What is the best way to swap out every instance of an array?

Question

What is the best way to swap out every instance of an array?

There are two arrays that I'm working with,

The first array is defined as var symbols = ['A', 'B'];, and the second array is defined as var num = ['3', 'A', '5', '4'];

I am looking for a solution to replace each element in the num array that matches an element in the symbols array by adding 10 to the index of that matching element in the symbol array.

So, the expected result would be:

num = ['3', '10', '5', '4']

How can I achieve this replacement for all occurrences?

javascript

Answer 1

Answer №1

Before asking for help, it's always best to attempt finding the solution on your own.

Nevertheless, here is a possible solution:

result = num.map((n) => {
    const index = symbols.indexOf(n);
    return index === -1 ? n : index + 10;
});

Answer 2

Before asking for help, it's always best to attempt finding the solution on your own.

Nevertheless, here is a possible solution:

result = num.map((n) => {
    const index = symbols.indexOf(n);
    return index === -1 ? n : index + 10;
});

Answer 3

Answer №2

There are several ways to achieve this goal, each with varying levels of efficiency.

const symbols = ['A', 'B'];
const numbers = ['3', 'A', '5', '4'];

// Create a new array by filtering out elements that are present in the symbols array
const numbersWithSymbolsRemoved = numbers.filter(item => !symbols.includes(item));
// Now numbersWithSymbolsRemoved = ['3', '10', '5', '4']

// Modify the existing array directly by index
for (let i = 0; i < numbers.length; i++) {
    if (symbols.includes(numbers[i])) {
        numbers.splice(i, 1); // Remove item at current index
    }
}
// Now numbers = ['3', '10', '5', '4']

Answer 4

There are several ways to achieve this goal, each with varying levels of efficiency.

const symbols = ['A', 'B'];
const numbers = ['3', 'A', '5', '4'];

// Create a new array by filtering out elements that are present in the symbols array
const numbersWithSymbolsRemoved = numbers.filter(item => !symbols.includes(item));
// Now numbersWithSymbolsRemoved = ['3', '10', '5', '4']

// Modify the existing array directly by index
for (let i = 0; i < numbers.length; i++) {
    if (symbols.includes(numbers[i])) {
        numbers.splice(i, 1); // Remove item at current index
    }
}
// Now numbers = ['3', '10', '5', '4']

Answer 5

Answer №3

If you want to create a map, consider using an array called symbols where you can store elements along with their corresponding indexes. Then utilize the symbolMap map to generate the desired result in an array named num.

let symbolMap = new Map();
symbols.forEach((symbol, index) => symbolMap.set(symbol, index));

num.forEach((n, index) => {
    if (symbolMap.has(n)) {
        num[index] = symbolMap.get(n) + 10
    }
})

In this process, I am altering the original array. Alternatively, you can opt for using .map() instead of .forEach() if you wish to avoid modifying the original array.

let newNum = num.map((n) => {
    if (symbolMap.has(n)) {
        return symbolMap.get(n) + 10
    }
   return n
})

Answer 6

If you want to create a map, consider using an array called symbols where you can store elements along with their corresponding indexes. Then utilize the symbolMap map to generate the desired result in an array named num.

let symbolMap = new Map();
symbols.forEach((symbol, index) => symbolMap.set(symbol, index));

num.forEach((n, index) => {
    if (symbolMap.has(n)) {
        num[index] = symbolMap.get(n) + 10
    }
})

In this process, I am altering the original array. Alternatively, you can opt for using .map() instead of .forEach() if you wish to avoid modifying the original array.

let newNum = num.map((n) => {
    if (symbolMap.has(n)) {
        return symbolMap.get(n) + 10
    }
   return n
})

What is the best way to swap out every instance of an array?

Answer №1

Answer №2

Answer №3

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