Obtain smallest value from array in Javascript?

Array justPrices has values such as:

[0] = 1.5
[1] = 4.5
[2] = 9.9.

How do I return the smallest value in the array?

Answers:

Answer

Jon Resig illustrated in this article how this could be achieved by extending the Array prototype and invoking the underlying Math.min method which unfortunately doesn't take an array but a variable number of arguments:

Array.min = function( array ){
    return Math.min.apply( Math, array );
};

and then:

var minimum = Array.min(array);
Answer

The tersest expressive code to find the minimum value is probably rest parameters:

const arr = [14, 58, 20, 77, 66, 82, 42, 67, 42, 4]
const min = Math.min(...arr)
console.log(min)


Rest parameters are essentially a convenient shorthand for Function.prototype.apply when you don't need to change the function's context:

var arr = [14, 58, 20, 77, 66, 82, 42, 67, 42, 4]
var min = Math.min.apply(Math, arr)
console.log(min)


This is also a great use case for Array.prototype.reduce:

const arr = [14, 58, 20, 77, 66, 82, 42, 67, 42, 4]
const min = arr.reduce((a, b) => Math.min(a, b))
console.log(min)

It may be tempting to pass Math.min directly to reduce, however the callback receives additional parameters:

callback (accumulator, currentValue, currentIndex, array)

In this particular case it may be a bit verbose. reduce is particularly useful when you have a collection of complex data that you want to aggregate into a single value:

const arr = [{name: 'Location 1', distance: 14}, {name: 'Location 2', distance: 58}, {name: 'Location 3', distance: 20}, {name: 'Location 4', distance: 77}, {name: 'Location 5', distance: 66}, {name: 'Location 6', distance: 82}, {name: 'Location 7', distance: 42}, {name: 'Location 8', distance: 67}, {name: 'Location 9', distance: 42}, {name: 'Location 10', distance: 4}]
const closest = arr.reduce(
  (acc, loc) =>
    acc.distance < loc.distance
      ? acc
      : loc
)
console.log(closest)


And of course you can always use classic iteration:

var arr,
  i,
  l,
  min

arr = [14, 58, 20, 77, 66, 82, 42, 67, 42, 4]
min = Number.POSITIVE_INFINITY
for (i = 0, l = arr.length; i < l; i++) {
  min = Math.min(min, arr[i])
}
console.log(min)

...but even classic iteration can get a modern makeover:

const arr = [14, 58, 20, 77, 66, 82, 42, 67, 42, 4]
let min = Number.POSITIVE_INFINITY
for (const value of arr) {
  min = Math.min(min, value)
}
console.log(min)

Answer

I find that the easiest way to return the smallest value of an array is to use the Spread Operator on Math.min() function.

return Math.min(...justPrices);
//returns 1.5 on example given 

The page on MDN helps to understand it better: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/min

A little extra: This also works on Math.max() function

return Math.max(...justPrices); //returns 9.9 on example given.

Hope this helps!

Answer

Imagine you have this array:

var arr = [1, 2, 3];

ES6 way:

var min = Math.min(...arr); //min=1

ES5 way:

var min = Math.min.apply(null, arr); //min=1

If you using D3.js, there is a handy function which does the same, but will ignore undefined values and also check the natural order:

d3.max(array[, accessor])

Returns the maximum value in the given array using natural order. If the array is empty, returns undefined. An optional accessor function may be specified, which is equivalent to calling array.map(accessor) before computing the maximum value.

Unlike the built-in Math.max, this method ignores undefined values; this is useful for ignoring missing data. In addition, elements are compared using natural order rather than numeric order. For example, the maximum of the strings [“20”, “3”] is “3”, while the maximum of the numbers [20, 3] is 20.

And this is the source code for D3 v4:

export default function(values, valueof) {
  var n = values.length,
      i = -1,
      value,
      max;

  if (valueof == null) {
    while (++i < n) { // Find the first comparable value.
      if ((value = values[i]) != null && value >= value) {
        max = value;
        while (++i < n) { // Compare the remaining values.
          if ((value = values[i]) != null && value > max) {
            max = value;
          }
        }
      }
    }
  }

  else {
    while (++i < n) { // Find the first comparable value.
      if ((value = valueof(values[i], i, values)) != null && value >= value) {
        max = value;
        while (++i < n) { // Compare the remaining values.
          if ((value = valueof(values[i], i, values)) != null && value > max) {
            max = value;
          }
        }
      }
    }
  }

  return max;
}
Answer

ES6 is the way of the future.

arr.reduce((a, b) => Math.min(a, b));

I prefer this form because it's easily generalized for other use cases

Answer

Possibly an easier way?

Let's say justPrices is mixed up in terms of value, so you don't know where the smallest value is.

justPrices[0] = 4.5
justPrices[1] = 9.9
justPrices[2] = 1.5

Use sort.

justPrices.sort();

It would then put them in order for you. (Can also be done alphabetically.) The array then would be put in ascending order.

justPrices[0] = 1.5
justPrices[1] = 4.5
justPrices[2] = 9.9

You can then easily grab by the first index.

justPrices[0]

I find this is a bit more useful than what's proposed above because what if you need the lowest 3 numbers as an example? You can also switch which order they're arranged, more info at http://www.w3schools.com/jsref/jsref_sort.asp

Answer

If you are using Underscore or Lodash you can get the minimal value using this kind of simple functional pipeline

_.chain([7, 6, -1, 3, 2]).sortBy().first().value()
// -1

You also have the .min function

_.min([7, 6, -1, 3, 2])
// -1
Answer

I think I have an easy-to-understand solution for this, using only the basics of javaScript.

function myFunction() {
            var i = 0;
            var smallestNumber = justPrices[0];
            for(i = 0; i < justPrices.length; i++) {
                if(justPrices[i] < smallestNumber) {
                    smallestNumber = justPrices[i];
                }
            }
            return smallestNumber;
        }

The variable smallestNumber is set to the first element of justPrices, and the for loop loops through the array (I'm just assuming that you know how a for loop works; if not, look it up). If an element of the array is smaller than the current smallestNumber (which at first is the first element), it will replace it's value. When the whole array has gone through the loop, smallestNumber will contain the smallest number in the array.

Answer

Update: use Darin's / John Resig answer, just keep in mind that you dont need to specifiy thisArg for min, so Math.min.apply(null, arr) will work just fine.


or you can just sort the array and get value #1: [2,6,7,4,1].sort()[0]

[!] But without supplying custom number sorting function, this will only work in one, very limited case: positive numbers less than 10. See how it would break:

var a = ['', -0.1, -2, -Infinity, Infinity, 0, 0.01, 2, 2.0, 2.01, 11, 1, 1e-10, NaN];

// correct: 
a.sort( function (a,b) { return a === b ? 0 : a < b ? -1: 1} );
//Array [NaN, -Infinity, -2, -0.1, 0, "", 1e-10, 0.01, 1, 2, 2, 2.01, 11, Infinity]

// incorrect:
a.sort();
//Array ["", -0.1, -2, -Infinity, 0, 0.01, 1, 11, 1e-10, 2, 2, 2.01, Infinity, NaN]

And, also, array is changed in-place, which might not be what you want.

Answer
var array =[2,3,1,9,8];
var minvalue = array[0]; 
for (var i = 0; i < array.length; i++) {
    if(array[i]<minvalue)
    {
        minvalue = array[i];
    }

}
  console.log(minvalue);
Answer

function smallest(){
  if(arguments[0] instanceof Array)
    arguments = arguments[0];

  return Math.min.apply( Math, arguments );
}
function largest(){
  if(arguments[0] instanceof Array)
    arguments = arguments[0];

  return Math.max.apply( Math, arguments );
}
var min = smallest(10, 11, 12, 13);
var max = largest([10, 11, 12, 13]);

console.log("Smallest: "+ min +", Largest: "+ max);

Answer

Here is code that will detect the lowest value in an array of numbers.

//function for finding smallest value in an array
function arrayMin(array){
    var min = array[0];
    for(var i = 0; i < array.length; i++){
        if(min < array[i]){
            min = min;
        }else if (min > array[i]){
            min = array[i + 1];
        }else if (min == array[i]){
            min = min;
        }
    }
    return min;
};

call it in this way:

var fooArray = [1,10,5,2];
var foo = arrayMin(fooArray);

(Just change the second else if result from: min = min to min = array[i] if you want numbers which reach the smallest value to replace the original number.)

Answer

Here is a recursive way on how to do it using ternary operators both for the recursion and decision whether you came across a min number or not.

const findMin = (arr, min, i) => arr.length === i ? min :
  findMin(arr, min = arr[i] < min ? arr[i] : min, ++i)

Code snippet:

const findMin = (arr, min, i) => arr.length === i ? min :
  findMin(arr, min = arr[i] < min ? arr[i] : min, ++i)
  
const arr = [5, 34, 2, 1, 6, 7, 9, 3];
const min = findMin(arr, arr[0], 0)
console.log(min);

Answer

Here’s a variant of Darin Dimitrov’s answer that doesn’t modify the Array prototype:

const applyToArray = (func, array) => func.apply(Math, array)

applyToArray(Math.min, [1,2,3,4]) // 1
applyToArray(Math.max, [1,2,3,4]) // 4
Answer
function tinyFriends() {
    let myFriends = ["Mukit", "Ali", "Umor", "sabbir"]
    let smallestFridend = myFriends[0];
    for (i = 0; i < myFriends.length; i++) {
        if (myFriends[i] < smallestFridend) {
            smallestFridend = myFriends[i];
        }
    }
    return smallestFridend   
}

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