The use of array.splice() gives me back the specific item I'm wanting to remove, instead of just the array without

Question

The use of array.splice() gives me back the specific item I'm wanting to remove, instead of just the array without

I attempted to eliminate an element from an array by utilizing the indexOf() combined with the splice() method as recommended. However, the outcome is not as expected.

let someArray: string[] = [first, second, third, fourth, fifth, sixth];
let newArray: string[] = someArray.splice(3, 1);

console.log(newArray);

//desired output = [first, second, third, fifth, sixth]

//actual result = [fourth]

This does not align with the information provided in most articles I have come across. Can anyone provide clarification on this issue?

UPDATE I encountered this discrepancy in my code when I received only a single result instead of the expected multiple results, leading me back to this particular point.

javascript typescript

Answer 1

Answer №1

When you splice an array, you are modifying it by changing the original array. The element being removed is stored in a new variable called "mine". For example:

var arr = [1, 2, 3, 4];

var mine = arr.splice(1, 1);
console.log(mine);
console.log(arr);

If we print "arr", it will show the original array without index one, and printing "mine" will display [2]. To achieve a different output without mutating the original array, you can iterate through the array and use splice differently like this:

var arr = [1, 2, 3, 4];

var mine = [];

for(var i = 0; i < arr.length; i++) {
    if(i !== 3) {
        mine.push(arr[i]);
    }
}

In this way, the original array remains unchanged as elements are pushed to a new array. If you simply want to mutate the original array, you can directly splice it:

var arr = [1, 2, 3, 4];

arr.splice(3, 1);
console.log(arr);

It's advisable not to mutate arrays outside of functions. Instead, return a value from the function and store it in a new variable:

var arr = [1, 2, 3, 4];


function deleteIndex(ar, i) {
    var a = [];
    ar.forEach(function(elt, index) {
        if(index === i) {

        }
        else {
            a.push(elt);
        }
    });
    return a;
}

var newArr = deleteIndex(arr, 3);
console.log(newArr);

This approach allows flexible deletion based on any criteria without altering the original array structure. Utilizing functional programming or underscore module functions can also assist in these operations.

Answer 2

When you splice an array, you are modifying it by changing the original array. The element being removed is stored in a new variable called "mine". For example:

var arr = [1, 2, 3, 4];

var mine = arr.splice(1, 1);
console.log(mine);
console.log(arr);

If we print "arr", it will show the original array without index one, and printing "mine" will display [2]. To achieve a different output without mutating the original array, you can iterate through the array and use splice differently like this:

var arr = [1, 2, 3, 4];

var mine = [];

for(var i = 0; i < arr.length; i++) {
    if(i !== 3) {
        mine.push(arr[i]);
    }
}

In this way, the original array remains unchanged as elements are pushed to a new array. If you simply want to mutate the original array, you can directly splice it:

var arr = [1, 2, 3, 4];

arr.splice(3, 1);
console.log(arr);

It's advisable not to mutate arrays outside of functions. Instead, return a value from the function and store it in a new variable:

var arr = [1, 2, 3, 4];


function deleteIndex(ar, i) {
    var a = [];
    ar.forEach(function(elt, index) {
        if(index === i) {

        }
        else {
            a.push(elt);
        }
    });
    return a;
}

var newArr = deleteIndex(arr, 3);
console.log(newArr);

This approach allows flexible deletion based on any criteria without altering the original array structure. Utilizing functional programming or underscore module functions can also assist in these operations.

The use of array.splice() gives me back the specific item I'm wanting to remove, instead of just the array without

Answer №1

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