在JavaScript中从数组中获取最大值和最小值

Question

I am creating the following array from data attributes and I need to be able to grab the highest and lowest value from it so I can pass it to another function later on.

var allProducts = $(products).children("li");
prices = []
$(allProducts).each(function () {
    var price = parseFloat($(this).data('price'));
    prices[price] = price;
});
console.log(prices[0]) <!-- this returns undefined

My list items look like this (I have cut down for readability):

<li data-price="29.97"><a href="#">Product</a></li>
<li data-price="31.00"><a href="#">Product</a></li>
<li data-price="19.38"><a href="#">Product</a></li>
<li data-price="20.00"><a href="#">Product</a></li>

A quick console.log on prices shows me my array which appears to be sorted so I could grab the first and last element I assume, but presently the names and values in the array are the same so whenever I try and do a prices[0], I get undefined

[]
19.38   19.38
20.00   20.00
29.97   29.97
31.00   31.00

Got a feeling this is a stupidly easy question, so please be kind :)

Answer 1

To get min/max value in array, you can use:

var _array = [1,3,2];
Math.max.apply(Math,_array); // 3
Math.min.apply(Math,_array); // 1

Answer 2

Why not store it as an array of prices instead of object?

prices = []
$(allProducts).each(function () {
    var price = parseFloat($(this).data('price'));
    prices.push(price);
});
prices.sort(function(a, b) { return a - b }); //this is the magic line which sort the array

That way you can just

prices[0]; // cheapest
prices[prices.length - 1]; // most expensive

Note that you can do shift() and pop() to get min and max price respectively, but it will take off the price from the array.

Even better alternative is to use Sergei solution below, by using Math.max and min respectively.

EDIT:

I realized that this would be wrong if you have something like [11.5, 3.1, 3.5, 3.7] as 11.5 is treated as a string, and would come before the 3.x in dictionary order, you need to pass in custom sort function to make sure they are indeed treated as float:

prices.sort(function(a, b) { return a - b });

Answer 3

Instead of .each, another (perhaps more concise) approach to getting all those prices might be:

var prices = $(products).children("li").map(function() {
    return $(this).prop("data-price");
}).get();

additionally you may want to consider filtering the array to get rid of empty or non-numeric array values in case they should exist:

prices = prices.filter(function(n){ return(!isNaN(parseFloat(n))) });

then use Sergey's solution above:

var max = Math.max.apply(Math,prices);
var min = Math.min.apply(Math,prices);

Answer 4

如果您有“分散”（不在数组内）值，您可以使用：

var max_value = Math.max(val1, val2, val3, val4, val5);

Answer 5

arr = [9,4,2,93,6,2,4,61,1];
ArrMax = Math.max.apply(Math, arr);

Answer 6

Find largest and smallest number in an array with lodash .

 var array = [1, 3, 2]; var func = _.over(Math.max, Math.min); var [max, min] = func(...array); // => [3, 1] console.log(max); console.log(min); 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script> 

Answer 7

use this and it works on both the static arrays and dynamically generated arrays.

var array = [12,2,23,324,23,123,4,23,132,23];
var getMaxValue = Math.max.apply(Math, array );

I had the issue when I use trying to find max value from code below

$('#myTabs').find('li.active').prevAll().andSelf().each(function () {
            newGetWidthOfEachTab.push(parseInt($(this).outerWidth()));
        });

        for (var i = 0; i < newGetWidthOfEachTab.length; i++) {
            newWidthOfEachTabTotal += newGetWidthOfEachTab[i];
            newGetWidthOfEachTabArr.push(parseInt(newWidthOfEachTabTotal));
        }

        getMaxValue = Math.max.apply(Math, array);

I was getting 'NAN' when I use

    var max_value = Math.max(12, 21, 23, 2323, 23);

with my code