I am trying to return the average. How can I get only the numbers and leave out false, "str"?
function average2 () {
console.log (`arguments passed: ${arguments.length}`)
console.log("All arguments:");
let sum = 0;
let numOnly = arguments.match(/\d+/)[0]
for (let i= 0; i < arguments.length; i++) {
sum += arguments[i]
}
return (sum) / arguments.length
}
console.log (average2 (5, 6, false, "str", 100))
You can filter numbers by using typeof operator, like this:
function average (...params) {
const nums = params.filter(n => typeof n == 'number');
return nums.reduce((acc,n) => acc+n) / nums.length;
}
console.log (average (5, 6, false, "str", 7))in this example I just calculated average of nums, if you want to calculate average based on arguments length, you can replace return nums.reduce((acc,n) => acc+n) / nums.length; with return nums.reduce((acc,n) => acc+n) / params.length;
You can use this:
function average() {
const numericArgs = [...arguments].filter(x => typeof x === 'number');
return numericArgs.reduce((a, b) => a + b, 0) / numericArgs.length;
}
And you can use this:
function average(...args) {
const numericArgs = args.filter(x => typeof x === 'number');
return numericArgs.reduce((a, b) => a + b, 0) / numericArgs.length;
}
Just add a type check before doing the addition:
function average2 () {
let sum = 0;
for (let i= 0; i < arguments.length; i++) {
if(typeof arguments[i] === "number"){
sum += arguments[i]
}
}
return (sum) / arguments.length
}
console.log (average2 (5, 6, false, "str", 100))