Find the min/max element of an Array in JavaScript

2018-12-31 01:11发布

How can I easily obtain the min or max element of a JavaScript Array?

Example Psuedocode:

let array = [100, 0, 50]

array.min() //=> 0
array.max() //=> 100

标签: javascript
30条回答
ら面具成の殇う
2楼-- · 2018-12-31 01:54

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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笑指拈花
3楼-- · 2018-12-31 01:54

tl;dr

var max = Math.max(...arrayOfNumbers);

Official Math.max() MDN documentation

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);
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骚的不知所云
4楼-- · 2018-12-31 01:54

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;
}
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步步皆殇っ
5楼-- · 2018-12-31 01:55

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);
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梦寄多情
6楼-- · 2018-12-31 01:55

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.

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听够珍惜
7楼-- · 2018-12-31 01:58

ChaosPandion's solution works if you're using protoype. If not, consider this:

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

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

The above will return NaN if an array value is not an integer so you should build some functionality to avoid that. Otherwise this will work.

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