Experiencing incorrect outcome when using the Number.isInteger() function in JavaScript

Question

Experiencing incorrect outcome when using the Number.isInteger() function in JavaScript

I have been experimenting with the Number.isInteger() method on the Chrome console. After running a for loop and checking the result using console.log(arr);, I noticed that the array only contains a single value of 1, like this: [1];


var arr = [1, 2, 3, 'some', 'string'];

for (var i = 0; i < arr.length; i++) {
    if (Number.isInteger(arr[i])) {
        arr.splice(arr.indexOf(arr[i], 1));
    }
}

Does anyone have any insight into what I might be doing wrong? Any help would be greatly appreciated. Thank you!

javascript arrays string numbers

Answer 1

Answer №1

You are facing a significant issue where you are continuously deleting items from an array while iterating through it. Each time an item is removed, you need to backtrack by one step (i--).

var arr = [1,2,3,'some','string'];

for (var i = 0; i < arr.length; i++) {
    if (!isNaN(arr[i])) { // To determine if it's a valid number, we have to invert the test since isNaN returns true for non-valid numbers
        arr.splice(i, 1); // If the element at index i is a valid number, remove it (no need to search for the index using indexOf, as we already have it with i)
        i--;              // After removing an element, we must move back one step (otherwise we will skip an item each time we remove another)
    }
}


console.log(arr);

Answer 2

You are facing a significant issue where you are continuously deleting items from an array while iterating through it. Each time an item is removed, you need to backtrack by one step (i--).

var arr = [1,2,3,'some','string'];

for (var i = 0; i < arr.length; i++) {
    if (!isNaN(arr[i])) { // To determine if it's a valid number, we have to invert the test since isNaN returns true for non-valid numbers
        arr.splice(i, 1); // If the element at index i is a valid number, remove it (no need to search for the index using indexOf, as we already have it with i)
        i--;              // After removing an element, we must move back one step (otherwise we will skip an item each time we remove another)
    }
}


console.log(arr);

Answer 3

Answer №2

Why not try utilizing the typeof operator instead?

var elements = [1,2,3,'some','string'];
for (var index = 0; index < elements.length; index++) {
  if (typeof elements[index] == 'number') {
    elements.splice(elements.indexOf(elements[index], 1));
  }
}

Answer 4

Why not try utilizing the typeof operator instead?

var elements = [1,2,3,'some','string'];
for (var index = 0; index < elements.length; index++) {
  if (typeof elements[index] == 'number') {
    elements.splice(elements.indexOf(elements[index], 1));
  }
}

Answer 5

Answer №3

Using the `.splice()' method directly modifies the original array itself.

Therefore, every time you iterate through the array, its core structure changes.

If your goal is to have only integers in the array:

    var arr = [1, 2, 3, 'some', 'string'];

    var newArray = [];

    arr.forEach(function(element) {
        if (Number.isInteger(element)){
            newArray.push(element);
        }
    });

    console.log(newArray);

Answer 6

Using the `.splice()' method directly modifies the original array itself.

Therefore, every time you iterate through the array, its core structure changes.

If your goal is to have only integers in the array:

    var arr = [1, 2, 3, 'some', 'string'];

    var newArray = [];

    arr.forEach(function(element) {
        if (Number.isInteger(element)){
            newArray.push(element);
        }
    });

    console.log(newArray);

Experiencing incorrect outcome when using the Number.isInteger() function in JavaScript

Answer №1

Answer №2

Answer №3

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