Removing an array of indexes from an given array

Question

So I am having a bit of difficulty trying to think of the best way of doing this in javascript. I want to remove an array of unsorted indexes from an array that was X number of elements. For example

var index = [ 0, 7, 10, 2, 5, 11] array = [{field0: 0}, {field1: 1}, {field2: 2}, ... {field5: 5}, {field6: 6}...]

So I tried using a nested for loop with splice, but then when I splice, my array loses its indexing and screws up.

The end result should come out to be like array = [{field1: 1}, {field3: 3}, {field4: 4}, .... {field6: 6} ...]

Any help would be greatly appreciated.

Is there a reason you cannot sort the indices before pulling the objects from the array? — Bennett Adams
– Bennett Adams, Commented Feb 6, 2016 at 20:05
No reason, I can sort just wanted to see if there would be any difference for sorted indexes vs non sorted. But lets just assume it is sorted. — Julius Doan
– Julius Doan, Commented Feb 6, 2016 at 20:07

georg · Accepted Answer · 2016-02-06 20:38:31Z

2

The solution to your problem is really a one-liner:

array = [00,11,22,33,44,55,66,77];
indexes = [1,7,5,3];

array = array.filter(function(_,i) { return indexes.indexOf(i) < 0 });

document.write('<pre>'+JSON.stringify(array,0,3));

edited Feb 6, 2016 at 20:38

answered Feb 6, 2016 at 20:32

georg

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Bennett Adams · Accepted Answer · 2016-02-06 20:34:54Z

1

If sorting the array of elements to remove is not an issue, you can just do the following:

  var array = [{field0: 0}, {field1: 1}, {field2: 2}, {field3: 3}, {field4: 4}, {field5: 5}, {field6: 6}, {field7: 7}, {field8: 8}, {field9: 9}];
  var indices = [0, 7, 2, 5];

  indices.sort(function(a, b) {
      if ( a > b ) { 
          return 1;
      } else if ( b > a ) { 
          return -1;
      } else {
          return 0;
      }
  });
  for (var i = 0, offset = 0; i < indices.length; i++, offset++) {   
      array.splice(indices[i] - offset, 1);
  }
  console.log(array);

And if for some reason you were unable to sort the array of indices to remove, you could keep track of the offset with the following:

  var array = [{field0: 0}, {field1: 1}, {field2: 2}, {field3: 3}, {field4: 4}, {field5: 5}, {field6: 6}, {field7: 7}, {field8: 8}, {field9: 9}];
  var indices = [0, 7, 2, 5];

  var removedIndices = [];
  function calcOffset(val) {
      var numRemoved = 0;
      for (var j = 0; j < removedIndices.length; j++) {
          if (val > removedIndices[j]) {
              numRemoved++;
          }
      }
      return numRemoved;
  }

  for (var i = 0, offset = 0; i < indices.length; i++, offset++) { 
      var offset = calcOffset(indices[i]);  
      array.splice(indices[i] - offset, 1);
      removedIndices.push(indices[i]);
  }

  console.log(array);

edited Feb 6, 2016 at 20:34

answered Feb 6, 2016 at 20:16

Bennett Adams

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2 Comments

Bennett Adams Over a year ago

@georg's answer is much better than this one :)

Julius Doan Over a year ago

I was hesitant to use it since I know the indexOf and the filter isn't in earlier versions of IE 8 or 9, but your choice lol.

Dave Bush · Accepted Answer · 2016-02-06 20:23:14Z

0

I would iterate through array (for/next loop) and at each element see if the number exist in index (index.indexOf(n)). If it does, push the element to a new temporary array.

when you are done, either copy the temp array back to array OR array.length = 0 and push the elements from the temp array back onto the original array.

something like this:

var index = [ 0, 7, 10, 2, 5, 11]
var array = [{field0: 0}, {field1: 1}, {field2: 2},{field3: 3}, {field4: 4}];
var tempArray = [];

for(var arrayIndex = 0;arrayIndex < array.length;arrayIndex++){
    if(index.indexOf(arrayIndex) === -1)
    {
        tempArray.push(array[arrayIndex]);
    }
}
array = tempArray;

edited Feb 6, 2016 at 20:23

answered Feb 6, 2016 at 20:08

Dave Bush

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3 Comments

Julius Doan Over a year ago

That just gets the results that are equal to index. I want the opposite. of removing the elements that are in the index array. Maybe some quick code may help me understand you better. Thanks

Dave Bush Over a year ago

OK.. sorry. if it doesn't, push the element. I'll work on some code.

Julius Doan Over a year ago

Thanks but was able to use Bennett's answer.

Sirjan Sharma · Accepted Answer · 2016-02-06 20:41:15Z

0

You can try deleting them with delete array[index[count]] by looping as long as undefined is not a problem.

Or, you can set loop through all the elements in the array and set any element whose index is not present to false. Then, loop through them again and copy only the ones which are not false to new array.

newArray = new Array();

for (var i = 0; i < array.length; i++) {
  for(var j=0; j<index.length; j++) {
     if (index[j] == i) {
        array[i] = false;
     }
  }
};

for (var i = 0; i < array.length; i++) {
     if (array[i] != false) {
        newArray[i] = array[i];
     }
  }

answered Feb 6, 2016 at 20:41

Sirjan Sharma

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2 Comments

Dave Bush Over a year ago

that's not what the delete keyword is for.

Dave Bush Over a year ago

not really, and it isn't future proof. The delete keyword removes properties from objects. developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

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