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I've been trying to calculate median but still I've got some mathematical issues I guess as I couldn't get the correct median value and couldn't figure out why. Here's the code;
class StatsCollector {
constructor() {
this.inputNumber = 0;
this.average = 0;
this.timeout = 19000;
this.frequencies = new Map();
for (let i of Array(this.timeout).keys()) {
this.frequencies.set(i, 0);
}
}
pushValue(responseTimeMs) {
let req = responseTimeMs;
if (req > this.timeout) {
req = this.timeout;
}
this.average = (this.average * this.inputNumber + req) / (this.inputNumber + 1);
console.log(responseTimeMs / 1000)
let groupIndex = Math.floor(responseTimeMs / 1000);
this.frequencies.set(groupIndex, this.frequencies.get(groupIndex) + 1);
this.inputNumber += 1;
}
getMedian() {
let medianElement = 0;
if (this.inputNumber <= 0) {
return 0;
}
if (this.inputNumber == 1) {
return this.average
}
if (this.inputNumber == 2) {
return this.average
}
if (this.inputNumber > 2) {
medianElement = this.inputNumber / 2;
}
let minCumulativeFreq = 0;
let maxCumulativeFreq = 0;
let cumulativeFreq = 0;
let freqGroup = 0;
for (let i of Array(20).keys()) {
if (medianElement <= cumulativeFreq + this.frequencies.get(i)) {
minCumulativeFreq = cumulativeFreq;
maxCumulativeFreq = cumulativeFreq + this.frequencies.get(i);
freqGroup = i;
break;
}
cumulativeFreq += this.frequencies.get(i);
}
return (((medianElement - minCumulativeFreq) / (maxCumulativeFreq - minCumulativeFreq)) + (freqGroup)) * 1000;
}
getAverage() {
return this.average;
}
}
Here's the snapshot of the results when I enter the values of
342,654,987,1093,2234,6243,7087,20123
The correct result should be;
Median: 1663.5
Change your median method to this:
function median(values){
if(values.length ===0) throw new Error("No inputs");
values.sort(function(a,b){
return a-b;
});
var half = Math.floor(values.length / 2);
if (values.length % 2)
return values[half];
return (values[half - 1] + values[half]) / 2.0;
}
fiddle
Here's another solution:
function median(numbers) {
const sorted = Array.from(numbers).sort((a, b) => a - b);
const middle = Math.floor(sorted.length / 2);
if (sorted.length % 2 === 0) {
return (sorted[middle - 1] + sorted[middle]) / 2;
}
return sorted[middle];
}
console.log(median([4, 5, 7, 1, 33]));
The solutions above - sort then find middle - are fine, but slow on large data sets. Sorting the data first has a complexity of n x log(n).
There is a faster median algorithm, which consists in segregating the array in two according to a pivot, then looking for the median in the larger set. Here is some javascript code, but here is a more detailed explanation
// Trying some array
alert(quickselect_median([7,3,5])); // 2300,5,4,0,123,2,76,768,28]));
function quickselect_median(arr) {
const L = arr.length, halfL = L/2;
if (L % 2 == 1)
return quickselect(arr, halfL);
else
return 0.5 * (quickselect(arr, halfL - 1) + quickselect(arr, halfL));
}
function quickselect(arr, k) {
// Select the kth element in arr
// arr: List of numerics
// k: Index
// return: The kth element (in numerical order) of arr
if (arr.length == 1)
return arr[0];
else {
const pivot = arr[0];
const lows = arr.filter((e)=>(e<pivot));
const highs = arr.filter((e)=>(e>pivot));
const pivots = arr.filter((e)=>(e==pivot));
if (k < lows.length) // the pivot is too high
return quickselect(lows, k);
else if (k < lows.length + pivots.length)// We got lucky and guessed the median
return pivot;
else // the pivot is too low
return quickselect(highs, k - lows.length - pivots.length);
}
}
Astute readers will notice a few things:
I simply transliterated Russel Cohen's Python solution into JS,
so all kudos to him.
There are several small optimisations worth
doing, but there's parallelisation worth doing, and the code as is
is easier to change in either a quicker single-threaded, or quicker
multi-threaded, version.
This is the average linear time
algorithm, there is more efficient a deterministic linear time version, see Russel's
post for details, including performance data.
ADDITION 19 Sept. 2019:
One comment asks whether this is worth doing in javascript. I ran the code in JSPerf and it gives interesting results.
if the array has an odd number of elements (one figure to find), sorting is 20% slower that this "fast median" proposition.
if there is an even number of elements, the "fast" algorithm is 40% slower, because it filters through the data twice, to find elements number k and k+1 to average. It is possible to write a version of fast median that doesn't do this.
The test used rather small arrays (29 elements in the jsperf test). The effect appears to be more pronounced as arrays get larger. A more general point to make is: it shows these kinds of optimisations are worth doing in Javascript. An awful lot of computation is done in JS, including with large amounts of data (think of dashboards, spreadsheets, data visualisations), and in systems with limited resources (think of mobile and embedded computing).
var arr = {
max: function(array) {
return Math.max.apply(null, array);
},
min: function(array) {
return Math.min.apply(null, array);
},
range: function(array) {
return arr.max(array) - arr.min(array);
},
midrange: function(array) {
return arr.range(array) / 2;
},
sum: function(array) {
var num = 0;
for (var i = 0, l = array.length; i < l; i++) num += array[i];
return num;
},
mean: function(array) {
return arr.sum(array) / array.length;
},
median: function(array) {
array.sort(function(a, b) {
return a - b;
});
var mid = array.length / 2;
return mid % 1 ? array[mid - 0.5] : (array[mid - 1] + array[mid]) / 2;
},
modes: function(array) {
if (!array.length) return [];
var modeMap = {},
maxCount = 1,
modes = [array[0]];
array.forEach(function(val) {
if (!modeMap[val]) modeMap[val] = 1;
else modeMap[val]++;
if (modeMap[val] > maxCount) {
modes = [val];
maxCount = modeMap[val];
}
else if (modeMap[val] === maxCount) {
modes.push(val);
maxCount = modeMap[val];
}
});
return modes;
},
variance: function(array) {
var mean = arr.mean(array);
return arr.mean(array.map(function(num) {
return Math.pow(num - mean, 2);
}));
},
standardDeviation: function(array) {
return Math.sqrt(arr.variance(array));
},
meanAbsoluteDeviation: function(array) {
var mean = arr.mean(array);
return arr.mean(array.map(function(num) {
return Math.abs(num - mean);
}));
},
zScores: function(array) {
var mean = arr.mean(array);
var standardDeviation = arr.standardDeviation(array);
return array.map(function(num) {
return (num - mean) / standardDeviation;
});
}
};
2022 TypeScript Approach
const median = (arr: number[]): number | undefined => {
if (!arr.length) return undefined;
const s = [...arr].sort((a, b) => a - b);
const mid = Math.floor(s.length / 2);
return s.length % 2 === 0 ? ((s[mid - 1] + s[mid]) / 2) : s[mid];
};
Notes:
The type in the function signature (number[]) ensures only an array of numbers can be passed to the function. It could possibly be empty though.
if (!arr.length) return undefined; checks for the possible empty array, which would not have a median.
[...arr] creates a copy of the passed-in array to ensure we don't overwrite the original.
.sort((a, b) => a - b) sorts the array of numbers in ascending order.
Math.floor(s.length / 2) finds the index of the middle element if the array has odd length, or the element just to the right of the middle if the array has even length.
s.length % 2 === 0 determines whether the array has an even length.
(s[mid - 1] + s[mid]) / 2 averages the two middle items of the array if the array's length is even.
s[mid] is the middle item of an odd-length array.
TypeScript Answer 2020:
// Calculate Median
const calculateMedian = (array: Array<number>) => {
// Check If Data Exists
if (array.length >= 1) {
// Sort Array
array = array.sort((a: number, b: number) => {
return a - b;
});
// Array Length: Even
if (array.length % 2 === 0) {
// Average Of Two Middle Numbers
return (array[(array.length / 2) - 1] + array[array.length / 2]) / 2;
}
// Array Length: Odd
else {
// Middle Number
return array[(array.length - 1) / 2];
}
}
else {
// Error
console.error('Error: Empty Array (calculateMedian)');
}
};
const median = (arr) => {
return arr.slice().sort((a, b) => a - b)[Math.floor(arr.length / 2)];
};
Short and sweet.
Array.prototype.median = function () {
return this.slice().sort((a, b) => a - b)[Math.floor(this.length / 2)];
};
Usage
[4, 5, 7, 1, 33].median()
Works with strings as well
["a","a","b","b","c","d","e"].median()
For better performance in terms of time complexity, use MaxHeap - MinHeap to find the median of stream of array.
Simpler & more efficient
const median = dataSet => {
if (dataSet.length === 1) return dataSet[0]
const sorted = ([ ...dataSet ]).sort()
const ceil = Math.ceil(sorted.length / 2)
const floor = Math.floor(sorted.length / 2)
if (ceil === floor) return sorted[floor]
return ((sorted[ceil] + sorted[floor]) / 2)
}
Simple solution:
function calcMedian(array) {
const {
length
} = array;
if (length < 1)
return 0;
//sort array asc
array.sort((a, b) => a - b);
if (length % 2) {
//length of array is odd
return array[(length + 1) / 2 - 1];
} else {
//length of array is even
return 0.5 * [(array[length / 2 - 1] + array[length / 2])];
}
}
console.log(2, calcMedian([1, 2, 2, 5, 6]));
console.log(3.5, calcMedian([1, 2, 2, 5, 6, 7]));
console.log(9, calcMedian([13, 9, 8, 15, 7]));
console.log(3.5, calcMedian([1, 4, 6, 3]));
console.log(5, calcMedian([5, 1, 11, 2, 8]));
Simpler, more efficient, and easy to read
cloned the data to avoid alterations to the original data.
sort the list of values.
get the middle point.
get the median from the list.
return the median.
function getMedian(data) {
const values = [...data];
const v = values.sort( (a, b) => a - b);
const mid = Math.floor( v.length / 2);
const median = (v.length % 2 !== 0) ? v[mid] : (v[mid - 1] + v[mid]) / 2;
return median;
}
const medianArr = (x) => {
let sortedx = x.sort((a,b)=> a-b);
let halfIndex = Math.floor(sortedx.length/2);
return (sortedx.length%2) ? (sortedx[Math.floor(sortedx.length/2)]) : ((sortedx[halfIndex-1]+sortedx[halfIndex])/2)
}
console.log(medianArr([1,2,3,4,5]));
console.log(medianArr([1,2,3,4,5,6]));
function Median(arr){
let len = arr.length;
arr = arr.sort();
let result = 0;
let mid = Math.floor(len/2);
if(len % 2 !== 0){
result += arr[mid];
}
if(len % 2 === 0){
result += (arr[mid] + arr[mid+1])/2
}
return result;
}
console.log(`The median is ${Median([0,1,2,3,4,5,6])}`)
function median(arr) {
let n = arr.length;
let med = Math.floor(n/2);
if(n % 2 != 0){
return arr[med];
} else{
return (arr[med -1] + arr[med])/ 2.0
}
}
console.log(median[1,2,3,4,5,6]);
The arr.sort() method sorts the elements of an array in place and returns the array. By default, it sorts the elements alphabetically, so if the array contains numbers, they will not be sorted in numerical order.
On the other hand, the arr.sort((a, b) => a - b) method uses a callback function to specify how the array should be sorted. The callback function compares the two elements a and b and returns a negative number if a should be sorted before b, a positive number if b should be sorted before a, and zero if the elements are equal. In this case, the callback function subtracts b from a, which results in a sorting order that is numerical in ascending order.
So, if you want to sort an array of numbers in ascending order, you should use arr.sort((a, b) => a - b), whereas if you want to sort an array of strings alphabetically, you can use arr.sort():
function median(numbers) {
const sorted = Array.from(numbers).sort((a, b) => a - b);
const middle = Math.floor(sorted.length / 2);
if (sorted.length % 2 === 0) {
return (sorted[middle - 1] + sorted[middle]) / 2;
}
return sorted[middle];
}
function findMedian(arr) {
arr.sort((a, b) => a - b)
let i = Math.floor(arr.length / 2)
return arr[i]
}
let result = findMedian([0, 1, 2, 4, 6, 5, 3])
console.log(result)
My function is taking in 2 integer parameters and returning the numbers between those two parameters. I cannot convert the numbers shown into an array.
As I'm supposed to buildArray function that takes two Numbers, and returns an Array filled with all numbers between the given number: buildArray(5, 10) should return [5, 6, 7, 8, 9, 10].
I have tried using split to convert the values into an array but I've failed many times.
My code prints each value separately whereas i need the values to be displayed in an array.
function buildArray(a, b) {
for (var i = 0; i <= b; i++) {
i = a;
console.log(a);
a = a + 1;
}
}
console.log(buildArray(5, 10));
Here is the simplest version using your original line of thoughts
Implementing a range function can be more complex than this, but since this works, I suggest it for now
function buildArray(a, b) {
var arr = [];
for (var i = a; i <= b; i++) {
arr.push(i);
}
return arr;
}
console.log(buildArray(5, 10));
You can construct a new array filled with zeros of the length you need and then use map() to set the values. I have updated to a generalized version for accept other types of ranges.
function buildArray(a, b)
{
let n = Math.abs(b - a);
return Array(n + 1).fill(0).map((x, idx) => a + ((b >= a) ? idx : -idx));
}
console.log(JSON.stringify(buildArray(5, 10)));
console.log(JSON.stringify(buildArray(12, 3)));
console.log(JSON.stringify(buildArray(1, -2)));
console.log(JSON.stringify(buildArray(-1, -1)));
Try the following (using Array.from):
function buildArray(min, max) {
return Array.from({ length: max - min + 1 }, (_, i) => min + i);
}
console.log(buildArray(5, 10));
I have attended a technical interview for a development company. They asked me the following:
Giving an array of numbers (n) find 2 numbers that sum gives (k) and print them.
e.g
Input: n = [2,6,4,5,7,1], k = 8
Output: result=(2,6),(7,1)
My solution:
function findSum(n,k){
let aux = []
for (let i = 0; i < n.length; i++) {
for (let j = i+1; j < n.length; j++) {
if (n[i] + n[j] == k) {
aux.push({ first: n[i], second: n[j] })
}
}
}
return aux;
}
They told me that, it is possible to make the exercise with some kind of key or mapping.
Does some one know how to do it with only one loop?
The key to solving a question like this with low time complexity is the ability to efficiently search the data structure. A lot of answers rearrange the array in a way where searching an array is optimized. Another approach is with a data structure that inherently has fast search.
Set and Map data structures have O(1) time complexity for searches, which make them good data structures where searching can be leveraged to increase performance.
I use a new Map and traverse the array while adding it as a key. I set the key to the number and the value to the number of times I see it. I use a map over a new Set because I can also keep track of the number of instances of that particular number.
I search for the number that would sum up to k, which is: (k - num). If I find that number, I add both numbers to my results data structure and decrement the value by 1, to show that it's been used.
Time complexity: O(n), memory complexity: O(2n). Twice the amount of space compared to the original array because I have a key and a value to store in my Map
function pairSums(arr, k){
const map = new Map
const matches = []
for (let num of arr) {
const search = k - num
if (map.get(search) > 0) {
matches.push([num, k - num])
map.set(search, map.get(search) - 1)
} else if (!map.has(num)){
map.set(num, 1)
} else {
map.set(num, map.get(num) + 1)
}
}
return matches
}
console.log(pairSums([2, 6, 6, 6, 2, 4, 4, 4, 5, 7, 1, 4, 2], 8))
Match a number x from array with a key Math.min(x, k - x). Then run through your array and store every number in a hash using mentioned key. When the key you are going to add already is in the hash - check if stored value and current number gives required sum.
function findSum(n, k){
let hash = {};
for(let i = 0; i < n.length; ++i){
let x = n[i], key = Math.min(x, k - x);
if((key in hash) && hash[key] + x == k)
return [hash[key], x];
else hash[key] = x;
}
}
A task like this can be as simple or as complicated as you want to make it. Here's one solution, for example:
function findPairs(n, k) {
return n.reduce((pairs, next, i) => pairs.concat(
n.slice(i + 1)
.filter(num => next + num === k)
.map(num => [ next, num ])
),
[]
);
}
For the inputs [2, 6, 4, 5, 7, 1] and 8 will output [ [2, 6], [7, 1] ].
From https://www.geeksforgeeks.org/write-a-c-program-that-given-a-set-a-of-n-numbers-and-another-number-x-determines-whether-or-not-there-exist-two-elements-in-s-whose-sum-is-exactly-x/:
Sort the array in non-decreasing order.
Initialize two index variables to find the candidate elements in the sorted array. Initialize l to the leftmost index: l = 0, Initialize r to the rightmost index: r = n.length - 1
Loop while l < r.
if (n[l] + n[r] == k) then return 1
else if( n[l] + n[r] < k ) then l++
else r--
No candidates in whole array - return 0
I think by sorting you can do that
var n = [2,6,4,5,7,1];
var k = 8 ;
n.sort();
let start = 0, end = n.length-1;
while(start < n.length && end >= 0) {
let current_sum = (n[start] + n[end]);
if(current_sum == k) {
console.log('Found sum with '+ n[start] +' and '+ n[end]);
break;
}
else if(current_sum > k) {
end--;
} else {
start++;
}
}
if(start == n.length || end < 0) {
console.log('Not Found');
}
but while writing this code I got one another approach also
const set = new Set([2,6,4,5,7,1]);
var k = 8;
let found = false;
for (let item of set) {
let another = k - item;
if(set.has(another)){
console.log('found with '+item +' and ' +another);
found = true;
break;
}
}
if(!found) {
console.log('Not found');
}
If numbers are non-negative and the target value is within JavaScript array limit:
function findsums(arr,k){
var ret=[];
var aux=[];
arr.forEach(function(i){
if(i<=k){
if(aux[k-i])
ret.push([k-i,i]);
aux[i]=true;
}
});
return ret;
}
console.log(findsums([2,6,4,5,7,1],8));
Similar approach could work with a bitfield (or even with a sparse array of bitfields) too.
Minified alternative similar to #Andrew's great answer, but assumes that all numbers are above 0 :
var pairs = (arr, k) => arr.reduce((a, n) =>
(a[n - k]-- ? a.push([n, k - n]) : a[-n] = a[-n] | 0 + 1, a), []);
console.log(JSON.stringify( pairs([2,6,4,5,7,1], 8) ));
I need to write a program that, when given a list of integers, it finds all 2-pairs of integers that have the same product. i.e. a 2-pair is 2 distinct pairs of integers lets say [(a,b),(c,d)] where a*b = c*d but a ≠ b ≠ c ≠ d.
The range of integers should be from 1 to 1024. What I would like to implement is that when the web page is opened the user is prompted by a pop up in which he will enter the array of integers, i.e [1,2,3,7,8,9,6] etc for instance from the input [1,2,3,7,8,9,6] the output should be [(9,2),(3,6)] since both evaluate to 18.
The coding I did so far is very basic and can be seen below. What I've done so far is the pop-up box alert the input etc, but can't seem to understand how to make the program check for the pairs and give the sum. Thanks in advance to this community who's helping me out to better understand and learn javascript!
I've done my fair bit of research below, definitely different question than mine but have gone through them.
Find a pair of elements from an array whose sum equals a given number
https://www.w3resource.com/javascript-exercises/javascript-array-exercise-26.php
Code:
function evaluate() {
const input = prompt("Please enter the array of integers in the form: 1,2,3,1")
.split(',')
.map(item => item.trim());
function pairs(items) {
}
if (input == "" || input == null) {
document.writeln("Sorry, there is nothing that can be calculated.");
} else {
document.writeln("Your calculation is: ");
document.writeln(pairs(input) + " with a starting input string of: " + input);
}
}
evaluate()
You could iterate the array and a copy of the array beginning by the actual index plus one for getting the products. Store the result in an object with product as key.
Then get the keys (products) of the object, filter it to get only the results with two or more products.
var array = [1, 2, 3, 7, 8, 9, 6],
result = {},
pairs;
array.forEach(function (a, i) {
array.slice(i + 1).forEach(function (b) {
(result[a * b] = (result[a * b] || [])).push([a, b]);
});
});
pairs = Object
.keys(result)
.filter(function (k) { return result[k].length >= 2; })
.map(function(k) { return result[k]; });
console.log(pairs);
We could mutate the equation:
a * b = c * d | : b
a = c * d : b
So actually we just need to get all different combinations of three numbers (b, c, d) and check if the result (a) is also in the given range:
while(true){
// shuffle
const [b, c, d] = items;
const a = c * d / b;
if(items.includes(a + ""))
return true;
}
return false;
Now you only need to shuffle the array to go through all different combinations. You can find an algorithm here
Assuming that you are given an array such as [1,2,3,7,8,9,6] and a value 18 and you need to find pairs that multiply to 18 then, use the following approach
Convert them to a map - O(n)
var inputArr = [1,2,3,7,8,9,6];
var map = inputArr.reduce( (acc, c) => {
acc[ c ] = true; //set any truthy value
return acc;
},{});
Iterate an inputArr and see if its compliment is available in the map - O(n)
var output = [];
var mulValue = 18;
inputArr.forEach( s => {
var remainder = mulValue/s;
if ( map[s] && map[remainder] )
{
output.push( [ s, remainder ] );
map[s] = false;
map[remainder] = false;
}
});
Demo
var inputArr = [1, 2, 3, 7, 8, 9, 6];
var map = inputArr.reduce((acc, c) => {
acc[c] = true; //set any truthy value
return acc;
}, {});
var output = [];
var mulValue = 18;
inputArr.forEach(s => {
var remainder = mulValue / s;
if (map[s] && map[remainder]) {
output.push([s, remainder]);
map[s] = false;
map[remainder] = false;
}
});
console.log(output);
You can try something like this:
Idea:
Loop over the array to compute product. Use this iterator(say i) as get first operand(say op1).
Now again loop over same array but the range will start from i+1. This is to reduce number of iteration.
Now create a temp variable that will hold product and operand.
On every iteration, add value to product in hashMap.
Now loop over hashMap and remove any value that has length that is less than 2.
function sameProductValues(arr) {
var hashMap = {};
for (var i = 0; i < arr.length - 1; i++) {
for (var j = i + 1; j < arr.length; j++) {
var product = arr[i] * arr[j];
hashMap[product] = hashMap[product] || [];
hashMap[product].push([arr[i], arr[j]]);
}
}
for(var key in hashMap) {
if( hashMap[key].length < 2 ) {
delete hashMap[key];
}
}
console.log(hashMap)
}
sameProductValues([1, 2, 3, 7, 8, 9, 6])
How can I easily obtain the min or max element of a JavaScript array?
Example pseudocode:
let array = [100, 0, 50]
array.min() //=> 0
array.max() //=> 100
How about augmenting the built-in Array object to use Math.max/Math.min instead:
Array.prototype.max = function() {
return Math.max.apply(null, this);
};
Array.prototype.min = function() {
return Math.min.apply(null, this);
};
let p = [35,2,65,7,8,9,12,121,33,99];
console.log(`Max value is: ${p.max()}` +
`\nMin value is: ${p.min()}`);
Here is a JSFiddle.
Augmenting the built-ins can cause collisions with other libraries (some see), so you may be more comfortable with just apply'ing Math.xxx() to your array directly:
var min = Math.min.apply(null, arr),
max = Math.max.apply(null, arr);
Alternately, assuming your browser supports ECMAScript 6, you can use spread syntax which functions similarly to the apply method:
var min = Math.min( ...arr ),
max = Math.max( ...arr );
var max_of_array = Math.max.apply(Math, array);
For a full discussion see:
http://aaroncrane.co.uk/2008/11/javascript_max_api/
Using spread operator (ES6)
Math.max(...array) // The same with "min" => Math.min(...array)
const array = [10, 2, 33, 4, 5];
console.log(
Math.max(...array)
)
For big arrays (~10⁷ elements), Math.min and Math.max both produces the following error in Node.js.
RangeError: Maximum call stack size exceeded
A more robust solution is to not add every element to the call stack, but to instead pass an array:
function arrayMin(arr) {
return arr.reduce(function (p, v) {
return ( p < v ? p : v );
});
}
function arrayMax(arr) {
return arr.reduce(function (p, v) {
return ( p > v ? p : v );
});
}
If you are concerned about speed, the following code is ~3 times faster then Math.max.apply is on my computer. See https://jsben.ch/JPOyL.
function arrayMin(arr) {
var len = arr.length, min = Infinity;
while (len--) {
if (arr[len] < min) {
min = arr[len];
}
}
return min;
};
function arrayMax(arr) {
var len = arr.length, max = -Infinity;
while (len--) {
if (arr[len] > max) {
max = arr[len];
}
}
return max;
};
If your arrays contains strings instead of numbers, you also need to coerce them into numbers. The below code does that, but it slows the code down ~10 times on my machine. See https://jsben.ch/uPipD.
function arrayMin(arr) {
var len = arr.length, min = Infinity;
while (len--) {
if (Number(arr[len]) < min) {
min = Number(arr[len]);
}
}
return min;
};
function arrayMax(arr) {
var len = arr.length, max = -Infinity;
while (len--) {
if (Number(arr[len]) > max) {
max = Number(arr[len]);
}
}
return max;
};
tl;dr
// For regular arrays:
var max = Math.max(...arrayOfNumbers);
// For arrays with tens of thousands of items:
let max = testArray[0];
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] > max) {
max = testArray[i];
}
}
MDN solution
The official MDN docs on Math.max() already covers this issue:
The following function uses Function.prototype.apply() to find the maximum element in a numeric array. getMaxOfArray([1, 2, 3]) is equivalent to Math.max(1, 2, 3), but you can use getMaxOfArray() on programmatically constructed arrays of any size.
function getMaxOfArray(numArray) {
return Math.max.apply(null, numArray);
}
Or with the new spread operator, getting the maximum of an array becomes a lot easier.
var arr = [1, 2, 3];
var max = Math.max(...arr);
Maximum size of an array
According to MDN the apply and spread solutions had a limitation of 65536 that came from the limit of the maximum number of arguments:
But beware: in using apply this way, you run the risk of exceeding the JavaScript engine's argument length limit. The consequences of applying a function with too many arguments (think more than tens of thousands of arguments) vary across engines (JavaScriptCore has hard-coded argument limit of 65536), because the limit (indeed even the nature of any excessively-large-stack behavior) is unspecified. Some engines will throw an exception. More perniciously, others will arbitrarily limit the number of arguments actually passed to the applied function. To illustrate this latter case: if such an engine had a limit of four arguments (actual limits are of course significantly higher), it would be as if the arguments 5, 6, 2, 3 had been passed to apply in the examples above, rather than the full array.
They even provide a hybrid solution which doesn't really have good performance compared to other solutions. See performance test below for more.
In 2019 the actual limit is the maximum size of the call stack. For modern Chromium based desktop browsers this means that when it comes to finding min/max with apply or spread, practically the maximum size for numbers only arrays is ~120000. Above this, there will be a stack overflow and the following error will be thrown:
RangeError: Maximum call stack size exceeded
With the script below (based on this blog post), by catching that error you can calculate the limit for your specific environment.
Warning! Running this script takes time and depending on the performance of your system it might slow or crash your browser/system!
let testArray = Array.from({length: 10000}, () => Math.floor(Math.random() * 2000000));
for (i = 10000; i < 1000000; ++i) {
testArray.push(Math.floor(Math.random() * 2000000));
try {
Math.max.apply(null, testArray);
} catch (e) {
console.log(i);
break;
}
}
Performance on large arrays
Based on the test in EscapeNetscape's comment I created some benchmarks that tests 5 different methods on a random number only array with 100000 items.
In 2019, the results show that the standard loop (which BTW doesn't have the size limitation) is the fastest everywhere. apply and spread comes closely after it, then much later MDN's hybrid solution then reduce as the slowest.
Almost all tests gave the same results, except for one where spread somewhy ended up being the slowest.
If you step up your array to have 1 million items, things start to break and you are left with the standard loop as a fast solution and reduce as a slower.
JSPerf benchmark
JSBen benchmark
JSBench.me benchmark
Benchmark source code
var testArrayLength = 100000
var testArray = Array.from({length: testArrayLength}, () => Math.floor(Math.random() * 2000000));
// ES6 spread
Math.min(...testArray);
Math.max(...testArray);
// reduce
testArray.reduce(function(a, b) {
return Math.max(a, b);
});
testArray.reduce(function(a, b) {
return Math.min(a, b);
});
// apply
Math.min.apply(Math, testArray);
Math.max.apply(Math, testArray);
// standard loop
let max = testArray[0];
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] > max) {
max = testArray[i];
}
}
let min = testArray[0];
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] < min) {
min = testArray[i];
}
}
// MDN hibrid soltuion
// Source: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/apply#Using_apply_and_built-in_functions
function minOfArray(arr) {
var min = Infinity;
var QUANTUM = 32768;
for (var i = 0, len = arr.length; i < len; i += QUANTUM) {
var submin = Math.min.apply(null, arr.slice(i, Math.min(i + QUANTUM, len)));
min = Math.min(submin, min);
}
return min;
}
minOfArray(testArray);
function maxOfArray(arr) {
var max = -Infinity;
var QUANTUM = 32768;
for (var i = 0, len = arr.length; i < len; i += QUANTUM) {
var submax = Math.max.apply(null, arr.slice(i, Math.max(i + QUANTUM, len)));
max = Math.max(submax, max);
}
return max;
}
maxOfArray(testArray);
If you're paranoid like me about using Math.max.apply (which could cause errors when given large arrays according to MDN), try this:
function arrayMax(array) {
return array.reduce(function(a, b) {
return Math.max(a, b);
});
}
function arrayMin(array) {
return array.reduce(function(a, b) {
return Math.min(a, b);
});
}
Or, in ES6:
function arrayMax(array) {
return array.reduce((a, b) => Math.max(a, b));
}
function arrayMin(array) {
return array.reduce((a, b) => Math.min(a, b));
}
The anonymous functions are unfortunately necessary (instead of using Math.max.bind(Math) because reduce doesn't just pass a and b to its function, but also i and a reference to the array itself, so we have to ensure we don't try to call max on those as well.
Alternative Methods
The Math.min and Math.max are great methods to get the minimum and maximum item out of a collection of items, however it's important to be aware of some cavities that can comes with it.
Using them with an array that contains large number of items (more than ~10⁷ items, depends on the user's browser) most likely will crash and give the following error message:
const arr = Array.from(Array(1000000).keys());
Math.min(arr);
Math.max(arr);
Uncaught RangeError: Maximum call stack size exceeded
UPDATE
Latest browsers might return NaN instead. That might be a better way to handle errors, however it doesn't solve the problem just yet.
Instead, consider using something like so:
function maxValue(arr) {
return arr.reduce((max, val) => max > val ? max : val)
}
Or with better run-time:
function maxValue(arr) {
let max = arr[0];
for (let val of arr) {
if (val > max) {
max = val;
}
}
return max;
}
Or to get both Min and Max:
function getMinMax(arr) {
return arr.reduce(({min, max}, v) => ({
min: min < v ? min : v,
max: max > v ? max : v,
}), { min: arr[0], max: arr[0] });
}
Or with even better run-time*:
function getMinMax(arr) {
let min = arr[0];
let max = arr[0];
let i = arr.length;
while (i--) {
min = arr[i] < min ? arr[i] : min;
max = arr[i] > max ? arr[i] : max;
}
return { min, max };
}
* Tested with 1,000,000 items:
Just for a reference, the 1st function run-time (on my machine) was 15.84ms vs 2nd function with only 4.32ms.
Two ways are shorter and easy:
let arr = [2, 6, 1, 0]
Way 1:
let max = Math.max.apply(null, arr)
Way 2:
let max = arr.reduce(function(a, b) {
return Math.max(a, b);
});
.apply is often used when the intention is to invoke a variadic function with a list of argument values, e.g.
The Math.max([value1[,value2, ...]]) function returns the largest of zero or more numbers.
Math.max(10, 20); // 20
Math.max(-10, -20); // -10
Math.max(-10, 20); // 20
The Math.max() method doesn't allow you to pass in an array. If you have a list of values of which you need to get the largest, you would normally call this function using Function.prototype.apply(), e.g.
Math.max.apply(null, [10, 20]); // 20
Math.max.apply(null, [-10, -20]); // -10
Math.max.apply(null, [-10, 20]); // 20
However, as of the ECMAScript 6 you can use the spread operator:
The spread operator allows an expression to be expanded in places where multiple arguments (for function calls) or multiple elements (for array literals) are expected.
Using the spread operator, the above can be rewritten as such:
Math.max(...[10, 20]); // 20
Math.max(...[-10, -20]); // -10
Math.max(...[-10, 20]); // 20
When calling a function using the variadic operator, you can even add additional values, e.g.
Math.max(...[10, 20], 50); // 50
Math.max(...[-10, -20], 50); // 50
Bonus:
Spread operator enables you to use the array literal syntax to create new arrays in situations where in ES5 you would need to fall back to imperative code, using a combination of push, splice, etc.
let foo = ['b', 'c'];
let bar = ['a', ...foo, 'd', 'e']; // ['a', 'b', 'c', 'd', 'e']
You do it by extending the Array type:
Array.max = function( array ){
return Math.max.apply( Math, array );
};
Array.min = function( array ){
return Math.min.apply( Math, array );
};
Boosted from here (by John Resig)
A simple solution to find the minimum value over an Array of elements is to use the Array prototype function reduce:
A = [4,3,-9,-2,2,1];
A.reduce((min, val) => val < min ? val : min, A[0]); // returns -9
or using JavaScript's built-in Math.Min() function (thanks #Tenflex):
A.reduce((min,val) => Math.min(min,val), A[0]);
This sets min to A[0], and then checks for A[1]...A[n] whether it is strictly less than the current min. If A[i] < min then min is updated to A[i]. When all array elements has been processed, min is returned as the result.
EDIT: Include position of minimum value:
A = [4,3,-9,-2,2,1];
A.reduce((min, val) => val < min._min ? {_min: val, _idx: min._curr, _curr: min._curr + 1} : {_min: min._min, _idx: min._idx, _curr: min._curr + 1}, {_min: A[0], _idx: 0, _curr: 0}); // returns { _min: -9, _idx: 2, _curr: 6 }
For a concise, modern solution, one can perform a reduce operation over the array, keeping track of the current minimum and maximum values, so the array is only iterated over once (which is optimal). Destructuring assignment is used here for succinctness.
let array = [100, 0, 50];
let [min, max] = array.reduce(([prevMin,prevMax], curr)=>
[Math.min(prevMin, curr), Math.max(prevMax, curr)], [Infinity, -Infinity]);
console.log("Min:", min);
console.log("Max:", max);
To only find either the minimum or maximum, we can use perform a reduce operation in much the same way, but we only need to keep track of the previous optimal value. This method is better than using apply as it will not cause errors when the array is too large for the stack.
const arr = [-1, 9, 3, -6, 35];
//Only find minimum
const min = arr.reduce((a,b)=>Math.min(a,b), Infinity);
console.log("Min:", min);//-6
//Only find maximum
const max = arr.reduce((a,b)=>Math.max(a,b), -Infinity);
console.log("Max:", max);//35
Others have already given some solutions in which they augment Array.prototype. All I want in this answer is to clarify whether it should be Math.min.apply( Math, array ) or Math.min.apply( null, array ). So what context should be used, Math or null?
When passing null as a context to apply, then the context will default to the global object (the window object in the case of browsers). Passing the Math object as the context would be the correct solution, but it won't hurt passing null either. Here's an example when null might cause trouble, when decorating the Math.max function:
// decorate Math.max
(function (oldMax) {
Math.max = function () {
this.foo(); // call Math.foo, or at least that's what we want
return oldMax.apply(this, arguments);
};
})(Math.max);
Math.foo = function () {
print("foo");
};
Array.prototype.max = function() {
return Math.max.apply(null, this); // <-- passing null as the context
};
var max = [1, 2, 3].max();
print(max);
The above will throw an exception because this.foo will be evaluated as window.foo, which is undefined. If we replace null with Math, things will work as expected and the string "foo" will be printed to the screen (I tested this using Mozilla Rhino).
You can pretty much assume that nobody has decorated Math.max so, passing null will work without problems.
One more way to do it:
var arrayMax = Function.prototype.apply.bind(Math.max, null);
Usage:
var max = arrayMax([2, 5, 1]);
I am surprised not one mentiond the reduce function.
var arr = [1, 10, 5, 11, 2]
var b = arr.reduce(function(previous,current){
return previous > current ? previous:current
});
b => 11
arr => [1, 10, 5, 11, 2]
https://developer.mozilla.org/ru/docs/Web/JavaScript/Reference/Global_Objects/Math/max
function getMaxOfArray(numArray) {
return Math.max.apply(null, numArray);
}
var arr = [100, 0, 50];
console.log(getMaxOfArray(arr))
this worked for me.
This may suit your purposes.
Array.prototype.min = function(comparer) {
if (this.length === 0) return null;
if (this.length === 1) return this[0];
comparer = (comparer || Math.min);
var v = this[0];
for (var i = 1; i < this.length; i++) {
v = comparer(this[i], v);
}
return v;
}
Array.prototype.max = function(comparer) {
if (this.length === 0) return null;
if (this.length === 1) return this[0];
comparer = (comparer || Math.max);
var v = this[0];
for (var i = 1; i < this.length; i++) {
v = comparer(this[i], v);
}
return v;
}
let array = [267, 306, 108]
let longest = Math.max(...array);
I thought I'd share my simple and easy to understand solution.
For the min:
var arr = [3, 4, 12, 1, 0, 5];
var min = arr[0];
for (var k = 1; k < arr.length; k++) {
if (arr[k] < min) {
min = arr[k];
}
}
console.log("Min is: " + min);
And for the max:
var arr = [3, 4, 12, 1, 0, 5];
var max = arr[0];
for (var k = 1; k < arr.length; k++) {
if (arr[k] > max) {
max = arr[k];
}
}
console.log("Max is: " + max);
For big arrays (~10⁷ elements), Math.min and Math.max procuces a RangeError (Maximum call stack size exceeded) in node.js.
For big arrays, a quick & dirty solution is:
Array.prototype.min = function() {
var r = this[0];
this.forEach(function(v,i,a){if (v<r) r=v;});
return r;
};
For an array containing objects instead of numbers:
arr = [
{ name: 'a', value: 5 },
{ name: 'b', value: 3 },
{ name: 'c', value: 4 }
]
You can use reduce to get the element with the smallest value (min)
arr.reduce((a, b) => a.value < b.value ? a : b)
// { name: 'b', value: 3 }
or the largest value (max)
arr.reduce((a, b) => a.value > b.value ? a : b)
// { name: 'a', value: 5 }
Aside using the math function max and min, another function to use is the built in function of sort(): here we go
const nums = [12, 67, 58, 30].sort((x, y) =>
x - y)
let min_val = nums[0]
let max_val = nums[nums.length -1]
I had the same problem, I needed to obtain the minimum and maximum values of an array and, to my surprise, there were no built-in functions for arrays. After reading a lot, I decided to test the "top 3" solutions myself:
discrete solution: a FOR loop to check every element of the array against the current max and/or min value;
APPLY solution: sending the array to the Math.max and/or Math.min internal functions using apply(null,array);
REDUCE solution: recursing a check against every element of the array using reduce(function).
The test code was this:
function GetMaxDISCRETE(A)
{ var MaxX=A[0];
for (var X=0;X<A.length;X++)
if (MaxX<A[X])
MaxX=A[X];
return MaxX;
}
function GetMaxAPPLY(A)
{ return Math.max.apply(null,A);
}
function GetMaxREDUCE(A)
{ return A.reduce(function(p,c)
{ return p>c?p:c;
});
}
The array A was filled with 100,000 random integer numbers, each function was executed 10,000 times on Mozilla Firefox 28.0 on an intel Pentium 4 2.99GHz desktop with Windows Vista. The times are in seconds, retrieved by performance.now() function. The results were these, with 3 fractional digits and standard deviation:
Discrete solution: mean=0.161s, sd=0.078
APPLY solution: mean=3.571s, sd=0.487
REDUCE solution: mean=0.350s, sd=0.044
The REDUCE solution was 117% slower than the discrete solution. The APPLY solution was the worse, 2,118% slower than the discrete solution. Besides, as Peter observed, it doesn't work for large arrays (about more than 1,000,000 elements).
Also, to complete the tests, I tested this extended discrete code:
var MaxX=A[0],MinX=A[0];
for (var X=0;X<A.length;X++)
{ if (MaxX<A[X])
MaxX=A[X];
if (MinX>A[X])
MinX=A[X];
}
The timing: mean=0.218s, sd=0.094
So, it is 35% slower than the simple discrete solution, but it retrieves both the maximum and the minimum values at once (any other solution would take at least twice that to retrieve them). Once the OP needed both values, the discrete solution would be the best choice (even as two separate functions, one for calculating maximum and another for calculating minimum, they would outperform the second best, the REDUCE solution).
Iterate through, keeping track as you go.
var min = null;
var max = null;
for (var i = 0, len = arr.length; i < len; ++i)
{
var elem = arr[i];
if (min === null || min > elem) min = elem;
if (max === null || max < elem) max = elem;
}
alert( "min = " + min + ", max = " + max );
This will leave min/max null if there are no elements in the array. Will set min and max in one pass if the array has any elements.
You could also extend Array with a range method using the above to allow reuse and improve on readability. See a working fiddle at http://jsfiddle.net/9C9fU/
Array.prototype.range = function() {
var min = null,
max = null,
i, len;
for (i = 0, len = this.length; i < len; ++i)
{
var elem = this[i];
if (min === null || min > elem) min = elem;
if (max === null || max < elem) max = elem;
}
return { min: min, max: max }
};
Used as
var arr = [3, 9, 22, -7, 44, 18, 7, 9, 15];
var range = arr.range();
console.log(range.min);
console.log(range.max);
You can use the following function anywhere in your project:
function getMin(array){
return Math.min.apply(Math,array);
}
function getMax(array){
return Math.max.apply(Math,array);
}
And then you can call the functions passing the array:
var myArray = [1,2,3,4,5,6,7];
var maximo = getMax(myArray); //return the highest number
The following code works for me :
var valueList = [10,4,17,9,3];
var maxValue = valueList.reduce(function(a, b) { return Math.max(a, b); });
var minValue = valueList.reduce(function(a, b) { return Math.min(a, b); });
array.sort((a, b) => b - a)[0];
Gives you the maximum value in an array of numbers.
array.sort((a, b) => a - b)[0];
Gives you the minimum value in an array of numbers.
let array = [0,20,45,85,41,5,7,85,90,111];
let maximum = array.sort((a, b) => b - a)[0];
let minimum = array.sort((a, b) => a - b)[0];
console.log(minimum, maximum)
let arr=[20,8,29,76,7,21,9]
Math.max.apply( Math, arr ); // 76
Simple stuff, really.
var arr = [10,20,30,40];
arr.max = function() { return Math.max.apply(Math, this); }; //attach max funct
arr.min = function() { return Math.min.apply(Math, this); }; //attach min funct
alert("min: " + arr.min() + " max: " + arr.max());
Here's one way to get the max value from an array of objects. Create a copy (with slice), then sort the copy in descending order and grab the first item.
var myArray = [
{"ID": 1, "Cost": 200},
{"ID": 2, "Cost": 1000},
{"ID": 3, "Cost": 50},
{"ID": 4, "Cost": 500}
]
maxsort = myArray.slice(0).sort(function(a, b) { return b.ID - a.ID })[0].ID;