Locate any instances of NaN and include them in the updated array

Question

Locate any instances of NaN and include them in the updated array

I've been tackling the issue of removing duplicates from an array, but I've hit a roadblock when it comes to handling NaN values. How can I identify NaN and ensure it's only added once to a new array?

Although my code works flawlessly in most cases, it struggles with NaN values.

removeDupReduce([1, 2, 1, 3, 1, 4]) // This works correctly
removeDupReduce([NaN, 2, NaN, 3, 1, NaN]) // Currently returns ['NaN', 2, 3, 1], but I'm aiming for [NaN, 2, 3, 1]. I know I converted NaN to a string using toString, but I'm open to hearing other suggestions.

function removeDupReduce(list) {
    return [...list].reduce((acc, curr) => {

        if (acc.indexOf(curr) === -1) {
            if (isNaN(curr)) {
                curr = curr.toString();
                if (acc.indexOf(curr) === -1) {
                    acc.push(curr);
                }
            } else {
                var a = acc.indexOf(curr);
                acc.push(curr);
            }
        }
        return acc;
    }, []);
}

javascript arrays

Answer 1

Answer №1

To locate NaN in an array, you can utilize the array.findIndex method as demonstrated below:

function eliminateDuplicates(arr) {
    return [...arr].reduce((acc, current) => {

        if (acc.indexOf(current) === -1) {
            if (isNaN(current)) {
                if (acc.findIndex(Number.isNaN) === -1) {
                    acc.push(current);
                }
            } else {
                var index = acc.indexOf(current);
                acc.push(current);
            }
        }
        return acc;
    }, []);
}
console.log("eliminateDuplicates([1, 2, 1, 3, 1, 4])");
console.log(eliminateDuplicates([1, 2, 1, 3, 1, 4]));
console.log("eliminateDuplicates([NaN, 2, NaN, 3, 1, NaN])");
console.log(eliminateDuplicates([NaN, 2, NaN, 3, 1, NaN]));

Answer 2

To locate NaN in an array, you can utilize the array.findIndex method as demonstrated below:

function eliminateDuplicates(arr) {
    return [...arr].reduce((acc, current) => {

        if (acc.indexOf(current) === -1) {
            if (isNaN(current)) {
                if (acc.findIndex(Number.isNaN) === -1) {
                    acc.push(current);
                }
            } else {
                var index = acc.indexOf(current);
                acc.push(current);
            }
        }
        return acc;
    }, []);
}
console.log("eliminateDuplicates([1, 2, 1, 3, 1, 4])");
console.log(eliminateDuplicates([1, 2, 1, 3, 1, 4]));
console.log("eliminateDuplicates([NaN, 2, NaN, 3, 1, NaN])");
console.log(eliminateDuplicates([NaN, 2, NaN, 3, 1, NaN]));

Answer 3

Answer №2

To identify NaN values, you can utilize either acc.find(a => a !== a) or acc.find(a => isNaN(a)), as both methods are effective.

removeDuplicateElements([1, 2, 1, 3, 1, 4]) // Working as expected
removeDuplicateElements([NaN, 2, NaN, 3, 1, NaN]) // The current output is ['NaN', 2, 3, 1], but I desire it to be [NaN, 2, 3, 1]. I am aware that I unintentionally converted them to strings using toString, so I'm open to alternative suggestions.

function removeDuplicateElements(list) {
    return [...list].reduce((acc,curr)=>{

        if(acc.indexOf(curr) === -1){
            if(isNaN(curr)){
                if(acc.find(a => a !== a))
                acc.push(curr)
            }
            else{
            var a = acc.indexOf(curr)
            acc.push(curr)
            }
        }
        return acc
    },[])
}

Answer 4

To identify NaN values, you can utilize either acc.find(a => a !== a) or acc.find(a => isNaN(a)), as both methods are effective.

removeDuplicateElements([1, 2, 1, 3, 1, 4]) // Working as expected
removeDuplicateElements([NaN, 2, NaN, 3, 1, NaN]) // The current output is ['NaN', 2, 3, 1], but I desire it to be [NaN, 2, 3, 1]. I am aware that I unintentionally converted them to strings using toString, so I'm open to alternative suggestions.

function removeDuplicateElements(list) {
    return [...list].reduce((acc,curr)=>{

        if(acc.indexOf(curr) === -1){
            if(isNaN(curr)){
                if(acc.find(a => a !== a))
                acc.push(curr)
            }
            else{
            var a = acc.indexOf(curr)
            acc.push(curr)
            }
        }
        return acc
    },[])
}

Answer 5

Answer №3

NaN is actually not equal to NaN. This fact may seem strange, but it is what it is.

indexOf method will compare items one by one to check if they are equal to your value. However, since NaN != NaN, it will always result in -1.

You can test this by running:

[NaN].indexOf(NaN) // this will give you -1

If you want to determine if an array contains NaN, you should use the find method along with isNaN.

if (acc.find(item => isNaN(item)) // this means we have NaN in the array.

Make sure to update your if statement accordingly:

if(acc.indexOf(curr) === -1 && (!isNaN(curr) || !acc.find(item => isNaN(item))) {
 acc.push(curr)
}

Answer 6

NaN is actually not equal to NaN. This fact may seem strange, but it is what it is.

indexOf method will compare items one by one to check if they are equal to your value. However, since NaN != NaN, it will always result in -1.

You can test this by running:

[NaN].indexOf(NaN) // this will give you -1

If you want to determine if an array contains NaN, you should use the find method along with isNaN.

if (acc.find(item => isNaN(item)) // this means we have NaN in the array.

Make sure to update your if statement accordingly:

if(acc.indexOf(curr) === -1 && (!isNaN(curr) || !acc.find(item => isNaN(item))) {
 acc.push(curr)
}

Answer 7

Answer №4

The reason for the error is that your if statement is not accurate. You should verify whether Boolean(curr) is true because Boolean(NaN) will return false. When you simply mention the variable curr within the condition, it automatically converts to Boolean(curr) and evaluates as either true or false. Additionally, you have a redundant condition in your code with acc.indexOf(). I recommend trying out my solution:

console.log(removeDupReduce([1, 2, 1, 3, 1, 4]))
console.log(removeDupReduce([NaN, 2, NaN, 3, 1, NaN]))
function removeDupReduce(list) {
    return [...list].reduce((acc,curr)=> curr && acc.indexOf(curr) === -1 ? [...acc, curr] : acc, [])
}

Answer 8

The reason for the error is that your if statement is not accurate. You should verify whether Boolean(curr) is true because Boolean(NaN) will return false. When you simply mention the variable curr within the condition, it automatically converts to Boolean(curr) and evaluates as either true or false. Additionally, you have a redundant condition in your code with acc.indexOf(). I recommend trying out my solution:

console.log(removeDupReduce([1, 2, 1, 3, 1, 4]))
console.log(removeDupReduce([NaN, 2, NaN, 3, 1, NaN]))
function removeDupReduce(list) {
    return [...list].reduce((acc,curr)=> curr && acc.indexOf(curr) === -1 ? [...acc, curr] : acc, [])
}

Locate any instances of NaN and include them in the updated array

Answer №1

Answer №2

Answer №3

Answer №4

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