631

I have the following for loop, and when I use splice() to remove an item, I then get that 'seconds' is undefined. I could check if it's undefined, but I feel there's probably a more elegant way to do this. The desire is to simply delete an item and keep on going.

for (i = 0, len = Auction.auctions.length; i < len; i++) {
    auction = Auction.auctions[i];
    Auction.auctions[i]['seconds'] --;
    if (auction.seconds < 0) { 
        Auction.auctions.splice(i, 1);
    }           
}
JJJ
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dzm
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18 Answers18

1062

The array is being re-indexed when you do a .splice(), which means you'll skip over an index when one is removed, and your cached .length is obsolete.

To fix it, you'd either need to decrement i after a .splice(), or simply iterate in reverse...

var i = Auction.auctions.length
while (i--) {
    ...
    if (...) { 
        Auction.auctions.splice(i, 1);
    } 
}

This way the re-indexing doesn't affect the next item in the iteration, since the indexing affects only the items from the current point to the end of the Array, and the next item in the iteration is lower than the current point.

Dan Dascalescu
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    Wondering if `length === 0` would end up in an infinite loop, I have tried this solution and (of course it does work), since it evaluates first the value of `i` and then decrements. However, `--` (and `++`) are so weird in their behaviour that a modern language like swift stopped supporting them. I think they are bad practise (at least in a such a context). – lukas_o Apr 22 '21 at 15:27
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    @lukas_o There is no weirdness or unexpected functionality if you simply understand what it means. `i++` mean evaluate the value, then increment it. `++i` means increment the value, then evaluate it. JS will never do anything but that. This is really easy to understand, and it's guaranteed to work exactly the same way every time, even if you use a different JS engine. – aggregate1166877 Jan 04 '22 at 09:52
250

This is a pretty common issue. The solution is to loop backwards:

for (var i = Auction.auctions.length - 1; i >= 0; i--) {
    Auction.auctions[i].seconds--;
    if (Auction.auctions[i].seconds < 0) { 
        Auction.auctions.splice(i, 1);
    }
}

It doesn't matter if you're popping them off of the end because the indices will be preserved as you go backwards.

frattaro
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59

Recalculate the length each time through the loop instead of just at the outset, e.g.:

for (i = 0; i < Auction.auctions.length; i++) {
      auction = Auction.auctions[i];
      Auction.auctions[i]['seconds'] --;
      if (auction.seconds < 0) { 
          Auction.auctions.splice(i, 1);
          i--; //decrement
      }
}

That way you won't exceed the bounds.

EDIT: added a decrement in the if statement.

Marc
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57

Although your question is about deleting elements from the array being iterated upon and not about removing elements (in addition to some other processing) efficiently, I think one should reconsider it if in similar situation.

The algorithmic complexity of this approach is O(n^2) as splice function and the for loop both iterate over the array (splice function shifts all elements of array in the worst case). Instead you can just push the required elements to the new array and then just assign that array to the desired variable (which was just iterated upon).

var newArray = [];
for (var i = 0, len = Auction.auctions.length; i < len; i++) {
    auction = Auction.auctions[i];
    auction.seconds--;
    if (!auction.seconds < 0) { 
        newArray.push(auction);
    }
}
Auction.auctions = newArray;

Since ES2015 we can use Array.prototype.filter to fit it all in one line:

Auction.auctions = Auction.auctions.filter(auction => --auction.seconds >= 0);
dhilt
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0xc0de
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28
Auction.auctions = Auction.auctions.filter(function(el) {
  return --el["seconds"] > 0;
});
Don Hatch
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Aesthete
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24

If you are e using ES6+ - why not just use Array.filter method?

Auction.auctions = Auction.auctions.filter((auction) => {
  auction['seconds'] --;
  return (auction.seconds > 0)
})  

Note that modifying the array element during filter iteration only works for objects and will not work for array of primitive values.

BuZZ-dEE
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Rubinsh
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17

Here is a simple linear time solution to this simple linear time problem.

When I run this snippet, with n = 1 million, each call to filterInPlace() takes .013 to .016 seconds. A quadratic solution (e.g. the accepted answer) would take a million times that, or so.

// Remove from array every item such that !condition(item).
function filterInPlace(array, condition) {
   var iOut = 0;
   for (var i = 0; i < array.length; i++)
     if (condition(array[i]))
       array[iOut++] = array[i];
   array.length = iOut;
}

// Try it out.  A quadratic solution would take a very long time.
var n = 1*1000*1000;
console.log("constructing array...");
var Auction = {auctions: []};
for (var i = 0; i < n; ++i) {
  Auction.auctions.push({seconds:1});
  Auction.auctions.push({seconds:2});
  Auction.auctions.push({seconds:0});
}
console.log("array length should be "+(3*n)+": ", Auction.auctions.length)
filterInPlace(Auction.auctions, function(auction) {return --auction.seconds >= 0; })
console.log("array length should be "+(2*n)+": ", Auction.auctions.length)
filterInPlace(Auction.auctions, function(auction) {return --auction.seconds >= 0; })
console.log("array length should be "+n+": ", Auction.auctions.length)
filterInPlace(Auction.auctions, function(auction) {return --auction.seconds >= 0; })
console.log("array length should be 0: ", Auction.auctions.length)

Note that this modifies the original array in place rather than creating a new array; doing it in place like this can be advantageous, e.g. in the case that the array is the program's single memory bottleneck; in that case, you don't want to create another array of the same size, even temporarily.

ChrisF
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Don Hatch
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15

The normal for loop is more familiar for me, I just need to decrement the index each time I remove an item from the array

//5 trues , 5 falses
var arr1 = [false, false, true, true, false, true, false, true, true, false];

//remove falses from array
for (var i = 0; i < arr1.length; i++){
    if (arr1[i] === false){
        arr1.splice(i, 1);
        i--;// decrement index if item is removed
    }
}
console.log(arr1);// should be 5 trues
Accountant م
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    I find this approach (splice and decrement) with *consideration for familiarity* as most portable/understandable when want to cut small edge case out of for-loop without rewriting loop into filter/reduce/flatmap etc or looping backwards (which may not notice 6 months later). The other solutions are better/smarter, but sometimes just need to patch out small loop edge case. – PotatoFarmer May 05 '22 at 03:03
14

Another simple solution to digest an array elements once:

while(Auction.auctions.length){
    // From first to last...
    var auction = Auction.auctions.shift();
    // From last to first...
    var auction = Auction.auctions.pop();

    // Do stuff with auction
}
Pablo
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11

Here is another example for the proper use of splice. This example is about to remove 'attribute' from 'array'.

for (var i = array.length; i--;) {
    if (array[i] === 'attribute') {
        array.splice(i, 1);
    }
}
daniel.szaniszlo
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1

why waste cpu cycles on .splice? that operation has to go through the whole loop again and again to remove an element in an array.

why not just use traditional 2 flags in one loop?

const elements = [1, 5, 5, 3, 5, 2, 4];
const remove = 5

i = 0

for(let j = 0; j < elements.length; j++){
  if (elements[j] !== remove) {
    elements[i] = elements[j]
    i++
  }
}
elements.length = i
0

Try to relay an array into newArray when looping:

var auctions = Auction.auctions;
var auctionIndex;
var auction;
var newAuctions = [];

for (
  auctionIndex = 0; 
  auctionIndex < Auction.auctions.length;
  auctionIndex++) {

  auction = auctions[auctionIndex];

  if (auction.seconds >= 0) { 
    newAuctions.push(
      auction);
  }    
}

Auction.auctions = newAuctions;
Zon
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0

Two examples that work:

Example ONE

// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    for (var l = temp_products_images.length; l--;) {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}

Example TWO

// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    let l = temp_products_images.length
    while (l--)
    {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}
OneCricketeer
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Fred
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0

Give this a try

RemoveItems.forEach((i, j) => {
    OriginalItems.splice((i - j), 1);
});
Rick
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0

Deleting Parameters

        oldJson=[{firstName:'s1',lastName:'v1'},
                 {firstName:'s2',lastName:'v2'},
                 {firstName:'s3',lastName:'v3'}]
        
        newJson = oldJson.map(({...ele}) => {
          delete ele.firstName;
          return ele;
          })

it deletes and and create new array and as we are using spread operator on each objects so the original array objects are also remains unharmed

Aayush Bhattacharya
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0
    const arr = [...]
    const indexes = [1,2,8,9,34,67]
    
    const result = arr.filter((_, index) => !indexes.includes(index));
    
    console.log(result);
rbrandino
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    Answer needs supporting information Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](https://stackoverflow.com/help/how-to-answer). – moken Aug 04 '23 at 13:38
-1

There are lot of wonderful answers on this thread already. However I wanted to share my experience when I tried to solve "remove nth element from array" in ES5 context.

JavaScript arrays have different methods to add/remove elements from start or end. These are:

arr.push(ele) - To add element(s) at the end of the array 
arr.unshift(ele) - To add element(s) at the beginning of the array
arr.pop() - To remove last element from the array 
arr.shift() - To remove first element from the array 

Essentially none of the above methods can be used directly to remove nth element from the array.

A fact worth noting is that this is in contrast with java iterator's using which it is possible to remove nth element for a collection while iterating.

This basically leaves us with only one array method Array.splice to perform removal of nth element (there are other things you could do with these methods as well, but in the context of this question I am focusing on removal of elements):

Array.splice(index,1) - removes the element at the index 

Here is the code copied from original answer (with comments):

var arr = ["one", "two", "three", "four"];
var i = arr.length; //initialize counter to array length 

while (i--) //decrement counter else it would run into IndexOutBounds exception
{
  if (arr[i] === "four" || arr[i] === "two") {
    //splice modifies the original array
    arr.splice(i, 1); //never runs into IndexOutBounds exception 
    console.log("Element removed. arr: ");

  } else {
    console.log("Element not removed. arr: ");
  }
  console.log(arr);
}

Another noteworthy method is Array.slice. However the return type of this method is the removed elements. Also this doesn't modify original array. Modified code snippet as follows:

var arr = ["one", "two", "three", "four"];
var i = arr.length; //initialize counter to array length 

while (i--) //decrement counter 
{
  if (arr[i] === "four" || arr[i] === "two") {
    console.log("Element removed. arr: ");
    console.log(arr.slice(i, i + 1));
    console.log("Original array: ");
    console.log(arr);
  }
}

Having said that, we can still use Array.slice to remove nth element as shown below. However it is lot more code (hence inefficient)

var arr = ["one", "two", "three", "four"];
var i = arr.length; //initialize counter to array length 

while (i--) //decrement counter 
{
  if (arr[i] === "four" || arr[i] === "two") {
    console.log("Array after removal of ith element: ");
    arr = arr.slice(0, i).concat(arr.slice(i + 1));
    console.log(arr);
  }

}

The Array.slice method is extremely important to achieve immutability in functional programming à la redux

bp4D
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  • Note that more code should not be a measure of a code's efficiency. – kano Apr 02 '20 at 06:43
  • @kano - please read above. It actually says, more code is inefficient. For ones, it's inefficient use of developer time writing it. – bp4D Dec 12 '22 at 19:14
-8

You can just look through and use shift()

Ivan
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