Generate list of items with 0 prefix using padStart - javascript

I implemented the way to generate a list of items with iterable counts with prefix 0. What is the best way to generate such kind of list?
Current behaviour:
const generateList = (length, n, i) => {
let b = n+i
return b.toString().padStart(length.toString().length + n.toString.length, 0)
}
Array(10).fill(null).map((x, i) => generateList(10,2, i))
Output result:
["002", "003", "004", "005", "006", "007", "008", "009", "010", "011"]
Do u have any idea to make it another way?

You could determine the number of characters needed at the start and used the predetermined value to format the output for the array.
function createList(startValue, endValue) {
let
// The minimum output length, for a single digit number, is 2 chars.
outputLength = 2,
testValue = 10,
// Create an empty array which has as many items as numbers we need to
// generate for the output. Add 1 to the end value as this is to be
// inclusive of the range to create. If the +1 is not done the resulting
// array is 1 item too small.
emptyArray = Array(endValue - startValue + 1);
// As long as test value is less than the end value, keep increasing the
// output size by 1 and continue to the next multiple of 10.
while (testValue <= endValue) {
outputLength++;
testValue = testValue * 10;
}
// Create a new array, with the same length as the empty array created
// earlier. For each position place a padded number into the output array.
return Array.from(emptyArray, (currentValue, index) => {
// Pad the current value to the determined max length.
return (startValue + index).toString().padStart(outputLength, '0');
});
}
function createListWithLength(length, startValue = 0) {
return createList(startValue, startValue + length);
}
console.log(createList(2,10));
console.log(createListWithLength(30));
console.log(createListWithLength(10, 995));

Have a look at generators:
function* range(from, to) {
for (var i=from; i<to; i++)
yield i;
}
function* paddedRange(from, to) {
const length = (to-1).toString(10) + 1 /* at least one pad */;
for (const i of range(from, to))
yield i.padStart(length, '0');
}
console.log(Array.from(paddedRange(2, 12)));
You can also inline the loop from range into paddedRange, or you can make it return an array directly:
function paddedRange(from, to) {
const length = (to-1).toString(10) + 1 /* at least one pad */;
return Array.from(range(from, to), i => i.padStart(length, '0'));
}
console.log(paddedRange(2, 12));
The main simplification is that you should compute the padding length only once and give it a denotative name, instead of computing it for every number again. Also ranges are usually given by their lower and upper end instead of their begin and a length, but you can easily switch back if you need the latter for some reason.

Not sure, but maybe something like this
const generateList = length => Array(length).fill('0').map((item, index) => item + index);
console.log(generateList(20));

Related

Given an array of integers elements, find such t that cyclic t-shift operation turns elements into a sorted array If there is no such t, return -1

The solution that i've been working on so far:
function solution(elements) {
let numOfShifts;
let shift = shiftedArray(elements);
for(i = 0; i < shift.length; i++){
//Here is where i'm getting stuck... how do i continue through the loop even after the first false is returned
//until all arrays are checked for equality?
if(areEqual(shift[i])){
numOfShifts = i
}else return -1;
}
return numOfShifts;
};
function shiftedArray(array){
let shiftedArrays = [];
for(let i = array.length -1 ; i >= 1; i--){
// first element is simply a formula to create chunk 1
let firstElement = array.length - (i % array.length);
//the result of first element.
let chunk1 = array.slice(firstElement);
//the remaining numbers
let chunk2 = array.slice(0, firstElement);
//Copy of both chunks now merged and pushed into shifted arrays
shiftedArrays.push([...chunk1, ...chunk2]);
};
return shiftedArrays;
}
function areEqual(shift){
let sortedShift = [...shift].sort((a ,b) => {
return a - b
});
//takes in a single shift and does a deep check to see if the array is equal to sorted version
if(sortedShift.length === shift.length){
return sortedShift.every((element, index) => {
if(element === shift[index]){
return true;
}
return false;
})
}
}
console.log(solution([1,4,2,3]));
console.log(solution([[2, 3, 4, 5, 6, 7, 8, 9, 10, 1]]))
How do I keep the for loop running even after the first false is returned from the areEqual() function?
A side note: I understand that this could probably use some refactoring... like, I was working on this with someone earlier and they mentioned how I can simply shift the array by writing a helper function like shiftArray(arr){return arr.push(arr.pop())} but I don't get how that implementation would work considering that all that's returned is the value that was taken out of the array, not the new shifted array.
You may be doing too much work.
Say you have an array A of n integers, zero indexed.
Parse the array from index 0 to n mod n (so 0 twice). Count the number of pairs where the latter integer is less than the former, and store the first index where this happens.
If the count is 1 and the location is k (so A[k] < A[k-1]), then a cyclic shift of either -k or n-k will convert A to a sorted array. If the count is greater than 1 then there's no solution.
E.g., [4,5,0,1,2,3] -- we see k=2 is the only index which has a value below its predecessor, and a cyclic shift of -2 or 4 forms [0,1,2,3,4,5] which is sorted.

Flipping 0's and 1's from a natural number

I want to create a javascript function to flip 1's to 0's in a natural number and I'm out of Ideas to achieve this,
Actually, I had a couple of URL's, and I replaced all 0's from a query parameter with 1's and now I no longer know the original parameter value, because there were few 1's in the original parameter value and now both are mixed, so basically I screwed myself,
The only solution for me is to try flipping each 1 to 0 and then 0's to 1's and test each number as the parameter.
This is the parameter value (after replacing 0's with 1's)
11422971
using above input I want to generate numbers as follows and test each of these
11422970
10422971
10422970
01422971
As you can see only 1's and 0's are changing, the change according to binary,
Each position in your string can be one of n characters:
A "0" can be either "0" or "1"
A "1" can be either "0" or "1"
Any other character c can only be c
We can store this in an array of arrays:
"11422971" -> [ ["0", "1"], ["0, "1"], ["4"], ... ]
To transform your string to this format, you can do a split and map:
const chars = "11422971"
.split("")
.map(c => c === "1" || c === "0" ? ["1", "0"] : [ c ]);
Once you got this format, the remaining logic is to create all possible combinations from this array. There are many ways to do so (search for "array combinations" or "permutations"). I've chosen to show a recursive pattern:
const chars = "11422971"
.split("")
.map(c =>
c === "1" || c === "0"
? ["1", "0"]
: [ c ]
);
const perms = ([ xs, ...others ], s = "", ps = []) =>
xs
? ps.concat(...xs.map(x => perms(others, s + x, ps)))
: ps.concat(s);
console.log(perms(chars));
you can do it with a number like a string, and after parse it, something like that
var number= "12551";
number= number.replace("1","0");
The result of number will be "02550"
after that parse number to int
This will generate all permutations.
const generatePermutations = (number) => {
let digits = number.split('');
// find out which digits can be flipped
let digitPositions = digits.reduce((acc, val, i) => {
if (val === '0' || val === '1') acc.push(i);
return acc;
}, []);
// we're going to be taking things in reverse order
digitPositions.reverse();
// how many digits can we flip
let noBits = digitPositions.length;
// number of permutations is 2^noBits i.e. 3 digits means 2^3 = 8 permutations.
let combinations = Math.pow(2, digitPositions.length);
let permutations = [];
// for each permutation
for (var p = 0; p < combinations; p++) {
// take a copy of digits for this permutation
permutations[p] = digits.slice();
// set each of the flippable bits according to the bit positions for this permutation
// i = 3 = 011 in binary
for (var i = 0; i < noBits; i++) {
permutations[p][digitPositions[i]] = '' + ((p >> i) & 1);
}
permutations[p] = permutations[p].join('');
}
return permutations;
};
console.log(generatePermutations('11422970'));
In case your looking for a recursive approach:
function recursive(acc, first, ...rest) {
if(!first) return acc;
if(first == '0' || first == '1') {
var acc0 = acc.map(x => x + '0');
var acc1 = acc.map(x => x + '1');
return recursive([].concat(acc0, acc1), ...rest);
} else {
return recursive(acc.map(x => x + first), ...rest);
}
}
recursive([''], ...'11422971')
// output ["00422970", "10422970", "01422970", "11422970", "00422971", "10422971", "01422971", "11422971"]
This just counts in binary and fills out a template for each value.
function getPossibleValues(str) {
function getResult(n) {
let nIndex = 0;
const strValue = n.toString(2).padStart(nDigits, '0');
return str.replace(rxMatch, () => strValue.charAt(nIndex++));
}
const rxMatch = /[01]/g;
const nDigits = str.length - str.replace(rxMatch, '').length;
const nMax = Math.pow(2, nDigits);
const arrResult = [];
for(let n = 0; n<nMax; n++) {
arrResult.push(getResult(n));
}
return arrResult;
}
console.log(getPossibleValues('11422970'));
Thank you all to respond, you saved my life, btw the approach I used was,
0- convert the number into a string. (so we can perform string operations like split())
1- count the number of 1's in the string (let's say the string is "11422971", so we get three 1's, I used split('1')-1 to count)
2- generate binary of three-digit length,(ie from 000 to 111). three came from step 1.
2- break down the string to single chars, (we'll get
array=['1','1','4','2','2','9','7','1'] )
3- take the first binary number (ie b=0b000)
4- replace first 1 from the character array with the first binary digit of b (ie replace 1 with 0), similarly replace second 1 with the second binary digit of b and so on.
5- we'll get the first combination (ie "00422970")
5- repeat step 3 and 4 for all binary numbers we generated in step 2.

Addition with carryover in arrays in pure Javascript

I want to create simple addition of array with carryover. Also need carryover and result value for display.
Something like this:-
e.g var input = [[0,0,9],[0,9,9]];
var carryover = [];
var result = [];
Thanks...
The two parts you might have been struggling with, I assume would be how you get the carry, and how you get the result..
result [diget] = t % 10;
The % 10 part is what is called modulus, here I'm doing a modulus by 10, so that gets you the 10's unit value.
carryover [diget] = Math.trunc(t / 10);
For the carryover, you just then divide by 10, and then we strip the decimals,. That's what Math.trunc does.
var input = [[0,0,0,9],[0,9,9]];
var carryover = [];
var result = [];
var digetlength = Math.max(input[0].length, input[1].length);
//lets padd inputs to be same size
input[0].unshift(
...new Array(digetlength - input[0].length).fill(0));
input[1].unshift(
...new Array(digetlength - input[1].length).fill(0));
for (var diget = digetlength - 1; diget >= 0; diget -= 1) {
var t = input[0][diget] + input[1][diget];
if (diget < digetlength - 1)
t += carryover[diget + 1];
result [diget] = t % 10;
carryover [diget] = Math.trunc(t / 10);
}
result.unshift(carryover[0]);
console.log('result: ' + result.join(', '));
console.log('carry: ' + carryover.join(', '));
1.turn both numbers into array of digits, reverse them.
2.determine the end index of the for-loop with max length of above 2 arrays.
3.create the 3rd carryover digits array of zeros (don't forget the extra digit).
4.Add the respective digits from step1 and step3,
as you iterate through each of digits from right to left,
4.1 if the sum is greater than 9 then add 1 into next carryover slot.
5. you should have array of carried over digits when the for-loop is done
count the number of 1s you have in them.
function numberOfCarryOperations(num1, num2) {
const dd1 = [...num1.toString()].reverse()
const dd2 = [...num2.toString()].reverse()
const end = Math.max(dd1.length, dd2.length)
const carry = Array(end+1).fill(0)
for (let i = 0; i < end; i++) {
//console.log(i,(Number(dd1[i]?dd1[i]:0)),Number(dd2[i]?dd2[i]:0),carry)
if (((Number(dd1[i]?dd1[i]:0)) + Number(dd2[i]?dd2[i]:0) + carry[i]) > 9) {
carry[i+1] = 1
}
//console.log('-----',carry)
}
//console.log(num1, num2,carry)
return carry.reduce((sum,curr)=>sum+curr)
}
Here is my attempt. It will accept the following as input:
Any number of input arrays
The input arrays don't all need to have the same number of items
I've added code comments to explain what goes on, I hope they're informative enough to explain the answer.
const
input = [
[0,0,9],
[0,9,9],
[1,0,9,9]
];
function getMaxArrayLength(values) {
// Determine the number of items in the longest array. Initialize the reduce with 0.
return values.reduce((maxLength, array) => {
// Return the largets number between the last largest number and the
// length of the current array.
return Math.max(maxLength, array.length);
}, 0);
}
function sumValues(values) {
const
// Determine the number of items in the longest array.
maxLength = getMaxArrayLength(values),
result = [],
carry = [];
// Loop over the length of the longest array. The value of index will be substracted from
// the length of the input arrays. Therefore it is easier to start at 1 as this will
// return a proper array index without needing to correct it.
for (let index = 1; index <= maxLength; index++) {
const
// Get the carryover value from the last sum or 0 in case there is no previous value.
carryValue = (carry.length === 0) ? 0 : carry[carry.length-1],
// Sum up all the values at the current index of all the input arrays. After summing up
// all the values, also add the carry over from the last sum.
sum = values.reduce((sum, array) => {
// Determine the index for the current array. Start at the end and substract the
// current index. This way the values in the array are processed last to first.
const
arrayIndex = array.length - index;
// It could be the current array doesn't have as many items as the longest array,
// when the arrayIndex is less than 0 just return the current result.
if (arrayIndex < 0) {
return sum;
}
// Return the accumulated value plus the value at the current index of the
// current source array.
return sum + array[arrayIndex];
}, 0) + carryValue;
// The carry over value is the number of times 10 fits into the sum. This should be rounded
// down so for instance 5/10=.5 becomes 0.
carry.push(Math.floor(sum / 10));
// Push the remainder of the sum divided by 10 into the result so 15 % 10 = 5.
result.push(sum % 10);
}
// Return the carry over and the result, reverse the arrays before returning them.
return {
carryOver: carry.reverse(),
result: result.reverse()
};
}
const
result = sumValues(input);
console.log(`Carry over: ${result.carryOver}`);
console.log(`Result: ${result.result}`);

javascript hackerranks sherlock and array performance issue

Watson gives Sherlock an array A of length N. Then he asks him to
determine if there exists an element in the array such that the sum of
the elements on its left is equal to the sum of the elements on its
right. If there are no elements to the left/right, then the sum is
considered to be zero. Formally, find an i, such that,
Input Format
The first line contains T, the number of test cases. For each test
case, the first line contains N, the number of elements in the array
A. The second line for each test case contains N space-separated
integers, denoting the array A.
Constraints
1<=T<=10
1<=N<=10^5
1<=Ai<=2*10^4
1<=i<=N
Output Format
For each test case print YES if there exists an element in the array,
such that the sum of the elements on its left is equal to the sum of
the elements on its right; otherwise print NO.
Sample Input
2
3
1 2 3
4
1 2 3 3
Sample Output
NO
YES
Explanation
For the first test case, no such index exists. For the second test
case,
therefore index 3 satisfies the given conditions.
I'm having timeout issues on 3 of the test cases
function check(input) {
var result = "NO";
var sum=0;
input.map(function(data){
sum=sum+(+data);
})
sumLeft=0;
sumRight=sum-(+input[0]);
for(var i=1;i<input.length;i++){
sumLeft=sumLeft+(+input[i-1]);
sumRight=sumRight-(+input[i])
if(sumLeft==sumRight)
{
console.log("YES");
return;
}
}
console.log("NO");
}
function processData(input) {
//Enter your code here
var lines = input.split("\r\n");
for (var m = 2; m < lines.length; m = m + 2) {
check(lines[m].split(" "));
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function(input) {
_input += input;
});
process.stdin.on("end", function() {
processData(_input);
});
Loop over the array once to find the sum. Declare two variables: sumLeft and sumRight. sumLeft should have an initial value of 0 and sumRight should be totalSum-arr[0].
Iterate over the array again and increment sumLeft by the (n-1) element and decrement sumRight by the nth element. Keep comparing the two variables to check if they equal each other. You cut your time complexity down to O(n)
The below code passed the test on https://www.hackerrank.com/challenges/sherlock-and-array . The tricky part was setting up default responses for when the array length was 1. I will admit that #trincot 's answer was more efficient (n as opposed to 2n) for arrays containing only positive integers.
function check(input) {
var result = "NO";
var sum=0;
if(input.length == 1){
console.log("YES");
return;
}
input.map(function(data){
sum=sum+(+data);
})
sumLeft=0;
sumRight=sum-(+input[0]);
for(var i=1;i<input.length-1;i++){
sumLeft=sumLeft+(+input[i-1]);
sumRight=sumRight-(+input[i])
if(sumLeft==sumRight)
{
console.log("YES");
return;
}else if (sumLeft>sumRight) { ///worked both with and without this optimization
console.log("NO");
return;
}
}
console.log("NO");
}
function processData(input) {
//var lines = input.split("\r\n");
var lines = input.split(/\r|\n/)
for (var m = 2; m < lines.length; m = m + 2) {
check(lines[m].split(" "));
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function(input) {
_input += input;
});
process.stdin.on("end", function() {
processData(_input);
});
You could go through the array from both ends in inwards direction using two pointers (indices). Keep a balance, starting with 0, as follows:
When the balance is negative move the left pointer one step to the right while increasing the balance with the value you leave behind. When the balance is positive, move the right pointer one step to the left while decreasing the balance with the value you leave behind.
When the two pointers meet each other, check the balance. If it is zero, you have success.
Here is the algorithm in ES6 code, together with a text area where you can adapt the input according to the required input format:
function hasMiddle(a) {
var balance = 0, i = 0, j = a.length-1;
while (i < j) balance += balance > 0 ? -a[j--] : a[i++];
return !balance;
}
// I/O: event handling, parsing input, formatting output
var input = document.querySelector('textarea');
var output = document.querySelector('pre');
input.oninput = function() {
var lines = this.value.trim().split(/[\r\n]+/).filter(s => s.trim().length);
// Strip the case count and array element counts:
lines = lines.slice(1).filter( (s, i) => i % 2 );
// Call function for each test case, returning array of booleans:
var results = lines.map( line => hasMiddle(line.match(/\d+/g).map(Number)) );
// Output results
output.textContent = results.map( pos => pos ? 'YES' : 'NO' ).join('\n');
}
// Evaluate input immediately
input.oninput();
Input:<br>
<textarea style="width:100%; height:120px">2
3
1 2 3
4
1 2 3 3
</textarea>
<pre></pre>
This algorithm requires your input array to consist of non-negative numbers.
If you need to support negative numbers in your array, then the algorithm needs to go through the array first to calculate the sum, and then go through the array again to find the point where the balance reaches 0:
function hasMiddle(a) {
var balance = a.reduce( (sum, v) => sum + v );
return !a.every ( (v, i) => balance -= v + (i ? a[i-1] : 0) );
}
// I/O for snippet
var input = document.querySelector('textarea');
var output = document.querySelector('pre');
input.oninput = function() {
var lines = this.value.trim().split(/[\r\n]+/).filter(s => s.trim().length);
// Strip the case count and array element counts:
lines = lines.slice(1).filter( (s, i) => i % 2 );
// Call function for each test case, returning array of booleans:
var results = lines.map( line => hasMiddle(line.match(/[\d-]+/g).map(Number)));
// Output results
output.textContent = results.map( pos => pos ? 'YES' : 'NO' ).join('\n');
}
// Evaluate input immediately
input.oninput();
Input:<br>
<textarea style="width:100%; height:120px">2
3
1 2 3
4
1 2 3 3
</textarea>
<pre></pre>
Given that we have a proper array one might do as follows
var arr = [...Array(35)].map(_ => ~~(Math.random()*10)+1),
sum = arr.reduce((p,c) => p+c),
half = Math.floor(sum/2),
ix;
console.log(JSON.stringify(arr));
midix = arr.reduce((p,c,i,a) => { (p+=c) < half ? p : !ix && (ix = i);
return i < a.length - 1 ? p : ix;
},0);
console.log("best possible item in the middle # index", midix,": with value:",arr[midix]);
console.log("sums around midix:",
arr.slice(0,midix)
.reduce((p,c) => p+c),
":",
arr.slice(midix+1)
.reduce((p,c) => p+c));
Of course for randomly populated arrays as above, we can not always get a perfect middle index.

How to compute the sum and average of elements in an array? [duplicate]

Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
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);

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