Removing object from one array if present in another array based on value

Question

Last question did not get well received as was a possible duplicate. Writing again to explain how this is slightly different as can't get the solution I want.

I have an array called _USERS and another one called newArray.

newArray looks like this:

newArray = ["BLRlXKIeWkanHiSCXbBCFRTIaqk1", "sF4gWbZvAMPmrbeHsKzln7LowOx2"]

_USERS is an array of objects and the objects have a property/attribute called useruid which equals a string. How can I remove an object from _USERS if the useruid string matches ANY string found in the newArray.

Solutions I have tried include:

for (var k = 0; k < newArray.length; k++){
 if (_USERS[j].useruid == newArray[k]){
 _USERS.splice(newArray[k])
 }
var result = _.differenceWith(_USERS, newArray, _.isEqual);

neither of these have worked and I just cant quite put my finger on the missing piece

INITIAL _USERS CODE:

console.log(_USERS) => [Object, Object, Object, Object]

Each object has

gender: "male", name: "Rich", username: "[email protected]", useruid: "BLRlXKIeWkanHiSCXbBCFRTIaqk1"

newArray = ["BLRlXKIeWkanHiSCXbBCFRTIaqk1", "sF4gWbZvAMPmrbeHsKzln7LowOx2"]

newArray[0] = a string. This string matches the useruid in the Rich object. therefore I would like that to be deleted and then the below to happen

console.log(_USERS) => [Object, Object, Object]

• Denys Séguret: Is a [lodash] tag missing ?
• Xotic750: Do you have any sample _USERS data? What was wrong with the result from the code that you have posted?
• Waterman1: @Xotic750 the code just gave me back my same initial _USERS array
• Waterman1: @DenysSéguret nope, i have installed lodash
• Denys Séguret: @Waterman1 What I mean is that if your code uses lodash, then you should add the lodash tag to that question.

Answer 1

It looks like a simple filter to me:

let filteredUsers = _USERS.filter(u => !newArray.includes(u.useruid))

Here it is in action:

var newArray = ["BLRlXKIeWkanHiSCXbBCFRTIaqk1", "sF4gWbZvAMPmrbeHsKzln7LowOx2"]
var _USERS = [{ gender: "male", name: "Rich", username: "[email protected]", useruid: "BLRlXKIeWkanHiSCXbBCFRTIaqk1" }]
let filteredUsers = _USERS.filter(u => !newArray.includes(u.useruid));
/*Printing*/
/*Before*/
document.write("Before: <br>" + JSON.stringify(_USERS) + "<br><br>");
/*After*/
document.write("After <br>" + JSON.stringify(filteredUsers));

• Xotic750: It should be noted that this does not not mutate the original array, rather creates a new one.

Answer 2

Here's a simple solution if I understand your question:

_USERS = _USERS.filter(u => newArray.indexOf(u.useruid)===-1);

• Waterman1: when i console.log this it still gives me the original _USERS array
• Denys Séguret: Then you have a problem with your data. Can you build a MCVE ?
• Waterman1: the problem is, I am pulling data from firebase so it's kinda hard to show. what would you like to see?

Answer 3

There are several clean ways to do this, but just to comment on your splice implementation. Splice expects the first parameter to be an index of an element not an object, so if you call splice(1) then it will delete from the object at index 1 to the end of the array. The second parameter is the number of elements that should be deleted, so splice(1, 1) will delete only the element at index 1. Here is some sample code that may help:

var newArray = ["xyz", "def"];
var _USERS = [
{useruid: "abc"},
{useruid: "def"},
{useruid: "ghi"}
];
for(var i = 0; i < _USERS.length; i++)
{
 for(var j = 0; j < newArray.length; j++)
 {
 if(_USERS[i].useruid == newArray[j])
 {
 _USERS.splice(i, 1); //remove element at index i
 i--; //decrement i because we now removed an element
 break; //move to next element
 }
 }
}

JSFiddle For Reference

• Waterman1: this should work but it still returns the same _USERS. what level should I console.log out to check?
• Xotic750: Forward running loops that mutate the array over which they are looping have issues where some items may not get iterated, hence the i-- in this solution,

Answer 4

To mutate your original array with a loop and Array#splice you need to run your loop in reverse as the length of the array and indexes of items in your array will change when you remove an item, and hence you will miss some of the items. Examples in ES6.

With a forward running loop, incorrect result.

function customRemove(users, unwanted) {
 for (let user of users.entries()) {
 if (unwanted.includes(user[1].useruid)) {
 users.splice(user[0], 1);
 }
 }
}
const newArray = ['BLRlXKIeWkanHiSCXbBCFRTIaqk1', 'sF4gWbZvAMPmrbeHsKzln7LowOx2'];
const _USERS = [{
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk1'
}, {
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk1'
}, {
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk2'
}, {
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk1'
}];
console.log(JSON.stringify(_USERS));
customRemove(_USERS, newArray);
console.log(JSON.stringify(_USERS));

With a reverse running loop, correct result.

// Generator creates index number in reverse
function* reverseIndexes(arr) {
 let key = arr.length - 1;
 while (key >= 0) {
 yield key;
 key -= 1;
 }
}
// Generator like Array#entries but in reverse
function* reverseEntries(arr) {
 for (let key of reverseIndexes(arr)) {
 yield [key, arr[key]];
 }
}
function customRemove(users, unwanted) {
 // Reversed loop removing unwanted matches with Array#splice
 for (let user of reverseEntries(users)) {
 if (unwanted.includes(user[1].useruid)) {
 users.splice(user[0], 1);
 }
 }
}
const newArray = ['BLRlXKIeWkanHiSCXbBCFRTIaqk1', 'sF4gWbZvAMPmrbeHsKzln7LowOx2'];
const _USERS = [{
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk1'
}, {
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk1'
}, {
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk2'
}, {
 useruid: 'BLRlXKIeWkanHiSCXbBCFRTIaqk1'
}];
console.log(JSON.stringify(_USERS));
customRemove(_USERS, newArray);
console.log(JSON.stringify(_USERS));

Answer 5

Well just to expand on @Denys answer. I have taken the .filter approach and added it to a jsfiddle. I've only included useruid for a user but that should be enough to filter as required.