Simplify array from many values to fewer - javascript

I have an array with many numbers (165) and want to 'simplify' the array to fewer numbers (50). The reduced array should still be representative for the 165 original numbers.
For example:
If I had this array with 8 values [2, 4, 3, 8, 1, 4, 9, 3] and would ask to reduce it to 4 values, I want to receive this array [3, 5.5, 2.5, 6]
Currently I have a function that works when the reduced number and the original number can be divided (as with 8 and 4) but when I try using it with let's say 165 values that should be simplified to 50 values it returns me 54 values:
const array = [28, 71, 64, 116, 128, 8, 78, 172, 142, 96, 12 ... 165 values]
const valuesToSum = Math.floor(array.length / 50); // 50 is the wanted array length
for (let i = 0; i < array.length; i++) {
if (i % valuesToSum !== 0 || i === 0) {
sum += array[i];
} else {
returnArray.push(sum / valuesToSum);
sum = 0;
}
}
return returnArray;
In the end this should be a JavaScript function but if someone could explain me this on a mathematical level it would help me a lot.


In order to get the exact number of groups you want, you can't round off the number of elements to group. For instance if you want to reduce from 165 to 50, some of the groups will have 3 elements, some will have 4.
To do this, use nested loops. The outer loop increments by the size of the group, while the inner loop increments by 1 within the group. The rounding happens when you convert this inner index to the array index.
const array = [28, 71, 64, 116, 128, 8, 78, 172, 142, 96, 12]
function reduceArray(array, newSize) {
const returnArray = [];
const valuesToSum = array.length / newSize;
for (let i = 0; i < array.length; i += valuesToSum) {
let sum = 0;
let j;
let start_i = Math.floor(i);
for (j = start_i; j < Math.min(start_i + valuesToSum, array.length); j++) {
sum += array[j];
}
returnArray.push(sum/(j - start_i));
}
return returnArray;
}
console.log(reduceArray(array, 4));
const bigArray = [];
for (let i = 0; i < 165; i++) {
bigArray.push(Math.floor(Math.random() * 200));
}
let result = reduceArray(bigArray, 50);
console.log(result.length);
console.log(result);

Related

How to generate list of n sized lists from given list

Say we are given a list,
var l = [50, 55, 56, 57, 58, 60]
and a random number n.
I need to create a function that return all possible lists of length n.
If n = 3,
[50,55,57],[50,55,58],[50,55,60],[50,57,58],[50,57,60],[50,58,60],[55,57,58],[55,57,60],[55,58,60],[57,58,60]
EDIT
I could not add the complete output array, because I don't know how to generate it.

Assuming n is less than the size of the input array, you could do this:
var soln = []
function generateListsFromSize(array, n, offset=0, current=[]) {
if (current.length == n){
soln.push(current.map(elem => elem))
return
}
for (var i=offset; i < array.length; i++) {
current.push(array[i]);
generateListsFromSize(array, n, i + 1, current);
current.pop();
}
}
generateListsFromSize([50, 55, 57, 58, 60], 3)
console.log(soln)
The idea is to maintain an array, in this case, current, and try all the different scenarios by backtracking.

That's called recursion:
const arr = [50, 55, 56, 57, 58]
const list = function (n, index, tempArr) {
// Return of recursion
if (tempArr.length == n) {
console.log(tempArr)
return
}
// Recursion
while (index < arr.length) {
tempArr.push(arr[index])
list(n, ++index, tempArr)
tempArr.pop()
}
}
list(3, 0, [])

Finding the closest level by experience [duplicate]

I have a number from minus 1000 to plus 1000 and I have an array with numbers in it. Like this:
[2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
I want that the number I've got changes to the nearest number of the array.
For example I get 80 as number I want it to get 82.

ES5 Version:
var counts = [4, 9, 15, 6, 2],
goal = 5;
var closest = counts.reduce(function(prev, curr) {
return (Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});
console.log(closest);

Here's the pseudo-code which should be convertible into any procedural language:
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
number = 112
print closest (number, array)
def closest (num, arr):
curr = arr[0]
foreach val in arr:
if abs (num - val) < abs (num - curr):
curr = val
return curr
It simply works out the absolute differences between the given number and each array element and gives you back one of the ones with the minimal difference.
For the example values:
number = 112 112 112 112 112 112 112 112 112 112
array = 2 42 82 122 162 202 242 282 322 362
diff = 110 70 30 10 50 90 130 170 210 250
|
+-- one with minimal absolute difference.
As a proof of concept, here's the Python code I used to show this in action:
def closest (num, arr):
curr = arr[0]
for index in range (len (arr)):
if abs (num - arr[index]) < abs (num - curr):
curr = arr[index]
return curr
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
number = 112
print closest (number, array)
And, if you really need it in Javascript, see below for a complete HTML file which demonstrates the function in action:
<html>
<head></head>
<body>
<script language="javascript">
function closest (num, arr) {
var curr = arr[0];
var diff = Math.abs (num - curr);
for (var val = 0; val < arr.length; val++) {
var newdiff = Math.abs (num - arr[val]);
if (newdiff < diff) {
diff = newdiff;
curr = arr[val];
}
}
return curr;
}
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
number = 112;
alert (closest (number, array));
</script>
</body>
</html>
Now keep in mind there may be scope for improved efficiency if, for example, your data items are sorted (that could be inferred from the sample data but you don't explicitly state it). You could, for example, use a binary search to find the closest item.
You should also keep in mind that, unless you need to do it many times per second, the efficiency improvements will be mostly unnoticable unless your data sets get much larger.
If you do want to try it that way (and can guarantee the array is sorted in ascending order), this is a good starting point:
<html>
<head></head>
<body>
<script language="javascript">
function closest (num, arr) {
var mid;
var lo = 0;
var hi = arr.length - 1;
while (hi - lo > 1) {
mid = Math.floor ((lo + hi) / 2);
if (arr[mid] < num) {
lo = mid;
} else {
hi = mid;
}
}
if (num - arr[lo] <= arr[hi] - num) {
return arr[lo];
}
return arr[hi];
}
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
number = 112;
alert (closest (number, array));
</script>
</body>
</html>
It basically uses bracketing and checking of the middle value to reduce the solution space by half for each iteration, a classic O(log N) algorithm whereas the sequential search above was O(N):
0 1 2 3 4 5 6 7 8 9 <- indexes
2 42 82 122 162 202 242 282 322 362 <- values
L M H L=0, H=9, M=4, 162 higher, H<-M
L M H L=0, H=4, M=2, 82 lower/equal, L<-M
L M H L=2, H=4, M=3, 122 higher, H<-M
L H L=2, H=3, difference of 1 so exit
^
|
H (122-112=10) is closer than L (112-82=30) so choose H
As stated, that shouldn't make much of a difference for small datasets or for things that don't need to be blindingly fast, but it's an option you may want to consider.

ES6 (ECMAScript 2015) Version:
const counts = [4, 9, 15, 6, 2];
const goal = 5;
const output = counts.reduce((prev, curr) => Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
console.log(output);
For reusability you can wrap in a curry function that supports placeholders (http://ramdajs.com/0.19.1/docs/#curry or https://lodash.com/docs#curry). This gives lots of flexibility depending on what you need:
const getClosest = _.curry((counts, goal) => {
return counts.reduce((prev, curr) => Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});
const closestToFive = getClosest(_, 5);
const output = closestToFive([4, 9, 15, 6, 2]);
console.log(output);
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.20/lodash.min.js"></script>

Working code as below:
var array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
function closest(array, num) {
var i = 0;
var minDiff = 1000;
var ans;
for (i in array) {
var m = Math.abs(num - array[i]);
if (m < minDiff) {
minDiff = m;
ans = array[i];
}
}
return ans;
}
console.log(closest(array, 88));

Works with unsorted arrays
While there were some good solutions posted here, JavaScript is a flexible language that gives us tools to solve a problem in many different ways.
It all comes down to your style, of course. If your code is more functional, you'll find the reduce variation suitable, i.e.:
arr.reduce(function (prev, curr) {
return (Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});
However, some might find that hard to read, depending on their coding style. Therefore I propose a new way of solving the problem:
var findClosest = function (x, arr) {
var indexArr = arr.map(function(k) { return Math.abs(k - x) })
var min = Math.min.apply(Math, indexArr)
return arr[indexArr.indexOf(min)]
}
findClosest(80, [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]) // Outputs 82
Contrary to other approaches finding the minimum value using Math.min.apply, this one doesn't require the input array arr to be sorted. We don't need to care about the indexes or sort it beforehand.
I'll explain the code line by line for clarity:
arr.map(function(k) { return Math.abs(k - x) }) Creates a new array, essentially storing the absolute values of the given numbers (number in arr) minus the input number (x). We'll look for the smallest number next (which is also the closest to the input number)
Math.min.apply(Math, indexArr) This is a legit way of finding the smallest number in the array we've just created before (nothing more to it)
arr[indexArr.indexOf(min)] This is perhaps the most interesting part. We have found our smallest number, but we're not sure if we should add or subtract the initial number (x). That's because we used Math.abs() to find the difference. However, array.map creates (logically) a map of the input array, keeping the indexes in the same place. Therefore, to find out the closest number we just return the index of the found minimum in the given array indexArr.indexOf(min).
I've created a bin demonstrating it.

All of the solutions are over-engineered.
It is as simple as:
const needle = 5;
const haystack = [1, 2, 3, 4, 5, 6, 7, 8, 9];
haystack.sort((a, b) => {
return Math.abs(a - needle) - Math.abs(b - needle);
})[0];
// 5

For sorted arrays (linear search)
All answers so far concentrate on searching through the whole array.
Considering your array is sorted already and you really only want the nearest number this is probably the easiest (but not fastest) solution:
var a = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
var target = 90000;
/**
* Returns the closest number from a sorted array.
**/
function closest(arr, target) {
if (!(arr) || arr.length == 0)
return null;
if (arr.length == 1)
return arr[0];
for (var i = 1; i < arr.length; i++) {
// As soon as a number bigger than target is found, return the previous or current
// number depending on which has smaller difference to the target.
if (arr[i] > target) {
var p = arr[i - 1];
var c = arr[i]
return Math.abs(p - target) < Math.abs(c - target) ? p : c;
}
}
// No number in array is bigger so return the last.
return arr[arr.length - 1];
}
// Trying it out
console.log(closest(a, target));
Note that the algorithm can be vastly improved e.g. using a binary tree.

ES6
Works with sorted and unsorted arrays
Numbers Integers and Floats, Strings welcomed
/**
* Finds the nearest value in an array of numbers.
* Example: nearestValue(array, 42)
*
* #param {Array<number>} arr
* #param {number} val the ideal value for which the nearest or equal should be found
*/
const nearestValue = (arr, val) => arr.reduce((p, n) => (Math.abs(p) > Math.abs(n - val) ? n - val : p), Infinity) + val
Examples:
let values = [1,2,3,4,5]
console.log(nearestValue(values, 10)) // --> 5
console.log(nearestValue(values, 0)) // --> 1
console.log(nearestValue(values, 2.5)) // --> 2
values = [100,5,90,56]
console.log(nearestValue(values, 42)) // --> 56
values = ['100','5','90','56']
console.log(nearestValue(values, 42)) // --> 56

This solution uses ES5 existential quantifier Array#some, which allows to stop the iteration, if a condition is met.
Opposit of Array#reduce, it does not need to iterate all elements for one result.
Inside the callback, an absolute delta between the searched value and actual item is taken and compared with the last delta. If greater or equal, the iteration stops, because all other values with their deltas are greater than the actual value.
If the delta in the callback is smaller, then the actual item is assigned to the result and the delta is saved in lastDelta.
Finally, smaller values with equal deltas are taken, like in the below example of 22, which results in 2.
If there is a priority of greater values, the delta check has to be changed from:
if (delta >= lastDelta) {
to:
if (delta > lastDelta) {
// ^^^ without equal sign
This would get with 22, the result 42 (Priority of greater values).
This function needs sorted values in the array.
Code with priority of smaller values:
function closestValue(array, value) {
var result,
lastDelta;
array.some(function (item) {
var delta = Math.abs(value - item);
if (delta >= lastDelta) {
return true;
}
result = item;
lastDelta = delta;
});
return result;
}
var data = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
console.log(21, closestValue(data, 21)); // 2
console.log(22, closestValue(data, 22)); // 2 smaller value
console.log(23, closestValue(data, 23)); // 42
console.log(80, closestValue(data, 80)); // 82
Code with priority of greater values:
function closestValue(array, value) {
var result,
lastDelta;
array.some(function (item) {
var delta = Math.abs(value - item);
if (delta > lastDelta) {
return true;
}
result = item;
lastDelta = delta;
});
return result;
}
var data = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
console.log(21, closestValue(data, 21)); // 2
console.log(22, closestValue(data, 22)); // 42 greater value
console.log(23, closestValue(data, 23)); // 42
console.log(80, closestValue(data, 80)); // 82

Other answers suggested the you would need to iterate through the entire array:
calculate the deviation for each element
keep track of the smallest deviation and its element
finally, after iterating through the entire array, return that element with that smallest deviation.
If the array is already sorted, that does not make sense. There is no need to calculate all deviations. e.g. in an ordered collection of 1 million elements, you only need to calculate ~19 deviations (at most) to find your match. You can accomplish this with a binary-search approach:
function findClosestIndex(arr, element) {
let from = 0, until = arr.length - 1
while (true) {
const cursor = Math.floor((from + until) / 2);
if (cursor === from) {
const diff1 = element - arr[from];
const diff2 = arr[until] - element;
return diff1 <= diff2 ? from : until;
}
const found = arr[cursor];
if (found === element) return cursor;
if (found > element) {
until = cursor;
} else if (found < element) {
from = cursor;
}
}
}
Result:
console.log(findClosestIndex([0, 1, 2, 3.5, 4.5, 5], 4));
// output: 3
console.log(findClosestIndex([0, 1, 2, 3.49, 4.5, 5], 4));
// output: 4
console.log(findClosestIndex([0, 1, 2, 3.49, 4.5, 5], 90));
// output: 5
console.log(findClosestIndex([0, 1, 2, 3.49, 4.5, 5], -1));
// output: 0

A simpler way with O(n) time complexity is to do this in one iteration of the array. This method is intended for unsorted arrays.
Following is a javascript example, here from the array we find the number which is nearest to "58".
var inputArr = [150, 5, 200, 50, 30];
var search = 58;
var min = Math.min();
var result = 0;
for(i=0;i<inputArr.length;i++) {
let absVal = Math.abs(search - inputArr[i])
if(min > absVal) {
min=absVal;
result = inputArr[i];
}
}
console.log(result); //expected output 50 if input is 58
This will work for positive, negative, decimal numbers as well.
Math.min() will return Infinity.
The result will store the value nearest to the search element.

I don't know if I'm supposed to answer an old question, but as this post appears first on Google searches, I hoped that you would forgive me adding my solution & my 2c here.
Being lazy, I couldn't believe that the solution for this question would be a LOOP, so I searched a bit more and came back with filter function:
var myArray = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
var myValue = 80;
function BiggerThan(inArray) {
return inArray > myValue;
}
var arrBiggerElements = myArray.filter(BiggerThan);
var nextElement = Math.min.apply(null, arrBiggerElements);
alert(nextElement);
That's all !

My answer to a similar question is accounting for ties too and it is in plain Javascript, although it doesn't use binary search so it is O(N) and not O(logN):
var searchArray= [0, 30, 60, 90];
var element= 33;
function findClosest(array,elem){
var minDelta = null;
var minIndex = null;
for (var i = 0 ; i<array.length; i++){
var delta = Math.abs(array[i]-elem);
if (minDelta == null || delta < minDelta){
minDelta = delta;
minIndex = i;
}
//if it is a tie return an array of both values
else if (delta == minDelta) {
return [array[minIndex],array[i]];
}//if it has already found the closest value
else {
return array[i-1];
}
}
return array[minIndex];
}
var closest = findClosest(searchArray,element);
https://stackoverflow.com/a/26429528/986160

I like the approach from Fusion, but there's a small error in it. Like that it is correct:
function closest(array, number) {
var num = 0;
for (var i = array.length - 1; i >= 0; i--) {
if(Math.abs(number - array[i]) < Math.abs(number - array[num])){
num = i;
}
}
return array[num];
}
It it also a bit faster because it uses the improved for loop.
At the end I wrote my function like this:
var getClosest = function(number, array) {
var current = array[0];
var difference = Math.abs(number - current);
var index = array.length;
while (index--) {
var newDifference = Math.abs(number - array[index]);
if (newDifference < difference) {
difference = newDifference;
current = array[index];
}
}
return current;
};
I tested it with console.time() and it is slightly faster than the other function.

The most efficient would be a binary search. However even simple solutions can exit when the next number is a further match from the current. Nearly all solutions here are not taking into account that the array is ordered and iterating though the whole thing :/
const closest = (orderedArray, value, valueGetter = item => item) =>
orderedArray.find((item, i) =>
i === orderedArray.length - 1 ||
Math.abs(value - valueGetter(item)) < Math.abs(value - valueGetter(orderedArray[i + 1])));
var data = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
console.log('21 -> 2', closest(data, 21) === 2);
console.log('22 -> 42', closest(data, 22) === 42); // equidistant between 2 and 42, select highest
console.log('23 -> 42', closest(data, 23) === 42);
console.log('80 -> 82', closest(data, 80) === 82);
This can be run on non-primitives too e.g. closest(data, 21, item => item.age)
Change find to findIndex to return the index in the array.

If the array is sorted like in your example, you can use a Binary Search for a better time complexity of O(log n).
const myArray = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
const binaryClosestIdx = (arr, target) => {
let start = 0;
let end = arr.length - 1;
let mid = Math.floor((start + end) / 2);
while (1) {
if (arr[mid] === target) {
return mid;
}
else if (start >= end) {
break;
}
else if (arr[mid] > target) {
end = mid - 1;
} else {
start = mid + 1;
}
mid = Math.floor((start + end) / 2);
}
// Return the closest between the last value checked and it's surrounding neighbors
const first = Math.max(mid - 1, 0);
const neighbors = arr.slice(first, mid + 2);
const best = neighbors.reduce((b, el) => Math.abs(el - target) < Math.abs(b - target) ? el : b);
return first + neighbors.indexOf(best);
}
const closestValue = myArray[binaryClosestIdx(myArray, 80)];
console.log(closestValue);
How it works :
It compares the target value to the middle element of the array. If the middle element is bigger we can ignore every element after it as they are going to be even bigger. The same goes if the middle element is smaller, we can ignore every element before it.
If the target value is found we return it, otherwise we compare the last value tested with its surrounding neighbors as the closest value can only be between those 3 values.

Another variant here we have circular range connecting head to toe and accepts only min value to given input. This had helped me get char code values for one of the encryption algorithm.
function closestNumberInCircularRange(codes, charCode) {
return codes.reduce((p_code, c_code)=>{
if(((Math.abs(p_code-charCode) > Math.abs(c_code-charCode)) || p_code > charCode) && c_code < charCode){
return c_code;
}else if(p_code < charCode){
return p_code;
}else if(p_code > charCode && c_code > charCode){
return Math.max.apply(Math, [p_code, c_code]);
}
return p_code;
});
}

To Find Two Closest Number in array
function findTwoClosest(givenList, goal) {
var first;
var second;
var finalCollection = [givenList[0], givenList[1]];
givenList.forEach((item, firtIndex) => {
first = item;
for (let i = firtIndex + 1; i < givenList.length; i++) {
second = givenList[i];
if (first + second < goal) {
if (first + second > finalCollection[0] + finalCollection[1]) {
finalCollection = [first, second];
}
}
}
});
return finalCollection;
}
var counts = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
var goal = 80;
console.log(findTwoClosest(counts, goal));

You can use below logic to find closest number without using reduce function
let arr = [0, 80, 10, 60, 20, 50, 0, 100, 80, 70, 1];
const n = 2;
let closest = -1;
let closeDiff = -1;
for (let i = 0; i < arr.length; i++) {
if (Math.abs(arr[i] - n) < closeDiff || closest === -1) {
closeDiff = Math.abs(arr[i] - n);
closest = arr[i];
}
}
console.log(closest);

#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
class CompareFunctor
{
public:
CompareFunctor(int n) { _n = n; }
bool operator()(int & val1, int & val2)
{
int diff1 = abs(val1 - _n);
int diff2 = abs(val2 - _n);
return (diff1 < diff2);
}
private:
int _n;
};
int Find_Closest_Value(int nums[], int size, int n)
{
CompareFunctor cf(n);
int cn = *min_element(nums, nums + size, cf);
return cn;
}
int main()
{
int nums[] = { 2, 42, 82, 122, 162, 202, 242, 282, 322, 362 };
int size = sizeof(nums) / sizeof(int);
int n = 80;
int cn = Find_Closest_Value(nums, size, n);
cout << "\nClosest value = " << cn << endl;
cin.get();
}

For a small range, the simplest thing is to have a map array, where, eg, the 80th entry would have the value 82 in it, to use your example. For a much larger, sparse range, probably the way to go is a binary search.
With a query language you could query for values some distance either side of your input number and then sort through the resulting reduced list. But SQL doesn't have a good concept of "next" or "previous", to give you a "clean" solution.

Split 100 numbers into N same length parts

Specify how many numbers you want to show.
E.g. if
you specify 2, the result should be 0 and 100. If you specify 3, the numbers should be 0, 50 and 100. If you specify 4, the numbers should be 0, 33, 67, 100 etc.
There should always be the same length between every displayed number.

The for-loop solution:
const run = (max, num) => {
let result = [];
let part = max / (num - 1);
for(let i = 0; i < max; i += part) {
result.push(Math.round(i));
}
result.push(max);
return (result);
};
console.log(run(100, 4)); // [0, 33, 67, 100]
console.log(run(100, 5)); // [0, 25, 50, 75, 100]
console.log(run(100, 7)); // [0, 17, 33, 50, 67, 83, 100]

If you can use es6 this is a nice functional one-liner to do it:
const fn = (n, l) => [...Array(n)].map((i, id) => Math.floor(l/(n-1) * id))
console.log(fn(4,100))
console.log(fn(2,100))
Of course you can't always have the exact distance between numbers when you want integers — you need to round somewhere when the number doesn't evenly divide.

function numbers(number){
var max=100;
spacing = Math.floor( max/ (number-1) );
var returnMe = [0];
var cur=spacing;
while(cur <= max){
returnMe.push( cur );
cur+=spacing;
}
return returnMe;
}

Every n times, skip n items and increase n by 1

This is probably an odd question since I have a solution (below), but was hoping someone could show me a more succinct or readable way to do this:
I created a loop that outputs the following array:
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91]
the gaps between numbers get progressively larger:
1-0 = 1
3-1 = 2
6-3 = 3
10-6 = 4
...
91-78 = 13
etc.
I did it by creating two variables, step keeps track of the gap size and count keeps track of the current 'position' in the gap. count counts down to zero, then increases step by one.
var output = [];
var step = 0;
var count = 0;
for (var i = 0; i < 100; i++) {
if (count == 0){
step += 1;
count = step;
output.push(i);
}
count -= 1;
}

You can try the following:
var output = [];
var total = 0;
for (var i=1; i < 100; i++) {
output.push(total);
total += i;
}
The gaps between numbers simply increase by one for each step, so a for loop should be able to track this change.

You should skip useless iterations. If you want a sequence of 100 numbers, use
var output = [];
var step = 0;
for (var i = 0; i < 100; i++) {
step += i;
output.push(step);
}
If you want the general term,
aₙ = ∑ⁿᵢ₌₀ i = n*(n+1)/2
So you can also do
var output = [];
for (var i = 0; i < 100; i++) {
output.push(i * (i+1) / 2);
}

You can save the total helper variable with this solution:
var output = [0]
for (var i = 1; i < 14; i++) {
output.push(output[i - 1] + i)
}
console.log(output) // [ 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91 ]
This solution takes into account that the value to add the counter value to is already present at the last position in the array.
A recursive version is also possible:
output = (function f(x) {
return x.length == 14 ? x : f(x.concat([x[x.length - 1] + x.length]))
})([0])
console.log(output); // [ 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91 ]
Here is no additional counter variable is needed. I use concat because it returns an array what I need for the recursive call, where push returns the new array length. The argument for concat is an array with one element with the new value to add.
Try online

Get the closest number out of an array

I have a number from minus 1000 to plus 1000 and I have an array with numbers in it. Like this:
[2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
I want that the number I've got changes to the nearest number of the array.
For example I get 80 as number I want it to get 82.

ES5 Version:
var counts = [4, 9, 15, 6, 2],
goal = 5;
var closest = counts.reduce(function(prev, curr) {
return (Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});
console.log(closest);

Here's the pseudo-code which should be convertible into any procedural language:
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
number = 112
print closest (number, array)
def closest (num, arr):
curr = arr[0]
foreach val in arr:
if abs (num - val) < abs (num - curr):
curr = val
return curr
It simply works out the absolute differences between the given number and each array element and gives you back one of the ones with the minimal difference.
For the example values:
number = 112 112 112 112 112 112 112 112 112 112
array = 2 42 82 122 162 202 242 282 322 362
diff = 110 70 30 10 50 90 130 170 210 250
|
+-- one with minimal absolute difference.
As a proof of concept, here's the Python code I used to show this in action:
def closest (num, arr):
curr = arr[0]
for index in range (len (arr)):
if abs (num - arr[index]) < abs (num - curr):
curr = arr[index]
return curr
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
number = 112
print closest (number, array)
And, if you really need it in Javascript, see below for a complete HTML file which demonstrates the function in action:
<html>
<head></head>
<body>
<script language="javascript">
function closest (num, arr) {
var curr = arr[0];
var diff = Math.abs (num - curr);
for (var val = 0; val < arr.length; val++) {
var newdiff = Math.abs (num - arr[val]);
if (newdiff < diff) {
diff = newdiff;
curr = arr[val];
}
}
return curr;
}
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
number = 112;
alert (closest (number, array));
</script>
</body>
</html>
Now keep in mind there may be scope for improved efficiency if, for example, your data items are sorted (that could be inferred from the sample data but you don't explicitly state it). You could, for example, use a binary search to find the closest item.
You should also keep in mind that, unless you need to do it many times per second, the efficiency improvements will be mostly unnoticable unless your data sets get much larger.
If you do want to try it that way (and can guarantee the array is sorted in ascending order), this is a good starting point:
<html>
<head></head>
<body>
<script language="javascript">
function closest (num, arr) {
var mid;
var lo = 0;
var hi = arr.length - 1;
while (hi - lo > 1) {
mid = Math.floor ((lo + hi) / 2);
if (arr[mid] < num) {
lo = mid;
} else {
hi = mid;
}
}
if (num - arr[lo] <= arr[hi] - num) {
return arr[lo];
}
return arr[hi];
}
array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
number = 112;
alert (closest (number, array));
</script>
</body>
</html>
It basically uses bracketing and checking of the middle value to reduce the solution space by half for each iteration, a classic O(log N) algorithm whereas the sequential search above was O(N):
0 1 2 3 4 5 6 7 8 9 <- indexes
2 42 82 122 162 202 242 282 322 362 <- values
L M H L=0, H=9, M=4, 162 higher, H<-M
L M H L=0, H=4, M=2, 82 lower/equal, L<-M
L M H L=2, H=4, M=3, 122 higher, H<-M
L H L=2, H=3, difference of 1 so exit
^
|
H (122-112=10) is closer than L (112-82=30) so choose H
As stated, that shouldn't make much of a difference for small datasets or for things that don't need to be blindingly fast, but it's an option you may want to consider.

ES6 (ECMAScript 2015) Version:
const counts = [4, 9, 15, 6, 2];
const goal = 5;
const output = counts.reduce((prev, curr) => Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
console.log(output);
For reusability you can wrap in a curry function that supports placeholders (http://ramdajs.com/0.19.1/docs/#curry or https://lodash.com/docs#curry). This gives lots of flexibility depending on what you need:
const getClosest = _.curry((counts, goal) => {
return counts.reduce((prev, curr) => Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});
const closestToFive = getClosest(_, 5);
const output = closestToFive([4, 9, 15, 6, 2]);
console.log(output);
<script src="https://cdn.jsdelivr.net/npm/lodash#4.17.20/lodash.min.js"></script>

Working code as below:
var array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
function closest(array, num) {
var i = 0;
var minDiff = 1000;
var ans;
for (i in array) {
var m = Math.abs(num - array[i]);
if (m < minDiff) {
minDiff = m;
ans = array[i];
}
}
return ans;
}
console.log(closest(array, 88));

Works with unsorted arrays
While there were some good solutions posted here, JavaScript is a flexible language that gives us tools to solve a problem in many different ways.
It all comes down to your style, of course. If your code is more functional, you'll find the reduce variation suitable, i.e.:
arr.reduce(function (prev, curr) {
return (Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});
However, some might find that hard to read, depending on their coding style. Therefore I propose a new way of solving the problem:
var findClosest = function (x, arr) {
var indexArr = arr.map(function(k) { return Math.abs(k - x) })
var min = Math.min.apply(Math, indexArr)
return arr[indexArr.indexOf(min)]
}
findClosest(80, [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]) // Outputs 82
Contrary to other approaches finding the minimum value using Math.min.apply, this one doesn't require the input array arr to be sorted. We don't need to care about the indexes or sort it beforehand.
I'll explain the code line by line for clarity:
arr.map(function(k) { return Math.abs(k - x) }) Creates a new array, essentially storing the absolute values of the given numbers (number in arr) minus the input number (x). We'll look for the smallest number next (which is also the closest to the input number)
Math.min.apply(Math, indexArr) This is a legit way of finding the smallest number in the array we've just created before (nothing more to it)
arr[indexArr.indexOf(min)] This is perhaps the most interesting part. We have found our smallest number, but we're not sure if we should add or subtract the initial number (x). That's because we used Math.abs() to find the difference. However, array.map creates (logically) a map of the input array, keeping the indexes in the same place. Therefore, to find out the closest number we just return the index of the found minimum in the given array indexArr.indexOf(min).
I've created a bin demonstrating it.

All of the solutions are over-engineered.
It is as simple as:
const needle = 5;
const haystack = [1, 2, 3, 4, 5, 6, 7, 8, 9];
haystack.sort((a, b) => {
return Math.abs(a - needle) - Math.abs(b - needle);
})[0];
// 5

For sorted arrays (linear search)
All answers so far concentrate on searching through the whole array.
Considering your array is sorted already and you really only want the nearest number this is probably the easiest (but not fastest) solution:
var a = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
var target = 90000;
/**
* Returns the closest number from a sorted array.
**/
function closest(arr, target) {
if (!(arr) || arr.length == 0)
return null;
if (arr.length == 1)
return arr[0];
for (var i = 1; i < arr.length; i++) {
// As soon as a number bigger than target is found, return the previous or current
// number depending on which has smaller difference to the target.
if (arr[i] > target) {
var p = arr[i - 1];
var c = arr[i]
return Math.abs(p - target) < Math.abs(c - target) ? p : c;
}
}
// No number in array is bigger so return the last.
return arr[arr.length - 1];
}
// Trying it out
console.log(closest(a, target));
Note that the algorithm can be vastly improved e.g. using a binary tree.

ES6
Works with sorted and unsorted arrays
Numbers Integers and Floats, Strings welcomed
/**
* Finds the nearest value in an array of numbers.
* Example: nearestValue(array, 42)
*
* #param {Array<number>} arr
* #param {number} val the ideal value for which the nearest or equal should be found
*/
const nearestValue = (arr, val) => arr.reduce((p, n) => (Math.abs(p) > Math.abs(n - val) ? n - val : p), Infinity) + val
Examples:
let values = [1,2,3,4,5]
console.log(nearestValue(values, 10)) // --> 5
console.log(nearestValue(values, 0)) // --> 1
console.log(nearestValue(values, 2.5)) // --> 2
values = [100,5,90,56]
console.log(nearestValue(values, 42)) // --> 56
values = ['100','5','90','56']
console.log(nearestValue(values, 42)) // --> 56

This solution uses ES5 existential quantifier Array#some, which allows to stop the iteration, if a condition is met.
Opposit of Array#reduce, it does not need to iterate all elements for one result.
Inside the callback, an absolute delta between the searched value and actual item is taken and compared with the last delta. If greater or equal, the iteration stops, because all other values with their deltas are greater than the actual value.
If the delta in the callback is smaller, then the actual item is assigned to the result and the delta is saved in lastDelta.
Finally, smaller values with equal deltas are taken, like in the below example of 22, which results in 2.
If there is a priority of greater values, the delta check has to be changed from:
if (delta >= lastDelta) {
to:
if (delta > lastDelta) {
// ^^^ without equal sign
This would get with 22, the result 42 (Priority of greater values).
This function needs sorted values in the array.
Code with priority of smaller values:
function closestValue(array, value) {
var result,
lastDelta;
array.some(function (item) {
var delta = Math.abs(value - item);
if (delta >= lastDelta) {
return true;
}
result = item;
lastDelta = delta;
});
return result;
}
var data = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
console.log(21, closestValue(data, 21)); // 2
console.log(22, closestValue(data, 22)); // 2 smaller value
console.log(23, closestValue(data, 23)); // 42
console.log(80, closestValue(data, 80)); // 82
Code with priority of greater values:
function closestValue(array, value) {
var result,
lastDelta;
array.some(function (item) {
var delta = Math.abs(value - item);
if (delta > lastDelta) {
return true;
}
result = item;
lastDelta = delta;
});
return result;
}
var data = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
console.log(21, closestValue(data, 21)); // 2
console.log(22, closestValue(data, 22)); // 42 greater value
console.log(23, closestValue(data, 23)); // 42
console.log(80, closestValue(data, 80)); // 82

Other answers suggested the you would need to iterate through the entire array:
calculate the deviation for each element
keep track of the smallest deviation and its element
finally, after iterating through the entire array, return that element with that smallest deviation.
If the array is already sorted, that does not make sense. There is no need to calculate all deviations. e.g. in an ordered collection of 1 million elements, you only need to calculate ~19 deviations (at most) to find your match. You can accomplish this with a binary-search approach:
function findClosestIndex(arr, element) {
let from = 0, until = arr.length - 1
while (true) {
const cursor = Math.floor((from + until) / 2);
if (cursor === from) {
const diff1 = element - arr[from];
const diff2 = arr[until] - element;
return diff1 <= diff2 ? from : until;
}
const found = arr[cursor];
if (found === element) return cursor;
if (found > element) {
until = cursor;
} else if (found < element) {
from = cursor;
}
}
}
Result:
console.log(findClosestIndex([0, 1, 2, 3.5, 4.5, 5], 4));
// output: 3
console.log(findClosestIndex([0, 1, 2, 3.49, 4.5, 5], 4));
// output: 4
console.log(findClosestIndex([0, 1, 2, 3.49, 4.5, 5], 90));
// output: 5
console.log(findClosestIndex([0, 1, 2, 3.49, 4.5, 5], -1));
// output: 0

A simpler way with O(n) time complexity is to do this in one iteration of the array. This method is intended for unsorted arrays.
Following is a javascript example, here from the array we find the number which is nearest to "58".
var inputArr = [150, 5, 200, 50, 30];
var search = 58;
var min = Math.min();
var result = 0;
for(i=0;i<inputArr.length;i++) {
let absVal = Math.abs(search - inputArr[i])
if(min > absVal) {
min=absVal;
result = inputArr[i];
}
}
console.log(result); //expected output 50 if input is 58
This will work for positive, negative, decimal numbers as well.
Math.min() will return Infinity.
The result will store the value nearest to the search element.

I don't know if I'm supposed to answer an old question, but as this post appears first on Google searches, I hoped that you would forgive me adding my solution & my 2c here.
Being lazy, I couldn't believe that the solution for this question would be a LOOP, so I searched a bit more and came back with filter function:
var myArray = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
var myValue = 80;
function BiggerThan(inArray) {
return inArray > myValue;
}
var arrBiggerElements = myArray.filter(BiggerThan);
var nextElement = Math.min.apply(null, arrBiggerElements);
alert(nextElement);
That's all !

My answer to a similar question is accounting for ties too and it is in plain Javascript, although it doesn't use binary search so it is O(N) and not O(logN):
var searchArray= [0, 30, 60, 90];
var element= 33;
function findClosest(array,elem){
var minDelta = null;
var minIndex = null;
for (var i = 0 ; i<array.length; i++){
var delta = Math.abs(array[i]-elem);
if (minDelta == null || delta < minDelta){
minDelta = delta;
minIndex = i;
}
//if it is a tie return an array of both values
else if (delta == minDelta) {
return [array[minIndex],array[i]];
}//if it has already found the closest value
else {
return array[i-1];
}
}
return array[minIndex];
}
var closest = findClosest(searchArray,element);
https://stackoverflow.com/a/26429528/986160

I like the approach from Fusion, but there's a small error in it. Like that it is correct:
function closest(array, number) {
var num = 0;
for (var i = array.length - 1; i >= 0; i--) {
if(Math.abs(number - array[i]) < Math.abs(number - array[num])){
num = i;
}
}
return array[num];
}
It it also a bit faster because it uses the improved for loop.
At the end I wrote my function like this:
var getClosest = function(number, array) {
var current = array[0];
var difference = Math.abs(number - current);
var index = array.length;
while (index--) {
var newDifference = Math.abs(number - array[index]);
if (newDifference < difference) {
difference = newDifference;
current = array[index];
}
}
return current;
};
I tested it with console.time() and it is slightly faster than the other function.

The most efficient would be a binary search. However even simple solutions can exit when the next number is a further match from the current. Nearly all solutions here are not taking into account that the array is ordered and iterating though the whole thing :/
const closest = (orderedArray, value, valueGetter = item => item) =>
orderedArray.find((item, i) =>
i === orderedArray.length - 1 ||
Math.abs(value - valueGetter(item)) < Math.abs(value - valueGetter(orderedArray[i + 1])));
var data = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
console.log('21 -> 2', closest(data, 21) === 2);
console.log('22 -> 42', closest(data, 22) === 42); // equidistant between 2 and 42, select highest
console.log('23 -> 42', closest(data, 23) === 42);
console.log('80 -> 82', closest(data, 80) === 82);
This can be run on non-primitives too e.g. closest(data, 21, item => item.age)
Change find to findIndex to return the index in the array.

If the array is sorted like in your example, you can use a Binary Search for a better time complexity of O(log n).
const myArray = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
const binaryClosestIdx = (arr, target) => {
let start = 0;
let end = arr.length - 1;
let mid = Math.floor((start + end) / 2);
while (1) {
if (arr[mid] === target) {
return mid;
}
else if (start >= end) {
break;
}
else if (arr[mid] > target) {
end = mid - 1;
} else {
start = mid + 1;
}
mid = Math.floor((start + end) / 2);
}
// Return the closest between the last value checked and it's surrounding neighbors
const first = Math.max(mid - 1, 0);
const neighbors = arr.slice(first, mid + 2);
const best = neighbors.reduce((b, el) => Math.abs(el - target) < Math.abs(b - target) ? el : b);
return first + neighbors.indexOf(best);
}
const closestValue = myArray[binaryClosestIdx(myArray, 80)];
console.log(closestValue);
How it works :
It compares the target value to the middle element of the array. If the middle element is bigger we can ignore every element after it as they are going to be even bigger. The same goes if the middle element is smaller, we can ignore every element before it.
If the target value is found we return it, otherwise we compare the last value tested with its surrounding neighbors as the closest value can only be between those 3 values.

Another variant here we have circular range connecting head to toe and accepts only min value to given input. This had helped me get char code values for one of the encryption algorithm.
function closestNumberInCircularRange(codes, charCode) {
return codes.reduce((p_code, c_code)=>{
if(((Math.abs(p_code-charCode) > Math.abs(c_code-charCode)) || p_code > charCode) && c_code < charCode){
return c_code;
}else if(p_code < charCode){
return p_code;
}else if(p_code > charCode && c_code > charCode){
return Math.max.apply(Math, [p_code, c_code]);
}
return p_code;
});
}

To Find Two Closest Number in array
function findTwoClosest(givenList, goal) {
var first;
var second;
var finalCollection = [givenList[0], givenList[1]];
givenList.forEach((item, firtIndex) => {
first = item;
for (let i = firtIndex + 1; i < givenList.length; i++) {
second = givenList[i];
if (first + second < goal) {
if (first + second > finalCollection[0] + finalCollection[1]) {
finalCollection = [first, second];
}
}
}
});
return finalCollection;
}
var counts = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
var goal = 80;
console.log(findTwoClosest(counts, goal));

You can use below logic to find closest number without using reduce function
let arr = [0, 80, 10, 60, 20, 50, 0, 100, 80, 70, 1];
const n = 2;
let closest = -1;
let closeDiff = -1;
for (let i = 0; i < arr.length; i++) {
if (Math.abs(arr[i] - n) < closeDiff || closest === -1) {
closeDiff = Math.abs(arr[i] - n);
closest = arr[i];
}
}
console.log(closest);

#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
class CompareFunctor
{
public:
CompareFunctor(int n) { _n = n; }
bool operator()(int & val1, int & val2)
{
int diff1 = abs(val1 - _n);
int diff2 = abs(val2 - _n);
return (diff1 < diff2);
}
private:
int _n;
};
int Find_Closest_Value(int nums[], int size, int n)
{
CompareFunctor cf(n);
int cn = *min_element(nums, nums + size, cf);
return cn;
}
int main()
{
int nums[] = { 2, 42, 82, 122, 162, 202, 242, 282, 322, 362 };
int size = sizeof(nums) / sizeof(int);
int n = 80;
int cn = Find_Closest_Value(nums, size, n);
cout << "\nClosest value = " << cn << endl;
cin.get();
}

For a small range, the simplest thing is to have a map array, where, eg, the 80th entry would have the value 82 in it, to use your example. For a much larger, sparse range, probably the way to go is a binary search.
With a query language you could query for values some distance either side of your input number and then sort through the resulting reduced list. But SQL doesn't have a good concept of "next" or "previous", to give you a "clean" solution.

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