I assume you used splice something like this?
Copyfor (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
}
}
All you need to do to fix the bug is decrement i for the next time around, then (and looping backwards is also an option):
Copyfor (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
i--;
}
}
To avoid linear-time deletions, you can write array elements you want to keep over the array:
Copyvar end = 0;
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) === -1) {
arrayOfObjects[end++] = obj;
}
}
arrayOfObjects.length = end;
and to avoid linear-time lookups in a modern runtime, you can use a hash set:
Copyconst setToDelete = new Set(listToDelete);
let end = 0;
for (let i = 0; i < arrayOfObjects.length; i++) {
const obj = arrayOfObjects[i];
if (setToDelete.has(obj.id)) {
arrayOfObjects[end++] = obj;
}
}
arrayOfObjects.length = end;
which can be wrapped up in a nice function:
Copyconst filterInPlace = (array, predicate) => {
let end = 0;
for (let i = 0; i < array.length; i++) {
const obj = array[i];
if (predicate(obj)) {
array[end++] = obj;
}
}
array.length = end;
};
const toDelete = new Set(['abc', 'efg']);
const arrayOfObjects = [{id: 'abc', name: 'oh'},
{id: 'efg', name: 'em'},
{id: 'hij', name: 'ge'}];
filterInPlace(arrayOfObjects, obj => !toDelete.has(obj.id));
console.log(arrayOfObjects);
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If you don’t need to do it in place, that’s Array#filter:
Copyconst toDelete = new Set(['abc', 'efg']);
const newArray = arrayOfObjects.filter(obj => !toDelete.has(obj.id));
Answer from Ry- on Stack OverflowI assume you used splice something like this?
Copyfor (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
}
}
All you need to do to fix the bug is decrement i for the next time around, then (and looping backwards is also an option):
Copyfor (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
i--;
}
}
To avoid linear-time deletions, you can write array elements you want to keep over the array:
Copyvar end = 0;
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) === -1) {
arrayOfObjects[end++] = obj;
}
}
arrayOfObjects.length = end;
and to avoid linear-time lookups in a modern runtime, you can use a hash set:
Copyconst setToDelete = new Set(listToDelete);
let end = 0;
for (let i = 0; i < arrayOfObjects.length; i++) {
const obj = arrayOfObjects[i];
if (setToDelete.has(obj.id)) {
arrayOfObjects[end++] = obj;
}
}
arrayOfObjects.length = end;
which can be wrapped up in a nice function:
Copyconst filterInPlace = (array, predicate) => {
let end = 0;
for (let i = 0; i < array.length; i++) {
const obj = array[i];
if (predicate(obj)) {
array[end++] = obj;
}
}
array.length = end;
};
const toDelete = new Set(['abc', 'efg']);
const arrayOfObjects = [{id: 'abc', name: 'oh'},
{id: 'efg', name: 'em'},
{id: 'hij', name: 'ge'}];
filterInPlace(arrayOfObjects, obj => !toDelete.has(obj.id));
console.log(arrayOfObjects);
Run code snippetEdit code snippet Hide Results Copy to answer Expand
If you don’t need to do it in place, that’s Array#filter:
Copyconst toDelete = new Set(['abc', 'efg']);
const newArray = arrayOfObjects.filter(obj => !toDelete.has(obj.id));
You can remove an item by one of its properties without using any 3rd party libs like this:
Copyvar removeIndex = array.map(item => item.id).indexOf("abc");
~removeIndex && array.splice(removeIndex, 1);
Videos
What a shame you have an array of integers, not an object where the keys are string equivalents of these integers.
I've looked through a lot of these answers and they all seem to use "brute force" as far as I can see. I haven't examined every single one, apologies if this is not so. For a smallish array this is fine, but what if you have 000s of integers in it?
Correct me if I'm wrong, but can't we assume that in a key => value map, of the kind which a JavaScript object is, that the key retrieval mechanism can be assumed to be highly engineered and optimised? (NB: if some super-expert tells me that this is not the case, I can suggest using ECMAScript 6's Map class instead, which certainly will be).
I'm just suggesting that, in certain circumstances, the best solution might be to convert your array to an object... the problem being, of course, that you might have repeating integer values. I suggest putting those in buckets as the "value" part of the key => value entries. (NB: if you are sure you don't have any repeating array elements this can be much simpler: values "same as" keys, and just go Object.values(...) to get back your modified array).
So you could do:
Copyconst arr = [ 1, 2, 55, 3, 2, 4, 55 ];
const f = function( acc, val, currIndex ){
// We have not seen this value before: make a bucket... NB: although val's typeof is 'number',
// there is seamless equivalence between the object key (always string)
// and this variable val.
! ( val in acc ) ? acc[ val ] = []: 0;
// Drop another array index in the bucket
acc[ val ].push( currIndex );
return acc;
}
const myIntsMapObj = arr.reduce( f, {});
console.log( myIntsMapObj );
Output:
Object [ <1 empty slot>, Array1, Array[2], Array1, Array1, <5 empty slots>, 46 more… ]
It is then easy to delete all the numbers 55.
Copydelete myIntsMapObj[ 55 ]; // Again, although keys are strings this works
You don't have to delete them all: index values are pushed into their buckets in order of appearance, so (for example):
CopymyIntsMapObj[ 55 ].shift(); // And
myIntsMapObj[ 55 ].pop();
will delete the first and last occurrence respectively. You can count frequency of occurrence easily, replace all 55s with 3s by transferring the contents of one bucket to another, etc.
Retrieving a modified int array from your "bucket object" is slightly involved but not so much: each bucket contains the index (in the original array) of the value represented by the (string) key. Each of these bucket values is also unique (each is the unique index value in the original array): so you turn them into keys in a new object, with the (real) integer from the "integer string key" as value... then sort the keys and go Object.values( ... ).
This sounds very involved and time-consuming... but obviously everything depends on the circumstances and desired usage. My understanding is that all versions and contexts of JavaScript operate only in one thread, and the thread doesn't "let go", so there could be some horrible congestion with a "brute force" method: caused not so much by the indexOf ops, but multiple repeated slice/splice ops.
Addendum If you're sure this is too much engineering for your use case surely the simplest "brute force" approach is
Copyconst arr = [ 1, 2, 3, 66, 8, 2, 3, 2 ];
const newArray = arr.filter( number => number !== 3 );
console.log( newArray )
(Yes, other answers have spotted Array.prototype.filter...)
Remove one value, using loose comparison, without mutating the original array, ES6
Copy/**
* Removes one instance of `value` from `array`, without mutating the original array. Uses loose comparison.
*
* @param {Array} array Array to remove value from
* @param {*} value Value to remove
* @returns {Array} Array with `value` removed
*/
export function arrayRemove(array, value) {
for(let i=0; i<array.length; ++i) {
if(array[i] == value) {
let copy = [...array];
copy.splice(i, 1);
return copy;
}
}
return array;
}
You can use several methods to remove item(s) from an Array:
//1
someArray.shift(); // first element removed
//2
someArray = someArray.slice(1); // first element removed
//3
someArray.splice(0, 1); // first element removed
//4
someArray.pop(); // last element removed
//5
someArray = someArray.slice(0, someArray.length - 1); // last element removed
//6
someArray.length = someArray.length - 1; // last element removed
If you want to remove element at position x, use:
someArray.splice(x, 1);
Or
someArray = someArray.slice(0, x).concat(someArray.slice(-x));
Reply to the comment of @chill182: you can remove one or more elements from an array using Array.filter, or Array.splice combined with Array.findIndex (see MDN).
See this Stackblitz project or the snippet below:
// non destructive filter > noJohn = John removed, but someArray will not change
let someArray = getArray();
let noJohn = someArray.filter( el => el.name !== "John" );
log(`let noJohn = someArray.filter( el => el.name !== "John")`,
`non destructive filter [noJohn] =`, format(noJohn));
log(`**someArray.length ${someArray.length}`);
// destructive filter/reassign John removed > someArray2 =
let someArray2 = getArray();
someArray2 = someArray2.filter( el => el.name !== "John" );
log("",
`someArray2 = someArray2.filter( el => el.name !== "John" )`,
`destructive filter/reassign John removed [someArray2] =`,
format(someArray2));
log(`**someArray2.length after filter ${someArray2.length}`);
// destructive splice /w findIndex Brian remains > someArray3 =
let someArray3 = getArray();
someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1);
someArray3.splice(someArray3.findIndex(v => v.name === "John"), 1);
log("",
`someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1),`,
`destructive splice /w findIndex Brian remains [someArray3] =`,
format(someArray3));
log(`**someArray3.length after splice ${someArray3.length}`);
// if you're not sure about the contents of your array,
// you should check the results of findIndex first
let someArray4 = getArray();
const indx = someArray4.findIndex(v => v.name === "Michael");
someArray4.splice(indx, indx >= 0 ? 1 : 0);
log("", `someArray4.splice(indx, indx >= 0 ? 1 : 0)`,
`check findIndex result first [someArray4] = (nothing is removed)`,
format(someArray4));
log(`**someArray4.length (should still be 3) ${someArray4.length}`);
// -- helpers --
function format(obj) {
return JSON.stringify(obj, null, " ");
}
function log(...txt) {
document.querySelector("pre").textContent += `${txt.join("\n")}\n`
}
function getArray() {
return [ {name: "Kristian", lines: "2,5,10"},
{name: "John", lines: "1,19,26,96"},
{name: "Brian", lines: "3,9,62,36"} ];
}
<pre>
**Results**
</pre>
The clean solution would be to use Array.filter:
var filtered = someArray.filter(function(el) { return el.Name != "Kristian"; });
The problem with this is that it does not work on IE < 9. However, you can include code from a Javascript library (e.g. underscore.js) that implements this for any browser.
Try like this
var id = 2;
var list = [{
Id: 1,
Name: 'a'
}, {
Id: 2,
Name: 'b'
}, {
Id: 3,
Name: 'c'
}];
var index = list.map(x => {
return x.Id;
}).indexOf(id);
list.splice(index, 1);
console.log(list);
JSFIDDLE
Or you can utilize .filter()
Like this
var id = 2;
var list = [{
Id: 1,
Name: 'a'
}, {
Id: 2,
Name: 'b'
}, {
Id: 3,
Name: 'c'
}];
var lists = list.filter(x => {
return x.Id != id;
})
console.log(lists);
DEMO
Two solutions, one evolve creating new instance and one changes the instance of your array.
Filter:
idToRemove = DESIRED_ID;
myArr = myArr.filter(function(item) {
return item.Id != idToRemove;
});
As you can see, the filter method returns new instance of the filtered array.
Second option is to find the index of the item and then remove it with splice:
idToRemove = DESIRED_ID;
index = myArr.map(function(item) {
return item.Id
}).indexOf(idToRemove);
myArr.splice(index, 1);
Need some help asap, please. I'm trying to loop through this array of objects and remove the entire object if the ID is '' or "ID".
That is, I need to get from this...
"newArr" [
{ ID: '16', Name: 'Abraham Lincoln', Start: '1861', End: '1865' },
{ ID: '', Name: undefined, Start: undefined, End: undefined },
{ ID: 'ID', Name: 'Name', Start: 'Start', End: 'End' },
{ ID: '15', Name: 'James Buchanan', Start: '1857', End: '1861' },
{ ID: '', Name: undefined, Start: undefined, End: undefined }
]
...to this:
[
{ ID: '16', Name: 'Abraham Lincoln', Start: '1861', End: '1865' },
{ ID: '15', Name: 'James Buchanan', Start: '1857', End: '1861' }
]
I either get "typeError: Cannot read properties of undefined (reading 'ID')" or it just does nothing and the object remains. Here's what I've tried:
for (let i=0; i<newArr.length; i++) {
for (let j=0; j<newArr[i].length; j++) {
if (newArr[i][j] == null || undefined || '') delete newArr[i];or
for (let i=0; i<newArr.length; i++) {
Object.keys(newArr[i]).forEach((k) => newArr[i][k] == null || undefined || '' && delete newArr[i]);Well splice works:
var arr = [{id:1,name:'serdar'}];
arr.splice(0,1);
// []
Do NOT use the delete operator on Arrays. delete will not remove an entry from an Array, it will simply replace it with undefined.
var arr = [0,1,2];
delete arr[1];
// [0, undefined, 2]
But maybe you want something like this?
var removeByAttr = function(arr, attr, value){
var i = arr.length;
while(i--){
if( arr[i]
&& arr[i].hasOwnProperty(attr)
&& (arguments.length > 2 && arr[i][attr] === value ) ){
arr.splice(i,1);
}
}
return arr;
}
Just an example below.
var arr = [{id:1,name:'serdar'}, {id:2,name:'alfalfa'},{id:3,name:'joe'}];
removeByAttr(arr, 'id', 1);
// [{id:2,name:'alfalfa'}, {id:3,name:'joe'}]
removeByAttr(arr, 'name', 'joe');
// [{id:2,name:'alfalfa'}]
If you have access to ES2015 functions, and you're looking for a more functional approach I'd go with something like:
const people = [
{ id: 1, name: 'serdar' },
{ id: 5, name: 'alex' },
{ id: 300, name: 'brittany' }
];
const idToRemove = 5;
const filteredPeople = people.filter((item) => item.id !== idToRemove);
// [
// { id: 1, name: 'serdar' },
// { id: 300, name: 'brittany' }
// [
Watch out though, filter() is non-mutating, so you'll get a new array back.
See the Mozilla Developer Network notes on Filter.