find min and max value using reduce es2015 - javascript

If I have array of object like this
[{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}]
How to find min and max using reduce? I know I can use generic loop and compare but I would like to explore reduce in es2015

You can use reduce like this to get the min and max numbers from each object in the array.
const arr = [{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}]
console.log(
arr.reduce((acc, x) => {
acc.min = Math.min(acc.min, x.min)
acc.max = Math.max(acc.max, x.max)
return acc
}, { min: Infinity, max: -Infinity })
)
// as Bergi suggested, we could just return a new Object literal
// from the reduction
console.log(
arr.reduce((acc, x) => ({
min: Math.min(acc.min, x.min),
max: Math.max(acc.max, x.max)
}), { min: Infinity, max: -Infinity })
)

Assuming that you want, for the minimum, to get the id of the item with the lowed min value, then this will do it.
const items = [{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}];
// Reduce to find the item with the lowest min
const min = items.reduce((res, item) => {
// No result yet, so the item must be the lowest seen
if (!res) return item;
// Use the lowest of the current lowest and the current item
return item.min < res.min ? item : res;
}, undefined).id;

const arr = [{min:5,max:10,id:1}, {min:50,max:3,id:2}, {min:1,max:40,id:3}];
const min = arr.reduce((m, o) => m < o.min? m: o.min, +Infinity),
max = arr.reduce((M, o) => M > o.max? M: o.max, -Infinity);
console.log("Min:", min);
console.log("Max:", max);
Explanation of min:
const min = arr.reduce((m, o) => { // for each object o in the array arr
return m < o.min? // if o.min is bigger than the minimum m
m: // then the minimum is still m
o.min; // otherwise, the new minimum is o.min
}, +Infinity); // pass in +Ifinity as the minimum (the initial value of m so whatever is arr[0].min is it will be smaller than m). You can pass in arr[0].min if you want instead of +Infinity

Related

create array of object from array of strings

I am trying to solve one issue in my code, so if anyone can help me here?
I have some values added below, So i have one array of string values, which has mac addresses and min & max which is constant values. I am trying to map over macValue and trying to create array of object as given sample below, but unfortunately getting some error there. If anyone can help me out here.
please check here i am trying to dynamically add property inside map.
let macValue = ["MO-CK-DR-01","02","03"]
let min = true
let max = true
// code i have tried
var print = macValue.map((item, i) => {
(item['macAddress'] = item), (item.minimum = min), (item.maximum = max);
return item;
});
trying to create array of object like this
[
{
macAddress: value, //01
mimimum: min,
maximum: max,
},
{
macvalue: value, // 02
mimimum: min,
maximum: max,
},
]
but didn't work and getting this error
As simple as:
let macValue = ["MO-CK-DR-01","02","03"]
let min = true
let max = true
const obj = macValue.map((value) => ({
macvalue: value,
minimum: min,
maximum: max,
}))
let macValue = ['MO-CK-DR-01', '02', '03'];
let min = true;
let max = true;
const result = macValue.map((value) => ({
macValue: value,
minimum: min,
maximum: max
}));
console.log(result);
let macValue = ['MO-CK-DR-01', '02', '03'];
let min = true;
let max = true;
const output = macValue.map(value => ({
macValue: value,
minimum: min,
maximum: max
}));
console.log(output);
As you mentioned, the properties are dynamic, I have used string properties. This should work -
let macValue = ['MO-CK-DR-01', '02', '03'];
let min = true;
let max = true;
const result = macValue.map((value) => ({
'macAddress': value,
'minimum': min,
'maximum': max
}));
console.log(result);

Find min(max) value amoung all object.properties and return result as object or (modify original object)

This a bit easy task at first look, with finding the min/max values in Object, has become a challenge for me.
I have the following object (with 3 fixed keys):
let object_ = {
vendor: 5, //(Number),
market: 2, //(Number)
derivative: {
price: 15 //(Number)
}
}
And I am trying to find not just the min/max value, but return it as an object / modify the original one, which should looks like that:
result for min: //as an object
{ market: 2}
result for max: //as an object
derivative: {
price: 15 //(Number)
}
As for now, I have the following code which solves this problem by creating an Array from original object's properties (by manually checking each property's name), like:
/** Create an array of objects from object_*/
let array = [
{market: 2},
{vendor: 5},
{derivative: 15}
]
/** Find the necessary values in array with map and reduce*/
And then finding min/max values from all objects inside array, but the problem is that I should convert result of my code back to the original schema like:
if (resultObject.property === 'derivative') {
resultObject.derivative = {
price: resultObject.derivative //15
}
}
So the question is:
Am I on the right path, should I create an Array of objects for this case? Cause it's important for me, return not just the number itself, but with a custom named property.
Maybe someone has more better/shorter solution with ES6 syntax or modifying the original object with delete not min (or max) properties. Or can point me a way in to it?
You could get the value and path to it, select min and max value and rebuild objects for min and max.
This approach works only for a single min and max values.
const
getValues = object => {
return Object
.entries(object)
.flatMap(([k, v]) => v && typeof v === 'object'
? getValues(v).map(([keys, v]) => [[k, ...keys], v])
: [[[k], v]]
);
},
setValue = (target, [keys, value]) => {
const last = keys.pop();
keys.reduce((o, k) => o[k] = o[k] || {}, target)[last] = value;
return target;
},
getMinMax = object => {
const
[min, max] = getValues(object)
.reduce((r, v, i) => {
if (!i) return [v, v];
if (v[1] < r[0][1]) r[0] = v;
if (v[1] > r[1][1]) r[1] = v;
return r;
}, [])
.map(v => setValue({}, v));
return { min, max };
},
object = { vendor: 5, market: 2, derivative: { price: 15 } },
result = getMinMax(object);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
At the start of the function set min index to 0. Then loop through every key in the object from index 1. Test if the value at the key is an object, and if it is call the current function for it, and set the value to the returned value.Then test if the value is less then one at the index. If it is set index to i. At the end return an object with the value.
function getMin(obj) {
let minIndex = 0;
let keys = Object.keys(obj); //Get all the keys in the object
for(let i = 0;i < keys.length;i++) { // Loop through the keys
let value = obj[keys[i]];
if(typeof value !== 'number') { // If it's not a number get the minimum value in the object
value = getMin(value);
}
if(value < obj[keys[minIndex]]) minIndex = i; //If the value is less than the value at the min index, set the min index to the current index
// You can just change the sign to greater for the max function
}
let returnObject = {};
returnObject[keys[minIndex]] = obj[keys[minIndex]]; // Sets the object so it doesn't only return the value
return returnObject;
}
My own solution that I am using right now. It does the job well, but it's a bit monstrous I guess. So I won't accept it as an answer on my own question. But it still an option anyway.
let object = {
market: 5,
vendor: 2,
derivative: {
price: 15
}
};
let bufferArray = [];
for (const property in object) {
if (typeof object[property] === 'object') {
bufferArray.push({name: property, value: object[property].price});
} else {
bufferArray.push({name: property, value: object[property]});
}
}
let {name, value} = bufferArray.reduce((prev, curr) => prev.value < curr.value ? prev : curr); //just change < to > for max value
let newObject = {[name]: value};
console.log(newObject); //{ vendor: 2 }

How to populate an array with integers

Please, how do you populate an array say ‘num’ with numbers not in a second array say ‘fig’? I’m trying to use a loop to have the values of the already populated array ‘fig’ compared to ‘num’ which is to be populated with integers not found in ‘fig’. I’m a bit confused.
If you need to do an array with n numbers you can use this two ways.
const arrayLength = 100;
const numberArray = [...new Array(arrayLength).keys()]
const anotherWay = new Array(arrayLength).fill().map((_, idx) => idx + 1);
console.log(numberArray, anotherWay)
so to do this we have to do a few things:
1) define an existing array with numbers to avoid
2) define length on new array
3) generate a random number and make it an integer
4) check to see if we need to avoid
5) if it's a new value add it to the second array
var first=[55,45,35,1,2,3,4,5];
var second = [];
var i = 7;
var x;
while (i != 0){
x = ~~(Math.random()*100);
var check = false;
for(let j=0; j<first.length;j++){
if(x == first[j]){
check = true;
}
}
if(!check){
second.push(x);
i--;
}
}
console.log(second);
const fig = [-21, 0, 3, 6, 7, 42]
const min = Math.min(...fig) // or fig[0] if the the array is already sorted
const max = Math.max(...fig) // or fig[fig.length - 1]
const num = Array.from({ length: max - min }, (_, i) => i + min)
.filter(el => !fig.includes(el))
or, saving one loop
const num = Array.from({ length: max - min }).reduce((acc, _, i) => {
const curr = i + min
if (!fig.includes(curr)) {
return acc.concat(curr)
}
return acc
}, [])
This is assuming your range is from the smallest number in fig to the largest in fig.

Getting keys with maximum value in JavaScript hashmap/object

Example:
hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2}
here the maximum key is both Apple and Mango. How do I write a function which gives both Apple and Mango as answer.
I tried something like this :
Object.keys(hash).reduce(function(a, b){ return hash[a] > hash[b] ? a : b });
But this gives only Apple as the answer.
You could first calculate the max value as a separate operation and then just filter:
const hash = {Apple: 2, Orange: 1, Mango: 2};
const max = Object.keys(hash).reduce((a, v) => Math.max(a, hash[v]), -Infinity);
const result = Object.keys(hash).filter(v => hash[v] === max);
console.log(result);
Simple and readable, but it requires and extra iteration, so it's not the most efficient implementation.
Your reduce can only return a single value, but you need an array of values. So create one as the initial value.
In your function body, first, check whether your next key has a larger value, in which case you clear out the array.
Then, if your array is empty, or if the first contained key (and thus all keys) have the same value, push that key into your array.
Also, it helps if you give your arguments more expressive names.
Object.keys(hash).reduce(function(longestKeys, key){
if(hash[key] > hash[longestKeys[0]]){
longestKeys.length = 0;
}
if(longestKeys.length === 0 || hash[key] === hash[longestKeys[0]]){
longestKeys.push(key);
}
return longestKeys;
}, []);
You might transform the object into one whose properties correspond to the count, whose values are the (original) property names in an array, and then get the value of the property with the maximum count:
const hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2};
const indexedByCount = Object.entries(hash).reduce((a, [key, val]) => {
if (!a[val]) a[val] = [];
a[val].push(key);
return a;
}, {});
console.log(
indexedByCount[Math.max(...Object.keys(indexedByCount))]
);
A less functional but more efficient method would be to keep track of a max variable corresponding to the maximum value found so far:
const hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2};
let max = -Infinity;
console.log(
Object.entries(hash).reduce((a, [key, val]) => {
if (val > max) {
max = val;
return [key];
}
if (val === max) a.push(key);
return a;
}, [])
);
This works. If max changes clear the result and set only the max value.
var hash = {'Apple':2, 'Orange':1 , 'Mango':2, "Jackfruit":10, "Pineapple":5, "Tomato":-4};
var max="";
var result = Object.keys(hash).reduce(function(acc, val){
if(max < hash[val]) (max=hash[val], acc={});
if(hash[val]==max) acc[val] = hash[val];
return acc;
},{});
console.log(result)
I believe the requirements are traverse the array only once and results is an array of keys
const hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2};
let max = 0;
let result = [];
Object.getOwnPropertyNames(hash).forEach(k => {
if (hash[k] > max) {
result = [k];
max = hash[k];
} else if (hash[k] === max) {
result.push(k);
}
});
console.log(result);

Find the min/max element of an array in JavaScript

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;

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