How to find the sum of an array of numbers

Question

Given an array [1, 2, 3, 4], how can I find the sum of its elements? (In this case, the sum would be 10.)

I thought $.each might be useful, but I'm not sure how to implement it.

Answer 1

This'd be exactly the job for reduce.

If you're using ECMAScript 2015 (aka ECMAScript 6):

const sum = [1, 2, 3].reduce((partialSum, a) => partialSum + a, 0);
console.log(sum); // 6

For older JS:

const sum = [1, 2, 3].reduce(add, 0); // with initial value to avoid when the array is empty

function add(accumulator, a) {
  return accumulator + a;
}

console.log(sum); // 6

Isn't that pretty? :-)

Answer 2

Recommended (reduce with default value)

Array.prototype.reduce can be used to iterate through the array, adding the current element value to the sum of the previous element values.

console.log(
  [1, 2, 3, 4].reduce((a, b) => a + b, 0)
)
console.log(
  [].reduce((a, b) => a + b, 0)
)

Without default value

You get a TypeError

console.log(
  [].reduce((a, b) => a + b)
)

Prior to ES6's arrow functions

console.log(
  [1,2,3].reduce(function(acc, val) { return acc + val; }, 0)
)

console.log(
  [].reduce(function(acc, val) { return acc + val; }, 0)
)

Non-number inputs

If non-numbers are possible inputs, you may want to handle that?

console.log(
  ["hi", 1, 2, "frog"].reduce((a, b) => a + b)
)

let numOr0 = n => isNaN(n) ? 0 : n

console.log(
  ["hi", 1, 2, "frog"].reduce((a, b) => 
    numOr0(a) + numOr0(b))
)

Non-recommended dangerous eval use

We can use eval to execute a string representation of JavaScript code. Using the Array.prototype.join function to convert the array to a string, we change [1,2,3] into "1+2+3", which evaluates to 6.

console.log(
  eval([1,2,3].join('+'))
)

//This way is dangerous if the array is built
// from user input as it may be exploited eg: 

eval([1,"2;alert('Malicious code!')"].join('+'))

Of course displaying an alert isn't the worst thing that could happen. The only reason I have included this is as an answer Ortund's question as I do not think it was clarified.

Answer 3

Why not reduce? It's usually a bit counter intuitive, but using it to find a sum is pretty straightforward:

var a = [1,2,3];
var sum = a.reduce(function(a, b) { return a + b; }, 0);

Answer 4

var arr = [1, 2, 3, 4];
var total = 0;
for (var i in arr) {
  total += arr[i];
}

Answer 5

var total = 0;
$.each(arr,function() {
    total += this;
});

Answer 6

Anyone looking for a functional oneliner like me?

Assuming:

const arr = [1, 2, 3, 4];

Here's the oneliner for modern JS:

sum = arr.reduce((a, b) => a + b, 0);

(If you happen to have to support ye olde IE without arrow functions:)

sum = arr.reduce(function (a, b) {return a + b;}, 0);

Note that 0 is the initial value here, so you can use that as offset if needed. Also note that this initial value is needed, otherwise calling the function with an empty array will error.

Answer 7

If you happen to be using Lodash you can use the sum function

array = [1, 2, 3, 4];
sum = _.sum(array); // sum == 10

Answer 8

This is possible by looping over all items, and adding them on each iteration to a sum-variable.

var array = [1, 2, 3];

for (var i = 0, sum = 0; i < array.length; sum += array[i++]);

JavaScript doesn't know block scoping, so sum will be accesible:

console.log(sum); // => 6

The same as above, however annotated and prepared as a simple function:

function sumArray(array) {
  for (
    var
      index = 0,              // The iterator
      length = array.length,  // Cache the array length
      sum = 0;                // The total amount
      index < length;         // The "for"-loop condition
      sum += array[index++]   // Add number on each iteration
  );
  return sum;
}

Answer 9

arr.reduce(function (a, b) {
    return a + b;
});

Reference: Array.prototype.reduce()

Answer 10

OK, imagine you have this array below:

const arr = [1, 2, 3, 4];

Let's start looking into many different ways to do it as I couldn't find any comprehensive answer here:

1) Using built-in reduce()

function total(arr) {
  if(!Array.isArray(arr)) return;
  return arr.reduce((a, v)=>a + v);
}

2) Using for loop

function total(arr) {
  if(!Array.isArray(arr)) return;
  let totalNumber = 0;
  for (let i=0,l=arr.length; i<l; i++) {
     totalNumber+=arr[i];
  }
  return totalNumber;
}

3) Using while loop

function total(arr) {
  if(!Array.isArray(arr)) return;
  let totalNumber = 0, i=-1;
  while (++i < arr.length) {
     totalNumber+=arr[i];
  }
  return totalNumber;
}

4) Using array forEach

function total(arr) {
  if(!Array.isArray(arr)) return;
  let sum=0;
  arr.forEach(each => {
    sum+=each;
  });
  return sum;
};

and call it like this:

total(arr); //return 10

It's not recommended to prototype something like this to Array...

Answer 11

You can try the following code:

[1, 2, 3, 4].reduce((pre,curr)=>pre+curr,0)

Answer 12

You can also use reduceRight.

[1,2,3,4,5,6].reduceRight(function(a,b){return a+b;})

which results output as 21.

Reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/ReduceRight

Answer 13

A standard JavaScript solution:

var addition = [];
addition.push(2);
addition.push(3);

var total = 0;
for (var i = 0; i < addition.length; i++)
{
    total += addition[i];
}
alert(total);          // Just to output an example
/* console.log(total); // Just to output an example with Firebug */

This works for me (the result should be 5). I hope there is no hidden disadvantage in this kind of solution.

Answer 14

TL;DR

If you care about performance, define a function that uses a for-loop.

function sum(arr) {
    var res = 0;
    for (var x of arr) {
        res += x;
    }
    return res;
}

Benchmark

I benchmarked a selection of implementations using benchmark.js (typescript version):

const arr = Array.from({ length: 100 }, () => Math.random());
const reducer = function (p: number, a: number) {
    return p + a;
};
const recursion = function (arr: number[], i: number) {
    if(i > 0) return arr[i] + recursion(arr, i - 1)
    else return 0
};
const recursion2 = function (arr: number[], i: number, len: number) {
    if(i < len) return arr[i] +  recursion2(arr, i + 1, len)
    else return 0
};
const recursion3 = function (arr: number[], i: number) {
    if(i < arr.length) return arr[i] + recursion3(arr, i + 1)
    else return 0
};
new Benchmark.Suite()
    .add("jquery", () => {
        let res = 0;
        $.each(arr, (_, x) => (res += x));
    })
    .add("lodash", ()=>_.sum(arr))
    .add("forEach", () => {
        let res = 0;
        arr.forEach((x) => (res += x));
    })
    .add("reduce", () => arr.reduce((p, a) => p + a, 0))
    .add("predefined reduce", () => arr.reduce(reducer, 0))
    .add("eval", () => eval(arr.join("+")))
    .add("recursion", () => recursion(arr, arr.length - 1))
    .add("recursion2", () => recursion2(arr, 0, arr.length))
    .add("recursion3", () => recursion3(arr, 0))
    .add("naive", () => (
        arr[0]+arr[1]+arr[2]+arr[3]+arr[4]+arr[5]+arr[6]+arr[7]+arr[8]+arr[9]+
        arr[10]+arr[11]+arr[12]+arr[13]+arr[14]+arr[15]+arr[16]+arr[17]+arr[18]+arr[19]+
        arr[20]+arr[21]+arr[22]+arr[23]+arr[24]+arr[25]+arr[26]+arr[27]+arr[28]+arr[29]+
        arr[30]+arr[31]+arr[32]+arr[33]+arr[34]+arr[35]+arr[36]+arr[37]+arr[38]+arr[39]+
        arr[40]+arr[41]+arr[42]+arr[43]+arr[44]+arr[45]+arr[46]+arr[47]+arr[48]+arr[49]+
        arr[50]+arr[51]+arr[52]+arr[53]+arr[54]+arr[55]+arr[56]+arr[57]+arr[58]+arr[59]+
        arr[60]+arr[61]+arr[62]+arr[63]+arr[64]+arr[65]+arr[66]+arr[67]+arr[68]+arr[69]+
        arr[70]+arr[71]+arr[72]+arr[73]+arr[74]+arr[75]+arr[76]+arr[77]+arr[78]+arr[79]+
        arr[80]+arr[81]+arr[82]+arr[83]+arr[84]+arr[85]+arr[86]+arr[87]+arr[88]+arr[89]+
        arr[90]+arr[91]+arr[92]+arr[93]+arr[94]+arr[95]+arr[96]+arr[97]+arr[98]+arr[99]))
    .add("loop with iterator", () => {
        let res = 0;
        for (const x of arr) res += x;
    })
    .add("traditional for loop", () => {
        let res = 0;
        // cache the length in case the browser can't do it automatically
        const len = arr.length;
        for (let i = 0; i < len; i++) res += arr[i];
    })
    .add("while loop", () => {
        let res = 0;
        let i = arr.length;
        while (i--) res += arr[i];
    })
    .add("loop in a function ", () => sum(arr))
    .on("cycle", (event) => console.log(String(event.target)))
    .run();

In chrome 104, the for-loop-based implementations are the fastest:

jquery               x  1,832,472 ops/sec ±1.35% (61 runs sampled)
lodash               x  2,079,009 ops/sec ±1.11% (68 runs sampled)
forEach              x  4,887,484 ops/sec ±2.35% (67 runs sampled)
reduce               x 21,762,391 ops/sec ±0.46% (69 runs sampled)
predefined reduce    x  2,026,411 ops/sec ±0.50% (68 runs sampled)
eval                 x     33,381 ops/sec ±2.54% (66 runs sampled)
recursion            x  2,252,353 ops/sec ±2.13% (62 runs sampled)
recursion2           x  2,301,516 ops/sec ±1.15% (65 runs sampled)
recursion3           x  2,395,563 ops/sec ±1.65% (66 runs sampled)
naive                x 31,244,240 ops/sec ±0.76% (66 runs sampled)
loop with iterator   x 29,554,762 ops/sec ±1.07% (66 runs sampled)
traditional for loop x 30,052,685 ops/sec ±0.67% (66 runs sampled)
while loop           x 18,624,045 ops/sec ±0.17% (69 runs sampled)
loop in a function   x 29,437,954 ops/sec ±0.54% (66 runs sampled)

Firefox 104 shows similar behaviour:

jquery               x  1,461,578 ops/sec ±1.58% (64 runs sampled)
lodash               x  4,931,619 ops/sec ±0.80% (66 runs sampled)
forEach              x  5,594,013 ops/sec ±0.51% (68 runs sampled)
reduce               x  3,731,232 ops/sec ±0.53% (66 runs sampled)
predefined reduce    x  2,633,652 ops/sec ±0.54% (66 runs sampled)
eval                 x    105,003 ops/sec ±0.88% (66 runs sampled)
recursion            x  1,194,551 ops/sec ±0.24% (67 runs sampled)
recursion2           x  1,186,138 ops/sec ±0.20% (68 runs sampled)
recursion3           x  1,191,921 ops/sec ±0.24% (68 runs sampled)
naive                x 21,610,416 ops/sec ±0.66% (66 runs sampled)
loop with iterator   x 15,311,298 ops/sec ±0.43% (67 runs sampled)
traditional for loop x 15,406,772 ops/sec ±0.59% (67 runs sampled)
while loop           x 11,513,234 ops/sec ±0.60% (67 runs sampled)
loop in a function   x 15,417,944 ops/sec ±0.32% (68 runs sampled)

Discussion

Implementations defining an anonymous function are generally slower because creating an anonymous function is a significant overhead. When running the benchmark with a large array, e.g., with length 1000 instead of 100, the difference between reduce and the for-loop-based implementations becomes insignificant in chrome.

Chrome's V8 engine knows how to inline simple anonymous functions in reduce since the reduce test case is much faster than the predefined reduce test case. Firefox seems to try something similar but is less efficient in doing so. Non-inlined function calls are pretty slow in js since the call stack is less efficient than the call stack in compiled software.

Similar to reduce, the forEach- and jquery-based implementations use anonymous functions and are relatively slow. lodash has a specialized sum implementation, but it is (as of v4.0.0) implemented as a special case of sumBy, which is relatively inefficient.

eval is the by far slowest test case. This makes sense since constructing the string using concatenations might cause several dynamic allocations (which are slow). Next, the parser has to be invoked and only then can the code be finally executed.

I've included some recursive implementations because some people on the internet claim that recursion is faster than loops in js. I can't reproduce their example - using benchmark.js, recursion is very slow, and when using console.time with a loop, both functions take the same time. When calculating the sum, as expected, recursion is much slower than loops, probably due to intense usage of the js call stack.

The naive implementation would be manually adding all 100 elements of the array. While being quite inconvenient, this is the fastest implementation. But, luckily, for-loops come very close. Adding a single function call around the loop doesn't harm the performance. Therefore, you can feel free to use the utility function from above.

I have no explanation why the while loop is slower than the for loop. Iterating the array in reverse doesn't seem to be the problem here.

Answer 15

Funny approach:

eval([1,2,3].join("+"))

Answer 16

I am a beginner with JavaScript and coding in general, but I found that a simple and easy way to sum the numbers in an array is like this:

    var myNumbers = [1,2,3,4,5]
    var total = 0;
    for(var i = 0; i < myNumbers.length; i++){
        total += myNumbers[i];
    }

Basically, I wanted to contribute this because I didn't see many solutions that don't use built-in functions, and this method is easy to write and understand.

Answer 17

Use a for loop:

const array = [1, 2, 3, 4];
let result = 0;

for (let i = 0; i < array.length - 1; i++) {
  result += array[i];
}

console.log(result); // Should give 10

Or even a forEach loop:

const array = [1, 2, 3, 4];
let result = 0;

array.forEach(number => {
  result += number;
})

console.log(result); // Should give 10

For simplicity, use reduce:

const array = [10, 20, 30, 40];
const add = (a, b) => a + b
const result = array.reduce(add);

console.log(result); // Should give 100

Answer 18

A few people have suggested adding a .sum() method to the Array.prototype. This is generally considered bad practice so I'm not suggesting that you do it.

If you still insist on doing it then this is a succinct way of writing it:

Array.prototype.sum = function() {return [].reduce.call(this, (a,i) => a+i, 0);}

then: [1,2].sum(); // 3

Note that the function added to the prototype is using a mixture of ES5 and ES6 function and arrow syntax. The function is declared to allow the method to get the this context from the Array that you're operating on. I used the => for brevity inside the reduce call.

Answer 19

ES6 for..of

let total = 0;

for (let value of [1, 2, 3, 4]) {
    total += value; 
}

Answer 20

A short piece of JavaScript code would do this job:

var numbers = [1,2,3,4];
var totalAmount = 0;

for (var x = 0; x < numbers.length; x++) {

    totalAmount += numbers[x];
}

console.log(totalAmount); //10 (1+2+3+4)

Answer 21

var totally = eval(arr.join('+'))

That way you can put all kinds of exotic things in the array.

var arr = ['(1/3)','Date.now()','foo','bar()',1,2,3,4]

I'm only half joking.

Answer 22

Use reduce

let arr = [1, 2, 3, 4];

let sum = arr.reduce((v, i) => (v + i));

console.log(sum);

Answer 23

No need to initial value! Because if no initial value is passed, the callback function is not invoked on the first element of the list, and the first element is instead passed as the initial value. Very cOOl feature :)

[1, 2, 3, 4].reduce((a, x) => a + x) // 10
[1, 2, 3, 4].reduce((a, x) => a * x) // 24
[1, 2, 3, 4].reduce((a, x) => Math.max(a, x)) // 4
[1, 2, 3, 4].reduce((a, x) => Math.min(a, x)) // 1

Answer 24

Here's an elegant one-liner solution that uses stack algorithm, though one may take some time to understand the beauty of this implementation.

const getSum = arr => (arr.length === 1) ? arr[0] : arr.pop() + getSum(arr);

getSum([1, 2, 3, 4, 5]) //15

Basically, the function accepts an array and checks whether the array contains exactly one item. If false, it pop the last item out of the stack and return the updated array.

The beauty of this snippet is that the function includes arr[0] checking to prevent infinite looping. Once it reaches the last item, it returns the entire sum.

Answer 25

Accuracy

Sort array and start sum form smallest numbers (snippet shows difference with nonsort)

[...arr].sort((a,b)=>a-b).reduce((a,c)=>a+c,0)

arr=[.6,9,.1,.1,.1,.1]

sum     =                       arr.reduce((a,c)=>a+c,0)
sortSum = [...arr].sort((a,b)=>a-b).reduce((a,c)=>a+c,0)

console.log('sum:     ',sum);
console.log('sortSum:',sortSum);
console.log('sum==sortSum :', sum==sortSum);

// we use .sort((a,b)=>a-b) instead .sort() because
// that second one treat elements like strings (so in wrong way)
// e.g [1,10,9,20,93].sort() -->  [1, 10, 20, 9, 93]

For multidimensional array of numbers use arr.flat(Infinity)

arr= [ [ [1,2,3,4],[1,2,3,4],[1,2,3,4] ],
       [ [1,2,3,4],[1,2,3,4],[1,2,3,4] ] ];
      
sum = arr.flat(Infinity).reduce((a,c)=> a+c,0);

console.log(sum);  // 60

Answer 26

Simplest answer to understand underlying process:

let array = [10, 20, 30, 40, 50]
let total = 0

for(let i in array)
{
    total += array[i]
}

console.log(total)

& if you're already familiar with underlying process then built-in method can save you time:

let array = [10, 20, 30, 40, 50]
let total = array.reduce((x, y) => x + y)
console.log(total)

Answer 27

Is there a reason not to just filter the array first to remove non-numbers? Seems simple enough:

[1, 2, 3, null, 'a'].filter((x) => !isNaN(x)).reduce((a, b) => a + b)

Answer 28

Those are really great answers, but just in case if the numbers are in sequence like in the question ( 1,2,3,4) you can easily do that by applying the formula (n*(n+1))/2 where n is the last number

Answer 29

You can combine reduce() method with lambda expression:

[1, 2, 3, 4].reduce((accumulator, currentValue) => accumulator + currentValue);

Answer 30

With reduce()

[1, 2, 3, 4].reduce((a, b) => a + b, 0); // 10

With forEach()

let sum = 0;
[1, 2, 3, 4].forEach(n => sum += n);
sum; // 10

With Parameter

function arrSum(arr) { 
  sum = 0;  
  arr.forEach(n => sum += n); 
  return sum; 
}

arrSum([1, 2, 3, 4]) // 10

Answer 31

i saw all answers going for 'reduce' solution

var array = [1,2,3,4]
var total = 0
for (var i = 0; i < array.length; i++) {
    total += array[i]
}
console.log(total)

Answer 32

very simple

step 1 we should have an array like :

const arrayNumber = [500,152,154,1555,12445];

step 2 (you can ignore this step if) step is to be sur that all values in table are number for that

let newArray = [];
for (let i = 0; i < arrayNumber.length; i++) {
        newArray.push(parseInt(arrayNumber[i], 10));
      }

step 3

const sumInArray = dataData.reduce( (a, b) => a + b);

finally

console.log(sumInArray);

Answer 33

getTotal = (arr) => {
    let total = 0
    for (let i = 0; i < arr.length; i++) {
        total += arr[i];
    }
    return total
}

getTotal([1, 2, 3, 4]) // 10
getTotal([1, 2, 3, 4, 5]) // 15

Answer 34

Cool tricks here, I've got a nit pick with a lot of the safe traditional answers not caching the length of the array.

function arraySum(array){
  var total = 0,
      len = array.length;

  for (var i = 0; i < len; i++){
    total += array[i];
  }

  return total;
};

var my_array = [1,2,3,4];

// Returns 10
console.log( arraySum( my_array ) );

Without caching the length of the array the JS compiler needs to go through the array with every iteration of the loop to calculate the length, it's unnecessary overhead in most cases. V8 and a lot of modern browsers optimize this for us, so it is less of a concern then it was, but there are older devices that benefit from this simple caching.

If the length is subject to change, caching's that could cause some unexpected side effects if you're unaware of why you're caching the length, but for a reusable function who's only purpose is to take an array and add the values together it's a great fit.

Here's a CodePen link for this arraySum function. http://codepen.io/brandonbrule/pen/ZGEJyV

It's possible this is an outdated mindset that's stuck with me, but I don't see a disadvantage to using it in this context.

Answer 35

Object.defineProperty(Object.prototype, 'sum', {
    enumerable:false,
    value:function() {
        var t=0;for(var i in this)
            if (!isNaN(this[i]))
                t+=this[i];
        return t;
    }
});

[20,25,27.1].sum()                 // 72.1
[10,"forty-two",23].sum()          // 33
[Math.PI,0,-1,1].sum()             // 3.141592653589793
[Math.PI,Math.E,-1000000000].sum() // -999999994.1401255

o = {a:1,b:31,c:"roffelz",someOtherProperty:21.52}
console.log(o.sum());              // 53.519999999999996

Answer 36

This is much easier

function sumArray(arr) {
    var total = 0;
    arr.forEach(function(element){
        total += element;
    })
    return total;
}

var sum = sumArray([1,2,3,4])

console.log(sum)

Answer 37

With ES6 rest parameter

let array = [1, 2, 3, 4]

function sum(...numbers) {
    let total = 0;
    for (const number of numbers) {
        total += number;
    }
    return total;
}

console.log(sum(...array));

Answer 38

A simple method example:

function add(array){
    var arraylength = array.length;
    var sum = 0;
    for(var timesToMultiply = 0; timesToMultiply<arraylength; timesToMultiply++){
        sum += array[timesToMultiply];
    }

    return sum;
}

console.log(add([1, 2, 3, 4]));

Answer 39

When the array consists of strings one has to alter the code. This can be the case, if the array is a result from a databank request. This code works:

alert(
["1", "2", "3", "4"].reduce((a, b) => Number(a) + Number(b), 0)
);

Here, ["1", "2", "3", "4"] ist the string array and the function Number() converts the strings to numbers.

Answer 40

You can get the sum using for of loop like following:

let arr = [1, 2, 3, 4];
let total = 0;
for (let i of arr) {
  total += i;
}
console.log(total)

Answer 41

try this...

function arrSum(arr){
    total = 0;  
    arr.forEach(function(key){
        total = total + key;            
    });
    return total;
}

Answer 42

Vanilla JavaScript is all you need:

> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; a.forEach(function(e){sum += e}); sum
10
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; a.forEach(e => sum += e); sum
10
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; for(e in a) sum += e; sum
"00123foobar"
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; for(e of a) sum += e; sum
10

Answer 43

This function can sum up all the numbers -

 function array(arr){
   var sum = 0;
   for (var i = 0; i< arr.length; i++){
    sum += arr[i];
   }
   console.log(sum);
 }
 array([5, 1, 3, 3])

Answer 44

Considering you have the array

const arr = [1, 2, 3, 4];

For me, the most intuitive way is using a for-in

let sum = 0;
for(var value in arr) {
    sum += arr[value];
}

console.log('Array:', arr);
console.log('Sum:', sum);

Yet, you can also use Arrow Functions and the reduce function

const sum = arr.reduce(function (a, b) {
    return a + b;
}, 0);

console.log('Array:', arr);
console.log('Sum:', sum);

They're both gonna output

Array: [ 1, 2, 3, 4]
Sum: 10

Answer 45

Just use this function:

function sum(pArray)
{
  pArray = pArray.reduce(function (a, b) {
      return a + b;
  }, 0);

  return pArray;
}

function sum(pArray)
{
  pArray = pArray.reduce(function (a, b) {
      return a + b;
  }, 0);

  return pArray;
}

var arr = [1, 4, 5];

console.log(sum(arr));

Answer 46

A "duplicate" question asked how to do this for a two-dimensional array, so this is a simple adaptation to answer that question. (The difference is only the six characters [2], 0, which finds the third item in each subarray and passes an initial value of zero):

const twoDimensionalArray = [
  [10, 10, 1],
  [10, 10, 2],
  [10, 10, 3],
];
const sum = twoDimensionalArray.reduce( (partial_sum, a) => partial_sum + a[2], 0 ) ; 
console.log(sum); // 6

Answer 47

You can try this:

var arr = [100,114,250,1200];
var total = 0; 
for(var i in arr){
  total += parseInt(arr[i]);
}

console.log(total);

Output will be: 1664

Or if value is Float, then try this:

var arr = [100.00,114.50,250.75,1200.00];
    var total = 0; 
    for(var i in arr){
      total += parseFloat(arr[i]);
    }
    
    console.log(total.toFixed(2));

Output will be: 1665.25

Answer 48

I´m aware that this is an old post. But I just wanted to share my approach to the problem.

let myArr = [1, 3, 4, 5, 6, 7, 8];
let counter = 0;
for(let i = 0; i < myArr.length; i++){
  counter = counter + myArr[i];
    console.log(counter);}

Answer 49

//Try this way

const arr = [10,10,20,60]; 
const sumOfArr = (a) =>{
    let sum=0;
    for(let i in a) { 
        sum += a[i];
    }
    return sum;
}
console.log(sumOfArr(arr))

Answer 50

No one mentioned functional programming, but it would be very clean approach to use Ramda in this case:

//Assuming you use nodejs, but can also be used in browser
const R = require('ramda');

let nums = [2, 4, 6, 8, 10];
console.log(R.sum(nums));

Answer 51

The truth is there is an old and funny classic solution (beside newbie 'foreach' and 'reduce'): the classic for of.

y = 0;
for (x of [1, 2, 3, 4]) y+=x;

Answer 52

I see that this is an old question, but hardly any of the answers actually use jQuery and specifically $.each as requested by the asker. There is a way to do this, but the native approaches as suggested by many (i.e. using Array.reduce()) are much better and easier.

Here's the jQuery way if you absolutely have to use it:

  var arr = [1,2,3,4];
  var sum = 0;
  
  $.each( arr, (i, v) => sum += v);

  alert(sum); // This will be 10

Answer 53

i think this is the simple way to find sum array of numbers

function sumArray(arr){

    let total = 0;

    for(i = 0; i<arr.length ; i++){

        total += arr[i];
    }
    return total;
}

console.log(sumArray([1,2,3,4,]));

Answer 54

const a = [1,'2',3,100,'200','text'];
var total= 0;
let b = a.filter( value => !Number.isNaN(Number(value)) );
for(let c in b)
{
  total += Number(b[c]);
}
console.log(total)
//output:306

Answer 55

// 1:
const numbers = new Array(1, 2, 3, 4, 5, 6, 7, 8, 9);
// Or const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9];
let sum = numbers.reduce(
  (accumulator, currentValue) => accumulator + currentValue
);
console.log(sum);
// 2:
sum = 0;
for (const number of numbers) {
  sum += number;
}
console.log(sum);
// 3:
sum = 0;
for (const index in numbers) {
  sum += numbers[index];
}
console.log(sum);
// 4:
sum = 0;
for (let index = 0; index < numbers.length; index++) {
  sum += numbers[index];
}
console.log(sum);
// 5:
sum = 0;
numbers.forEach((num) => (sum += num));
console.log(sum);
// ;)

Answer 56

Use recursion

var sum = (arr) => arr.length === 1 ? arr[0] : arr.shift() + sum(arr);
sum([1,2,3,4]) // 10

Answer 57

In addition, sum with es6 for simple array.

const sum = [1, 2, 3].reduce((partial_sum, a) => partial_sum + a,0);
 
console.log(sum); 

For object array with default initialize value

const totalAmount = obj => 
    Object.values(obj).reduce((acc, { order_qty, mrp_price }) => 
    acc + order_qty * mrp_price, 0);
    
    console.log(totalAmount); 

Answer 58

Someone also can use map() function for the summation of an array values.

function sumOfArrVal(arr){
    let sum=0;
    arr.map(val=>sum+=val)
    return sum
}

let result=sumOfArrVal([1,2,3,4])
console.log(result)

Answer 59

For really large numbers cycling or reducing could be process intensive. What about using Gauss?

sum = (n * (n+1))/2;

From mathcentral.

Answer 60

    <!DOCTYPE html>
    <html>
    <body>

      <p>Click the button to join two arrays.</p>
      <button onclick="myFunction()">Try it</button>
      <p id="demo"></p>
    <script>
var hege = [1, 2,4,6,7,8,8];
var stale = [1, 2,4,5];
function myFunction() {
    console.log((hege.length > stale.length))    
    var children  = (hege.length > stale.length)? abc1() :abc2();       document.getElementById("demo").innerHTML = children;
}
function abc1(){
    console.log(hege,"Abc1")    
    var abcd=hege.map(function (num, idx) {
        console.log(hege.length , idx)
        return stale.length>idx?num + stale[idx]:num;
    })
    return abcd;
}

function abc2(){

    console.log(hege,"Abc2",stale)    
    var abcd=stale.map(function (num, idx) {
        console.log(hege.length , idx)
        return hege.length>idx?num + hege[idx]:num;
    })
    return abcd;
}
</script>

</body>
</html>

Answer 61

Use map:

var sum = 0;
arr.map(function(item){
    sum += item;
});

// sum now contains the total.

You could potentially add the method to the Array prototype.

Array.prototype.sum = function(){
    var sum = 0;
    this.map(function(item){
        sum += item;
    });
    return sum;
}

Then you can use it on any Array like so:

arr.sum();