To swap array elements with JavaScript, we use destructuring.
For instance, we write
[arr[0], arr[1]] = [arr[1], arr[0]];
to ass arr[1] to arr[0]and assignarr[0]toarr[1]` with destructuring.
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To swap array elements with JavaScript, we use destructuring.
For instance, we write
[arr[0], arr[1]] = [arr[1], arr[0]];
to ass arr[1] to arr[0]and assignarr[0]toarr[1]` with destructuring.
To loop through an array containing objects and access their properties with JavaScript, we use the forEach method.
For instance, we write
yourArray.forEach((arrayItem) => {
const x = arrayItem.prop + 2;
console.log(x);
});
to call forEach with a callback that gets the prop property value from the arrayItem being looped through.
And then we add 2 to it assign the sum to x.
To reverse array in JavaScript without mutating original array, we copy the array before calling reverse.
For instance, we write
const newArray = [...array].reverse();
to copy the array with the spread operator.
And then we call reverse to reverse the copied array.
To sum a property value in an array with JavaScript, we use the map and reduce methods.
For instance, we write
const traveler = [
{ description: "Senior", amount: 50 },
{ description: "Senior", amount: 50 },
{ description: "Adult", amount: 75 },
{ description: "Child", amount: 35 },
{ description: "Infant", amount: 25 },
];
const sum = traveler
.map((item) => item.amount)
.reduce((prev, next) => prev + next);
to call traveler.map with a callback to get the amount property from each entry and put it in the returned array.
And then we call reduce with a callback to sum up the values from the array returned by map.
To count the occurrences / frequency of array elements with JavaScript, we use a for-of loop.
For instance, we write
const arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const counts = {};
for (const num of arr) {
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
to define the counts object.
Then we use a for-of loop to loop through the arr array and set counts[num] to counts[num] + 1 if it’s already set and 1 otherwise.