I need help creating the code to find the factorial of a number. The task is to
Create a variable to store your answer and initialize it to one
Create a loop that beings at the given value, fact
Check if fact is one or zero
multiply fact with your answer variable
At the end of the loop decrease fact
Print answer using console.log
The pseudocode is
while(factorial)
if factorial == 0 or factorial == 1
break
result => result * factorial
factorial => factorial - 1
My code below isn't complete because I'm confused by the pseudocode.
function nth_fact(nth){
var a = 1
while(nth_fact)
if (nth_fact == 0 || nth_fact == 1){
break;
result => result * nth_fact
nth_fact => nth - 1
console.log()
}
}
At first lets examine what went wrong:
var a = 1
What is a? Its definetly not a good name for a variable. Maybe name it to result ? The same applies to nth which should be named factorial and nth_fact which should rather be factorize or sth. You should also always use ; to end a statement.
while(nth_fact)
As your while loop contains multiple statements (the if and the two assignments) you need to open a block here by using { right after the condition. nth_fact refers to the function, you rather want to take factorial here.
if (nth_fact == 0 || nth_fact == 1){
break;
Now you open a block statement for the if, but you never close it. So you need another } after the break.
result => result * nth_fact
nth_fact => nth - 1
console.log()
=> is the arrow function expression, but you want the assignment operator =. Also you need to pass something to console.log, e.g. console.log(result)
All together:
function factorize(factorial){
var result = 1;
while(factorial){
if (factorial == 0 || factorial == 1){
break;
}
// ?
factorial = factorial - 1;
console.log(result);
}
return result;
}
That pseudocode is indeed confusing, because what it calls factorial is actually not the factorial -- it's the current value, which the result (which is actually the factorial we're looking for) is multiplied by. Also, if is superfluous, because while already checks for the same condition. So the correct pseudocode would be
currentValue = argument
factorial = 1
while (currentValue > 1)
factorial = factorial * currentValue
currentValue = currentValue - 1
// now, 'factorial' is the factorial of the 'argument'
Once you get this sorted out, here's a bonus assignment:
create a function range(a, b) that creates an array of numbers from a to b. For example, range(5, 8) => [5, 6, 7, 8]
create a function product(array) that multiples array elements by each other. For example, product([2, 3, 7]) => 42
write the factorial function using product and range
I solve this this way
function factorial(number) {
let num = 1;
let result = 1;
while (num <= number) {
result = result * num;
num++;
}
return result;
}
const myNumber = factorial(6);
console.log(myNumber);
function factorial(num) {
var result = 1
while (num) {
if ((num) == 0 || (num) == 1) {
break;
} else {
result = result * num;
num = num - 1;
}
}
return `The factorial of ${val} is ${result}`
}
let val = prompt("Please Enter the number : ", "0");
var x = parseInt(val);
console.log(factorial(x));
A Short And Clean Code is :
let number = 5;
let numberFactorial = number;
while(number > 1){
numberFactorial = numberFactorial * (number-1);
number--;
}
console.log(numberFactorial);
function factorize(factorial) {
if(factorial == 0 | factorial == 1) {
return 1
}
else{
var result = factorial;
while(factorial >= 1 ){
if(factorial-1 == 0) {
break
};
result = result * (factorial - 1);
factorial = factorial-1;
//DEBUG: console.log(factorial + ' ' + result);
};
return(result);
}
}
If you want more info about functions, can see in my GitHub, good learning!
Github: https://github.com/bennarthurdev/JavaScript/tree/main/FUNCOES
You used the right approach. Just the syntax was wrong. Here it is:
function nth_fact(nth){
var result = 1 ;
while(nth){
if ((nth) == 0 || (nth) == 1)
break ;
result = result * nth;
nth = nth - 1
}
console.log(result);
return result;
}
var toyProblem = function () {
var sol= 0;
var operators = ['+','-','*','/'];
console.log(sol)
for(var i in arguments){
for(var j in operators){
sol = eval(sol + (operators[j]) + arguments[i]);
}
}
return sol;
}
toyProblem(6, 0, 10, 3); //6 + 0 - 10 * 3 === -12)
I'm trying to loop through 4 math operators for an unknown number of input values. I'm thinking of using eval in a nest for loop as a way of going through both the unknown number of arguments while also changing the math operator. At the bottom is the solution that I want to arrive at. Is this a good way of going about this problem or am I barking up the wrong tree?
var toyProblem = function () {
var sol_str='';
var operators = ['+','-','*','/'];
for(var i in operators){
var prev_operator=(i-1);
if(sol_str!=''){sol_str+=operators[prev_operator];}
sol_str +=arguments[i];
}
console.log(sol_str);
return eval(sol_str);
}
console.log(toyProblem(6, 0, 10, 3));
Nesting the 2 loops will result in doing 6 + 6 - 6 * 6 / 6 + 0 - 0 * 0 / 0 + 10 - 10 * 10 / 10 + 3 - 3 * 3 / 3
I didn't find a way to do this without eval as looping through operations one by one would modify the operators priority so this is what I propose : Building an operation 'query' to be eval'd and returned.
Hope this helps
var toyProblem = function () {
var operation = '';
var operators = ['+','-','*','/'];
var args = Array.from(arguments);
args.forEach(function (arg, index) {
if (index > 0) {
operation += operators[index - 1];
}
operation += arg;
});
return eval(operation);
}
console.log(6 + 0 - 10 * 3);
console.log(toyProblem(6, 0, 10, 3)); //6 + 0 - 10 * 3 === -24)
Let's decompose the problem. You have a variadic function that accepts unknown number of arguments and applies an operator to each next argument depending on the index of that element.
Because the number of arguments can be greater than the number of operators, it's appropriate to use modulo operator to infinitely loop through the array of operators while going once through the list of arguments.
The eval operation takes a string, evaluates it, and returns the result of evaluation of the expression that string represents. So you're on the right track. But because eval function takes a string as the first argument, I'd recommend using template literals, it's supported in almost all browsers natively and doesn't need to be transpiled into good old ES5.
The function then would look like this:
function toyProblem(first = 0, ...args) {
const operators = ['+', '-', '*', '/'];
let sol = first;
for (let i in args) {
sol = eval(`${sol} ${operators[i % operators.length]} ${args[i]}`);
}
return sol;
}
However, as there is recommended in the comments, using eval isn't something you'd like to ship to users. Instead, I'd suggest using functions. Functions in Javascript are first-class citizens, so you can pass them as an argument.
Imagine that you have a function (a, b) => a + b instead of just a string "+". The code would then look like this:
function toyProblem(first = 0, ...args) {
const operations = [
(a, b) => a + b,
(a, b) => a - b,
(a, b) => a * b,
(a, b) => a / b,
];
let sol = first;
for (let i in args) {
sol = operations[i](sol, args[i]);
}
return sol;
}
You could go even further and make the function universal in terms of possible operations. Have fun!
Since you are hard coding the names of the operators you might as well hard code the functions and avoid eval. You put the functions into an array that will let you loop through. Then you can just reduce through the arguments with a simple one-liner, which will handle any amount of arguments:
const op = [
(a, b) => a + b,
(a, b) => a - b,
(a, b) => a * b,
(a, b) => a / b
]
function prob(...args){
return args.reduce((curr, acc, idx) => op[(idx - 1) % op.length](curr, acc))
}
console.log(prob(6, 0, 10, 3))
console.log(prob(6, 0, 10, 3, 20, 11, 15, 100))
To get product -12 the first three parts of the expression need to be evaluated within parentheses, else the result will be -24. You can use String.prototype.replace() to replace "," characters after calling .toString() on input array, replace the "," with the operator, return the expression (6 + 0 - 10) * 3 from Function() constructor
var toyProblem = function () {
var operators = ['+','-','*'];
var opts = operators.slice(0);
var n = [...arguments];
var res = n.toString().replace(/,/g, () => opts.shift());
var index = res.indexOf(operators[operators.length -1]);
return new Function(`return (${res.slice(0, index)})${res.slice(index)}`)();
}
var product = toyProblem(6, 0, 10, 3);
console.log(product);
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This question already has answers here:
How to find the sum of an array of numbers
(59 answers)
Closed 3 months ago.
I am having problems adding all the elements of an array as well as averaging them out. How would I do this and implement it with the code I currently have? The elements are supposed to be defined as I have it below.
<script type="text/javascript">
//<![CDATA[
var i;
var elmt = new Array();
elmt[0] = "0";
elmt[1] = "1";
elmt[2] = "2";
elmt[3] = "3";
elmt[4] = "4";
elmt[5] = "7";
elmt[6] = "8";
elmt[7] = "9";
elmt[8] = "10";
elmt[9] = "11";
// Problem here
for (i = 9; i < 10; i++){
document.write("The sum of all the elements is: " + /* Problem here */ + " The average of all the elements is: " + /* Problem here */ + "<br/>");
}
//]]>
</script>
A solution I consider more elegant:
const sum = times.reduce((a, b) => a + b, 0);
const avg = (sum / times.length) || 0;
console.log(`The sum is: ${sum}. The average is: ${avg}.`);
ES6
const average = arr => arr.reduce( ( p, c ) => p + c, 0 ) / arr.length;
const result = average( [ 4, 4, 5, 6, 6 ] ); // 5
console.log(result);
var sum = 0;
for( var i = 0; i < elmt.length; i++ ){
sum += parseInt( elmt[i], 10 ); //don't forget to add the base
}
var avg = sum/elmt.length;
document.write( "The sum of all the elements is: " + sum + " The average is: " + avg );
Just iterate through the array, since your values are strings, they have to be converted to an integer first. And average is just the sum of values divided by the number of values.
Calculating average (mean) using reduce and ES6:
const average = list => list.reduce((prev, curr) => prev + curr) / list.length;
const list = [0, 10, 20, 30]
average(list) // 15
Shortest one liner for Average
const avg = arr => arr.reduce((acc,v,i,a)=>(acc+v/a.length),0);
Shortest one liner for Sum
const sum = arr => arr.reduce((a,b)=>a+b);
Let's imagine we have an array of integers like this:
var values = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
The average is obtained with the following formula
A= (1/n)Σxi ( with i = 1 to n ) ... So: x1/n + x2/n + ... + xn/n
We divide the current value by the number of values and add the previous result to the returned value.
The reduce method signature is
reduce(callback[,default_previous_value])
The reduce callback function takes the following parameters:
p : Result
of the previous calculation
c : Current value (from the current index)
i : Current array element's index value
a : The current reduced Array
The second reduce's parameter is the default value ... (Used in case the array is empty ).
So the average reduce method will be:
var avg = values.reduce(function(p,c,i,a){return p + (c/a.length)},0);
If you prefer you can create a separate function
function average(p,c,i,a){return p + (c/a.length)};
function sum(p,c){return p + c)};
And then simply refer to the callback method signature
var avg = values.reduce(average,0);
var sum= values.reduce(sum,0);
Or Augment the Array prototype directly..
Array.prototype.sum = Array.prototype.sum || function (){
return this.reduce(function(p,c){return p+c},0);
};
It's possible to divide the value each time the reduce method is called..
Array.prototype.avg = Array.prototype.avg || function () {
return this.reduce(function(p,c,i,a){return p+(c/a.length)},0);
};
Or even better , using the previously defined Array.protoype.sum()
method, optimize the process my calling the division only once :)
Array.prototype.avg = Array.prototype.avg || function () {
return this.sum()/this.length;
};
Then on any Array object of the scope:
[2, 6].avg();// -> 4
[2, 6].sum();// -> 8
NB: an empty array with return a NaN wish is more correct than 0 in my point of view and can be useful in specific use cases.
generally average using one-liner reduce is like this
elements.reduce(function(sum, a,i,ar) { sum += a; return i==ar.length-1?(ar.length==0?0:sum/ar.length):sum},0);
specifically to question asked
elements.reduce(function(sum, a,i,ar) { sum += parseFloat(a); return i==ar.length-1?(ar.length==0?0:sum/ar.length):sum},0);
an efficient version is like
elements.reduce(function(sum, a) { return sum + a },0)/(elements.length||1);
Understand Javascript Array Reduce in 1 Minute
http://www.airpair.com/javascript/javascript-array-reduce
as gotofritz pointed out seems Array.reduce skips undefined values.
so here is a fix:
(function average(arr){var finalstate=arr.reduce(function(state,a) { state.sum+=a;state.count+=1; return state },{sum:0,count:0}); return finalstate.sum/finalstate.count})([2,,,6])
You can also use lodash, _.sum(array) and _.mean(array) in Math part (also have other convenient stuff).
_.sum([4, 2, 8, 6]);
// => 20
_.mean([4, 2, 8, 6]);
// => 5
Not the fastest, but the shortest and in one line is using map() & reduce():
var average = [7,14,21].map(function(x,i,arr){return x/arr.length}).reduce(function(a,b){return a + b})
I use these methods in my personal library:
Array.prototype.sum = Array.prototype.sum || function() {
return this.reduce(function(sum, a) { return sum + Number(a) }, 0);
}
Array.prototype.average = Array.prototype.average || function() {
return this.sum() / (this.length || 1);
}
EDIT:
To use them, simply ask the array for its sum or average, like:
[1,2,3].sum() // = 6
[1,2,3].average() // = 2
In ES6-ready browsers this polyfill may be helpful.
Math.sum = (...a) => Array.prototype.reduce.call(a,(a,b) => a+b)
Math.avg = (...a) => Math.sum(...a)/a.length;
You can share same call method between Math.sum,Math.avg and Math.max,such as
var maxOne = Math.max(1,2,3,4) // 4;
you can use Math.sum as
var sumNum = Math.sum(1,2,3,4) // 10
or if you have an array to sum up,you can use
var sumNum = Math.sum.apply(null,[1,2,3,4]) // 10
just like
var maxOne = Math.max.apply(null,[1,2,3,4]) // 4
One sneaky way you could do it although it does require the use of (the much hated) eval().
var sum = eval(elmt.join('+')), avg = sum / elmt.length;
document.write("The sum of all the elements is: " + sum + " The average of all the elements is: " + avg + "<br/>");
Just thought I'd post this as one of those 'outside the box' options. You never know, the slyness might grant you (or taketh away) a point.
Here is a quick addition to the “Math” object in javascript to add a “average” command to it!!
Math.average = function(input) {
this.output = 0;
for (this.i = 0; this.i < input.length; this.i++) {
this.output+=Number(input[this.i]);
}
return this.output/input.length;
}
Then i have this addition to the “Math” object for getting the sum!
Math.sum = function(input) {
this.output = 0;
for (this.i = 0; this.i < input.length; this.i++) {
this.output+=Number(input[this.i]);
}
return this.output;
}
So then all you do is
alert(Math.sum([5,5,5])); //alerts “15”
alert(Math.average([10,0,5])); //alerts “5”
And where i put the placeholder array just pass in your variable (The input if they are numbers can be a string because of it parsing to a number!)
I found Mansilla's answer to work fine with the extension of making sure that I am doing summation of floats and not concatonation of strings using parseFloat():
let sum = ourarray.reduce((a, b) => parseFloat(a) + parseFloat(b), 0);
let avg = (sum / ourarray.length) || 0;
console.log(sum); // print out sum
console.log(avg); // print out avg
set your for loop counter to 0.... you're getting element 9 and then you're done as you have it now. The other answers are basic math. Use a variable to store your sum (need to cast the strings to ints), and divide by your array length.
Start by defining all of the variables we plan on using. You'll note that for the numbers array, I'm using the literal notation of [] as opposed to the constructor method array(). Additionally, I'm using a shorter method to set multiple variables to 0.
var numbers = [], count = sum = avg = 0;
Next I'm populating my empty numbers array with the values 0 through 11. This is to get me to your original starting point. Note how I'm pushing onto the array count++. This pushing the current value of count, and then increments it for the next time around.
while ( count < 12 )
numbers.push( count++ );
Lastly, I'm performing a function "for each" of the numbers in the numbers array. This function will handle one number at a time, which I'm identifying as "n" within the function body.
numbers.forEach(function(n){
sum += n;
avg = sum / numbers.length;
});
In the end, we can output both the sum value, and the avg value to our console in order to see the result:
// Sum: 66, Avg: 5.5
console.log( 'Sum: ' + sum + ', Avg: ' + avg );
See it in action online at http://jsbin.com/unukoj/3/edit
I am just building on Abdennour TOUMI's answer. here are the reasons why:
1.) I agree with Brad, I do not think it is a good idea to extend object that we did not create.
2.) array.length is exactly reliable in javascript, I prefer Array.reduce beacuse a=[1,3];a[1000]=5; , now a.length would return 1001.
function getAverage(arry){
// check if array
if(!(Object.prototype.toString.call(arry) === '[object Array]')){
return 0;
}
var sum = 0, count = 0;
sum = arry.reduce(function(previousValue, currentValue, index, array) {
if(isFinite(currentValue)){
count++;
return previousValue+ parseFloat(currentValue);
}
return previousValue;
}, sum);
return count ? sum / count : 0;
};
Array.prototype.avg=function(fn){
fn =fn || function(e,i){return e};
return (this.map(fn).reduce(function(a,b){return parseFloat(a)+parseFloat(b)},0) / this.length ) ;
};
Then :
[ 1 , 2 , 3].avg() ; //-> OUT : 2
[{age:25},{age:26},{age:27}].avg(function(e){return e.age}); // OUT : 26
On evergreen browsers you can use arrow functions
avg = [1,2,3].reduce((a,b) => (a+b);
Running it 100,000 times, the time difference between the for loop approach and reduce is negligible.
s=Date.now();for(i=0;i<100000;i++){ n=[1,2,3]; a=n.reduce((a,b) => (a+b)) / n.length };
console.log("100k reduce took " + (Date.now()-s) + "ms.");
s=Date.now();for(i=0;i<100000;i++){n=[1,2,3]; nl=n.length; a=0; for(j=nl-1;j>0;j--){a=a+n[j];} a/nl };
console.log("100k for loop took " + (Date.now()-s) + "ms.");
s=Date.now();for(i=0;i<1000000;i++){n=[1,2,3]; nl=n.length; a=0; for(j=nl-1;j>0;j--){a=a+n[j];} a/nl };
console.log("1M for loop took " + (Date.now()-s) + "ms.");
s=Date.now();for(i=0;i<1000000;i++){ n=[1,2,3]; a=n.reduce((a,b) => (a+b)) / n.length };
console.log("1M reduce took " + (Date.now()-s) + "ms.");
/*
* RESULT on Chrome 51
* 100k reduce took 26ms.
* 100k for loop took 35ms.
* 10M for loop took 126ms.
* 10M reduce took 209ms.
*/
If you are in need of the average and can skip the requirement of calculating the sum, you can compute the average with a single call of reduce:
// Assumes an array with only values that can be parsed to a Float
var reducer = function(cumulativeAverage, currentValue, currentIndex) {
// 1. multiply average by currentIndex to find cumulative sum of previous elements
// 2. add currentValue to get cumulative sum, including current element
// 3. divide by total number of elements, including current element (zero-based index + 1)
return (cumulativeAverage * currentIndex + parseFloat(currentValue))/(currentIndex + 1)
}
console.log([1, 2, 3, 4, 5, 6, 7, 8, 9, 10].reduce(reducer, 0)); // => 5.5
console.log([].reduce(reducer, 0)); // => 0
console.log([0].reduce(reducer, 0)); // => 0
console.log([].reduce(reducer, 0)); // => 0
console.log([,,,].reduce(reducer, 0)); // => 0
console.log([].reduce(reducer, 0)); // => 0
If anyone ever needs it - Here is a recursive average.
In the context of the original question, you may want to use the recursive average if you allowed the user to insert additional values and, without incurring the cost of visiting each element again, wanted to "update" the existing average.
/**
* Computes the recursive average of an indefinite set
* #param {Iterable<number>} set iterable sequence to average
* #param {number} initAvg initial average value
* #param {number} initCount initial average count
*/
function average(set, initAvg, initCount) {
if (!set || !set[Symbol.iterator])
throw Error("must pass an iterable sequence");
let avg = initAvg || 0;
let avgCnt = initCount || 0;
for (let x of set) {
avgCnt += 1;
avg = avg * ((avgCnt - 1) / avgCnt) + x / avgCnt;
}
return avg; // or {avg: avg, count: avgCnt};
}
average([2, 4, 6]); //returns 4
average([4, 6], 2, 1); //returns 4
average([6], 3, 2); //returns 4
average({
*[Symbol.iterator]() {
yield 2; yield 4; yield 6;
}
}); //returns 4
How:
this works by maintaining the current average and element count. When a new value is to be included you increment count by 1, scale the existing average by (count-1) / count, and add newValue / count to the average.
Benefits:
you don't sum all the elements, which may result in large number that cannot be stored in a 64-bit float.
you can "update" an existing average if additional values become available.
you can perform a rolling average without knowing the sequence length.
Downsides:
incurs lots more divisions
not infinite - limited to Number.MAX_SAFE_INTEGER items unless you employ BigNumber
Having read the other choices, I will try to make a simpler version for the future viewers, elaborating on the existing code and not creating a more elegant one. First of all, you declared the numbers as strings. Apart from the .parseInt we can also do:
const numberConverter = elmt.map(Number);
So what map does is that it "returns a copy of the original array". But I convert its values to numbers. Then we can use the reduce method (It can also be simpler, but I am writing easy to read versions and I also have 2 average methods) What the reduce method does is it has an accumulator that gets bigger and bigger if you add values to it, as it iterates through the array and adds (in this case) the currentValue to it.:
var i;
const elmt = new Array();
elmt[0] = '0';
elmt[1] = '1';
elmt[2] = '2';
elmt[3] = '3';
elmt[4] = '4';
elmt[5] = '7';
elmt[6] = '8';
elmt[7] = '9';
elmt[8] = '10';
elmt[9] = '11';
console.log(elmt);
const numberConverter = elmt.map(Number);
const sum = numberConverter.reduce((accumulator, currentValue) => {
return accumulator + currentValue;
}, 0);
const average = numberConverter.reduce(
(accumulator, currentvalue, index, numArray) => {
return accumulator + currentvalue / numArray.length;
},
0
);
const average2 =
numberConverter.reduce(
(accumulator, currentValue) => accumulator + currentValue,
0
) / numberConverter.length;
for (i = 9; i < 10; i++) {
console.log(
`The sum of all the elements is: ${sum}. <br> The average of all the elements is: ${average2}`
);}
Unless I missed something, every solution up to this point uses the length of the list to calculate the average after summing the values.
There is a downside to this approach that a slightly modified, yet still simple algorithm will address without the downsides.
The downside is that you assuming that there won't be an overflow by summing all the numbers. If you have a lot of numbers that are very big, and you add them all up, they may exceed the maximum size that can fit into the data type.
A better approach is to simply calculate the average as you go, rather than summing it and then dividing with the length at the end:
function getAvg(values) {
return values.reduce((m, x, i) => m + (x - m) / (i + 1), 0)
}
Props to Knuth's "Art of Computer Programming" vol. 2.
just for fun
let avg = [81, 77, -88, 195, 6.8].reduce((a,e,i) => (a*i+e)/(i+1));
console.log(avg)
Just for kicks:
var elmt = [0, 1, 2,3, 4, 7, 8, 9, 10, 11], l = elmt.length, i = -1, sum = 0;
for (; ++i < l; sum += elmt[i])
;
document.body.appendChild(document.createTextNode('The sum of all the elements is: ' + sum + ' The average of all the elements is: ' + (sum / l)));
I think we can do like
var k=elmt.reduce(function(a,b){return parseFloat(a+parseFloat(b));})
var avg=k/elmt.length;
console.log(avg);
I am using parseFloat twice because
when
1) you add (a)9+b("1") number then result will be "91" but we want addition. so i used parseFloat
2)When addition of (a)9+parseFloat("1") happen though result will be "10" but it will be in string which we don't want so again i used parseFloat.
I hope i am clear. Suggestions are welcome
Here is my rookie way of simply finding the avg. Hope this helps somebody.
function numAvg(num){
var total = 0;
for(var i = 0;i < num.length; i++) {
total+=num[i];
}
return total/num.length;
}
here's your one liner:
var average = arr.reduce((sum,item,index,arr)=>index !== arr.length-1?sum+item:sum+item/arr.length,0)
I think this may be a direct solution to calculate the average with a for loop and function.
var elmts = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
function average(arr) {
var total = 0;
for (var i = 0; i < arr.length; i++) {
total += arr[i];
}
console.log(Math.round(total/arr.length));
}
average(elmts);
There seem to be an endless number of solutions for this but I found this to be concise and elegant.
const numbers = [1,2,3,4];
const count = numbers.length;
const reducer = (adder, value) => (adder + value);
const average = numbers.map(x => x/count).reduce(reducer);
console.log(average); // 2.5
Or more consisely:
const numbers = [1,2,3,4];
const average = numbers.map(x => x/numbers.length).reduce((adder, value) => (adder + value));
console.log(average); // 2.5
Depending on your browser you may need to do explicit function calls because arrow functions are not supported:
const r = function (adder, value) {
return adder + value;
};
const m = function (x) {
return x/count;
};
const average = numbers.map(m).reduce(r);
console.log(average); // 2.5
Or:
const average1 = numbers
.map(function (x) {
return x/count;
})
.reduce(function (adder, value) {
return adder + value;
});
console.log(average1);