How to calculate the sum and average value of elements in an array?

Question

I have problems with adding all elements of the array and their averaging. How would I do this and implement it using the code that I currently have? Elements should be defined, as I have 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> 

Answer 1

 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 over the array, since your values ​​are strings, they must first be converted to integers. And the average value is simply the sum of the values ​​divided by the number of values.

Answer 2

The solution that I find more elegant:

 var sum, avg = 0; // dividing by 0 will return Infinity // arr must contain at least 1 element to use reduce if (arr.length) { sum = arr.reduce(function(a, b) { return a + b; }); avg = sum / arr.length; } document.write("The sum is: " + sum + ". The average is: " + avg + "<br/>"); 

Answer 3

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); 

Answer 4

Calculation of average (average) using reduction and ES6:

 const average = list => list.reduce((prev, curr) => prev + curr) / list.length; const list = [0, 10, 20, 30] average(list) // 15 

Answer 5

usually the average value using decreasing one row is as follows

 elements.reduce(function(sum, a,i,ar) { sum += a; return i==ar.length-1?(ar.length==0?0:sum/ar.length):sum},0); 

in particular the 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); 

effective version is similar to

 elements.reduce(function(sum, a) { return sum + a },0)/(elements.length||1); 

Understand Javascript Array Decrease by 1 minute http://www.airpair.com/javascript/javascript-array-reduce

as gotofritz pointed out, it seems that Array.reduce is missing undefined values. so here is the 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]) 

Answer 6

Suppose we have an array of numbers like this:

 var values = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]; 

The average value is obtained by the following formula

A = (1 / n) Σxi (from 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 return value.

Signature Reduction Method

 reduce(callback[,default_previous_value]) 

The callback function reduces the following parameters:

• p : Result of previous calculation
• c : current value (from current index)
• i : current array element index value
• a : current reduced array

The second reduction parameter is the default value ... (used if the array is empty ).

Thus, the average reduction 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 just refer to the callback method signature

 var avg = values.reduce(average,0); var sum= values.reduce(sum,0); 

Or skip the prototype array directly.

 Array.prototype.sum = Array.prototype.sum || function (){ return this.reduce(function(p,c){return p+c},0); }; 

You can split the value each time you call the reduction method.

 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 ()

optimize the process causing division only once :)

 Array.prototype.avg = Array.prototype.avg || function () { return this.sum()/this.length; }; 

Then to any object of the array area:

 [2, 6].avg();// -> 4 [2, 6].sum();// -> 8 

NB: an empty array with the return of the NaN wish is rather 0 in my point of view and can be useful in specific cases of use.

Answer 7

You can also use lodash , _.sum (array) and _.mean (array) in the Math part (there are other convenient things too).

 _.sum([4, 2, 8, 6]); // => 20 _.mean([4, 2, 8, 6]); // => 5 

Answer 8

Not the fastest, but the shortest and in one line uses map () and reduce ():

 var average = [7,14,21].map(function(x,i,arr){return x/arr.length}).reduce(function(a,b){return a + b}) 

Answer 9

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 specify an array for its sum or average value, for example:

 [1,2,3].sum() // = 6 [1,2,3].average() // = 2 

Answer 10

In ES6 ready-made browsers, this polyfill can be useful.

 Math.sum = (...a) => Array.prototype.reduce.call(a,(a,b) => a+b) Math.avg = (...a) => this.sum(...a)/a.length; 

You can use the same calling method between Math.sum , Math.avg and Math.max , for example:

 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 summarize, you can use

 var sumNum = Math.sum.apply(null,[1,2,3,4]) // 10 

as

 var maxOne = Math.max.apply(null,[1,2,3,4]) // 4 

Answer 11

One sneaky way you could do this, though it does require the use of the (highly 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 would post this as one of these out-of-the-box options. You never know, cunning can give you (or take away) a point.

Answer 12

Here is a quick addition to the "Math" object in javascript to add the "middle" 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 add-on to the Math object to get 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 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 your variable (input, if they are numbers, can be a string because it is parsed into a number!)

Answer 13

set your loop counter to 0 .... you get element 9, and then you finish what you have now. Other answers are basic math. Use a variable to store your sum (you must pass the lines to int) and divide it by the length of the array.

Answer 14

Start by defining all the variables that we plan to use. You will notice that for the numbers array I use the literal notation [] , unlike the constructor method array() . In addition, I use a shorter method to set multiple variables to 0.

 var numbers = [], count = sum = avg = 0; 

Next, I fill my empty array of numbers with values ​​from 0 to 11. This should lead me to the starting point. Notice how I click on the count++ array. This results in the current value of the counter, and then increments it the next time.

 while ( count < 12 ) numbers.push( count++ ); 

Finally, I perform the function “for each” of numbers in an array of numbers. This function will process one number at a time, which I define as "n" inside the body of the function.

 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 to see the result:

 // Sum: 66, Avg: 5.5 console.log( 'Sum: ' + sum + ', Avg: ' + avg ); 

Watch it in action online http://jsbin.com/unukoj/3/edit

Answer 15

I will simply describe the answer of Abdennour TOUMI. that's why:

1.) I agree with Brad, I do not think it is a good idea to extend an object that we did not create.

2.) array.length absolutely reliable in javascript, I prefer Array.reduce beacuse a=[1,3];a[1000]=5; , now a.length will 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; }; 

Answer 16

In evergreen browsers, you can use the arrow functions avg = [1,2,3].reduce((a,b) => (a+b);

Performing it 100,000 times, the time difference between approaching and shortening the cycle for the cycle 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. */ 

Answer 17

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))); 

Answer 18

Here is a small little solution for a medium array that is very inefficient but has very specific applications:

 var average = elmt.reduce(function(x,y,z){ return ((x * z) + y) / (z + 1); }); 

Documents for the reduce method

Answer 19

 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 

Answer 20

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Answer 21

I think we can do as

 var k=elmt.reduce(function(a,b){return parseFloat(a+parseFloat(b));}) var avg=k/elmt.length; console.log(avg); 

I use parseFloat twice, because when 1) you add (a) 9 + b ("1"), then the result will be "91", but we want to add. so i used parseFloat

2) When the addition of (a) 9 + parseFloat ("1") occurs, the result will be "10", but it will be on the line that we do not want to use again parseFloat.

I hope I get it. Suggestions are welcome

Answer 22

Here is my newbie way to just find avg. Hope this helps someone.

 function numAvg(num){ var total = 0; for(var i = 0;i < num.length; i++) { total+=num[i]; } return total/num.length; } 

Answer 23

here is your one liner:

 var average = arr.reduce((sum,item,index,arr)=>index !== arr.length-1?sum+item:sum+item/arr.length,0) 

Answer 24

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); 

Answer 25

There seems to be an infinite number of solutions for this, but I found it 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 precisely:

 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 make explicit function calls, as 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); 

Answer 26

If you need an average value and you can skip the requirement for calculating the sum, you can calculate the average value with a single call to the Reduce method:

 // 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 

Answer 27

If someone needs it - here is a recursive mean.

In the context of the original question, you can use a recursive average if you allowed the user to insert additional values ​​and, while not 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 number of elements. When you need to add a new value, you increment the counter by 1, scale the existing average value (count-1)/count and add newValue/count to the average.

Benefits:

• You do not summarize all the elements, which can lead to a large number that cannot be stored in 64-bit floating point format.
• You can “update” the existing average if additional values ​​become available.
• You can perform a moving average without knowing the length of the sequence.

Disadvantages:

• entails much more units
• not infinite - limited by Number.MAX_SAFE_INTEGER elements if you are not using BigNumber

Answer 28

Average HTML content itens

With jQuery or Javascript querySelector , you have a direct binding to formatted data ... Example:

 <p>Elements for an average: <span class="m">2</span>, <span class="m">4</span>, <span class="m">2</span>, <span class="m">3</span>. </p> 

So with jQuery you have

 var A = $('.m') .map(function(idx) { return parseInt($(this).html()) }) .get(); var AVG = A.reduce(function(a,b){return a+b}) / A5.length; 

See 4 more ways (!) To access them and average it: http://jsfiddle.net/4fLWB/

Answer 29

var arr = [1,2,3,4,5]

 function avg(arr){ var sum = 0; for (var i = 0; i < arr.length; i++) { sum += parseFloat(arr[i]) } return sum / i; } 

avg (arr) ====== → 3

This works with strings as numbers or numbers in an array.

Answer 30

A simple code might look like this:

 var sum = eval("parseInt(" + elmt.join(") + parseInt(") + ");"); var avg = sum / elmt.length; document.write("The sum of all the elements is: " + sum + " The average of all the elements is: " + avg + "<br/>"); 

If you need a summary of floating point numbers, use parseFloat instead of parseInt .