console.log(result);

Shidersz · Accepted Answer · 2019-02-26 19:09:10Z

First you can create a Set() with the userID's of the owners array, and then you can use Array.filter() on the users array to filter the users whose userID does not belong to the previous created set.

const owners = [
  {userID: "58c4d7ac", username: "John.Doe", firstName: "John", lastName: "Doe", email: "[email protected]"},
  {userID: "68c4d7ac", username: "User2.Name2", firstName: "User2", lastName: "Name2", email: "[email protected]"}
];

const users = [
  {userID: "58c4d7ac", username: "John.Doe", firstName: "John", lastName: "Doe", email: "[email protected]"},
  {userID: "68c4d7ac", username: "User2.Name2", firstName: "User2", lastName: "Name2", email: "[email protected]"},
  {userID: "88c4d7ac", username: "User3.Name3", firstName: "User3", lastName: "Name3", email: "[email protected]"}
];

let ownerIdsSet = new Set(owners.map(x => x.userID));
let res = users.filter(x => !ownerIdsSet.has(x.userID));
console.log(res);

.as-console {background-color:black !important; color:lime;}

But why to construct a Set first?

In summary, it will improve the performance of the filtering process, particularly if the owners array is large. You should note that methods like findIndex(), find() and some() needs to traverse the array for check to the related condition while checking if the userID belongs to the Set is a O(1) calculation. However, of course, there will be an extra overload at initialization to create the mentioned Set.

Code Maniac · Accepted Answer · 2019-02-26 17:34:23Z

1

You can use filter and some

const owners = [{userID:"58c4d7ac",username:"John.Doe",firstName:"John",lastName:"Doe",email:"[email protected]"},{userID:"68c4d7ac",username:"User2.Name2",firstName:"User2",lastName:"Name2",email:"[email protected]"}]
const users = [{userID:"58c4d7ac",username:"John.Doe",firstName:"John",lastName:"Doe",email:"[email protected]"},{userID:"68c4d7ac",username:"User2.Name2",firstName:"User2",lastName:"Name2",email:"[email protected]"},{userID:"88c4d7ac",username:"User3.Name3",firstName:"User3",lastName:"Name3",email:"[email protected]"}]

const newUsers = users
                .filter(({userID}) => !owners.some(({userID:ownerID})=> ownerID === userID))

console.log(newUsers)

answered Feb 26, 2019 at 17:34

Code Maniac

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Comments

mhodges · Accepted Answer · 2019-02-26 17:34:41Z

You can just use a single .filter() function, like so:

let owners = [{userID: "58c4d7ac",username: "John.Doe",firstName: "John",lastName: "Doe",email: "[email protected]"},{userID: "68c4d7ac",username: "User2.Name2",firstName: "User2",lastName: "Name2",email: "[email protected]"}];

let users = [{userID: "58c4d7ac",username: "John.Doe",firstName: "John",lastName: "Doe",email: "[email protected]"},{userID: "68c4d7ac",username: "User2.Name2",firstName: "User2",lastName: "Name2",email: "[email protected]"},{userID: "88c4d7ac",username: "User3.Name3",firstName: "User3",lastName: "Name3",email: "[email protected]"}];

let newUsersItems = users.filter(user => owners.findIndex(owner => owner.userID === user.userID) === -1);

console.log(newUsersItems)

You can just use a single .filter() function, like so:

Maxime Girou · Accepted Answer · 2019-02-26 17:40:13Z

0

You can use the reduce function

    const diff = users.reduce((acc, user) => { 
      if(!owners.find(owner => owner.id === user.id){
     acc.push(user);
    }
    return acc;
     }, []);

to avoid use the find() function every loop you can store owner ids in a array with the map an just use includes()

 const ownersIds = owners.map(owner => owner.id);
 const diff = users.reduce((acc, user) => { 
   if(!ownersIds.includes(user.id){
     acc.push(user);
    }
    return acc;
     }, []);

edited Feb 26, 2019 at 17:40

answered Feb 26, 2019 at 17:32

Maxime Girou

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Comments

Ele · Accepted Answer · 2019-02-26 17:53:35Z

You can use the function some or find, this approach uses the function find

let owners = [   {     userID: "58c4d7ac",     username: "John.Doe",     firstName: "John",     lastName: "Doe",     email: "[email protected]"   },   {     userID: "68c4d7ac",     username: "User2.Name2",     firstName: "User2",     lastName: "Name2",     email: "[email protected]"   } ],
    users = [   {     userID: "58c4d7ac",     username: "John.Doe",     firstName: "John",     lastName: "Doe",     email: "[email protected]"   },   {     userID: "68c4d7ac",     username: "User2.Name2",     firstName: "User2",     lastName: "Name2",     email: "[email protected]"   },   {     userID: "88c4d7ac",     username: "User3.Name3",     firstName: "User3",     lastName: "Name3",     email: "[email protected]"   } ],
    result = users.filter(({userID}) => !owners.find(o => o.userID === userID));

console.log(result);

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