或者一个简单的排序!保持真实:)

array.sort((a,b)=>a.y<b.y)[0].y

求数组中所有对象的y的最大值:

    Math.max.apply(Math, array.map(function(o) { return o.y; }))

或者在更现代的JavaScript中:

    Math.max(...array.map(o => o.y))

比较三个处理负号大小写的oneliner(在数组中输入):

var maxA = a.reduce((a,b)=>a.y>b.y?a:b).y; // 30 chars time complexity:  O(n)

var maxB = a.sort((a,b)=>b.y-a.y)[0].y;    // 27 chars time complexity:  O(nlogn)
           
var maxC = Math.max(...a.map(o=>o.y));     // 26 chars time complexity: >O(2n)

这里是可编辑的示例。来自:maxA, maxB和maxC的想法(maxB的副作用是数组a被改变，因为排序是到位的)。

var a = [ {" x ": 8/11/2009”、“y”:026572007 {" x "}表示:“8/12/2009”、“y”:025057454} {" x ": 8/14/2009”、“y”:031004457 {" x "}表示:“8/13/2009”、“y”:024530916} ] var maxA = a.reduce((a,b)=>a.y>b.y?a:b) y; var maxC = Math.max(.. a.map(o=>o.y)); var maxB = a.sort((a,b)=>b.y-a.y)[0].y; 文件。身体。innerHTML = ' < pre - > maxA: $ {maxA} \ nmaxB: $ {maxB} nmaxC: $ {maxC} < / pre - > ';

对于较大的数组，Math.max…最大调用堆栈大小超过(Chrome 76.0.3809, Safari 12.1.2，日期2019-09-13)

let a =数组(400*400)。Fill ({"x": "8/11/2009"， "y": 0.026572007}); // Exception:超过最大调用堆栈大小 尝试{ let max1= Math.max。apply(Math, a.p ap(o => o.y)); } catch(e) {console.error('Math.max. error ')应用:'，e.message)} 尝试{ let max2= Math.max(…a.map(o=>o.y)); } catch(e) {console.error('Math. error ')Max-map:'， e.message)}

4元素数组的基准测试

ES6解决方案

Math.max(…array.map(函数(o){返回o.y;}))

详情见https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/max

每个数组，并得到最大值与数学。

data.reduce((max, b) => Math.max(max, b.costo), data[0].costo);

// Here is very simple way to go:

// Your DataSet.

let numberArray = [
  {
    "x": "8/11/2009",
    "y": 0.026572007
  },
  {
    "x": "8/12/2009",
    "y": 0.025057454
  },
  {
    "x": "8/13/2009",
    "y": 0.024530916
  },
  {
    "x": "8/14/2009",
    "y": 0.031004457
  }
]

// 1. First create Array, containing all the value of Y
let result = numberArray.map((y) => y)
console.log(result) // >> [0.026572007,0.025057454,0.024530916,0.031004457]

// 2.
let maxValue = Math.max.apply(null, result)
console.log(maxValue) // >> 0.031004457