Say you have a simple array with random numbers, given the following code, how would i reduce the line of code to use a ternary operator to render the following.
[1, 2, 1, 4, 6]
let arr = [1, 2, 3, 4, 6, 8, 9, 12, 13, 15];
let myArray = arr.map((val, i, arr) => {
if (val % 2 === 0) {
return val
} else {
return val % 2
}
});
console.log(myArray)Logical OR can do the trick for you.
let arr = [1,2,3,4,6,8,9,12,13,15]
let myArray = arr.map((val) => val % 2 || val);
console.log(myArray)You could take a bitwise AND & or (with logical OR ||) the value, instead of a conditional (ternary) operator ?:.
Bitwise AND with a value of one returns either one or zero, depending on the other operand.
var array = [1, 2, 3, 4, 6, 8, 9, 12, 13, 15],
result = array.map(v => v & 1 || v);
console.log(result);Just write the ternary condition as a single liner:
const arr = [1,2,3,4,6,8,9,12,13,15];
const myArray = arr.map((val) => val % 2 === 0 ? val: val % 2);
console.log(myArray);
const instead of let as you are not reassigning the arr nor the myArray . Also why add i and arr in the map if you are not using them ? also arr is used before so you shouldn't use 'shadow' naming.
return val % 2, notval * 2(though it's exactly the same in all other respects)