查找对象数组中属性的最大值

Question

我正在寻找一种非常快速，干净且高效的方法来获取以下JSON切片中的最大“y”值：

[
  {
    "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
  }
]

for循环是唯一可行的方法吗？我非常热衷于使用Math.max。

Answer 1

要在y中找到对象的最大array值：

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

Answer 2

在对象数组中找到属性“X”具有最大值的对象

一种方法是使用Array reduce ..

const max = data.reduce(function(prev, current) {
    return (prev.y > current.y) ? prev : current
}) //returns object

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce http://caniuse.com/#search=reduce（IE9及以上）

如果你不需要支持IE（只有Edge），或者可以使用像Babel这样的预编译器，你可以使用更简洁的语法。

const max = data.reduce((prev, current) => (prev.y > current.y) ? prev : current)

Answer 3

干净简单的ES6（Babel）

const maxValueOfY = Math.max(...arrayToSearchIn.map(o => o.y), 0);

如果arrayToSearchIn为空，则第二个参数应确保默认值。

Answer 4

好吧，首先你应该解析JSON字符串，以便你可以轻松访问它的成员：

var arr = $.parseJSON(str);

使用map方法提取值：

arr = $.map(arr, function(o){ return o.y; });

然后您可以使用max方法中的数组：

var highest = Math.max.apply(this,arr);

或者作为一个单行：

var highest = Math.max.apply(this,$.map($.parseJSON(str), function(o){ return o.y; }));

Answer 5

我想逐步解释terse accepted answer：

＆＃13;

＆＃13;

var objects = [{ x: 3 }, { x: 1 }, { x: 2 }];

// array.map lets you extract an array of attribute values
var xValues = objects.map(function(o) { return o.x; });
// es6
xValues = Array.from(objects, o => o.x);

// function.apply lets you expand an array argument as individual arguments
// So the following is equivalent to Math.max(3, 1, 2)
// The first argument is "this" but since Math.max doesn't need it, null is fine
var xMax = Math.max.apply(null, xValues);
// es6
xMax = Math.max(...xValues);

// Finally, to find the object that has the maximum x value (note that result is array):
var maxXObjects = objects.filter(function(o) { return o.x === xMax; });

// Altogether
xMax = Math.max.apply(null, objects.map(function(o) { return o.x; }));
var maxXObject = objects.filter(function(o) { return o.x === xMax; })[0];
// es6
xMax = Math.max(...Array.from(objects, o => o.x));
maxXObject = objects.find(o => o.x === xMax);


document.write('<p>objects: ' + JSON.stringify(objects) + '</p>');
document.write('<p>xValues: ' + JSON.stringify(xValues) + '</p>');
document.write('<p>xMax: ' + JSON.stringify(xMax) + '</p>');
document.write('<p>maxXObjects: ' + JSON.stringify(maxXObjects) + '</p>');
document.write('<p>maxXObject: ' + JSON.stringify(maxXObject) + '</p>');

＆＃13;

＆＃13;

＆＃13;

更多信息：

• Math.max.apply()
• Array.prototype.map()
• Array.from()
• Function.prototype.apply()
• Array.prototype.filter()
• Array.prototype.find()

Answer 6

这是最短的解决方案（单线） ES6 ：

export class DynamicRouteService implements CanActivate {

  constructor(private router: Router) { }

  canActivate(next: ActivatedRouteSnapshot, state: RouterStateSnapshot): Observable<boolean> | Promise<boolean> | boolean {

    return next.params.pipe(first()).pipe(map(param) => {
      console.log(param.id)

      return ...   // make any API call with param.id and get a response as promise

      .then( (response) => {

         ...    // do whatever you want to do on success
         return true;

      })
      .catch( (error) => {

         console.error(error);
         this.router.navigate(['404']);    // route to 404 on failure
         return false;

      }))

    }
  }
}

Answer 7

如果您（或此处有人）可以免费使用lodash实用程序库，则它具有 maxBy 功能，在您的情况下非常方便。

因此您可以这样使用：

_.maxBy(jsonSlice, 'y');

Answer 8

var data = [
  { 'name': 'Vins', 'age': 27 },
  { 'name': 'Jan', 'age': 38 },
  { 'name': 'Alex', 'age': 80 },
  { 'name': 'Carl', 'age': 25 },
  { 'name': 'Digi', 'age': 40 }
];
var max = data.reduce(function (prev, current) {
   return (prev.age > current.age) ? prev : current
});
//output = {'name': 'Alex', 'age': 80}

Answer 9

或者简单的排序！保持真实：）

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

Answer 10

处理负数大小写（在a数组中输入）的树 ONELINERS 的比较：

var maxA = Math.max(...a.map(o=>o.y),a[0].y); // 33chars time complexity: >O(2n)

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

var maxC = a.sort((a,b)=>b.y-a.y)[0].y;       // 27chars time complexity:  O(nlogn)

可编辑的示例here。来自maxA，maxB，maxC的想法（副作用：已更改a-sort就位）。

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

var maxA = Math.max(...a.map(o=>o.y),a[0].y);
var maxB = a.reduce((a,b)=>a.y>b.y?a:b).y;
var maxC = a.sort((a,b)=>b.y-a.y)[0].y;


document.body.innerHTML=`<pre>maxA: ${maxA}\nmaxB: ${maxB}\nmaxC: ${maxC}</pre>`;

Answer 11

ES6解决方案

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

有关详细信息，请参阅https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/max

Answer 12

每个数组并通过Math获得最大值。

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

Answer 13

var max = 0;                
jQuery.map(arr, function (obj) {
  if (obj.attr > max)
    max = obj.attr;
});

Answer 14

又快又脏：

Object.defineProperty(Array.prototype, 'min',
{
    value: function(f)
    {
        f = f || (v => v);
        return this.reduce((a, b) => (f(a) < f(b)) ? a : b);
    }
});

Object.defineProperty(Array.prototype, 'max',
{
    value: function(f)
    {
        f = f || (v => v);
        return this.reduce((a, b) => (f(a) > f(b)) ? a : b);
    }
});

console.log([1,2,3].max());
console.log([1,2,3].max(x => x*(4-x)));
console.log([1,2,3].min());
console.log([1,2,3].min(x => x*(4-x)));

Answer 15

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

Answer 16

很简单

     const array1 = [
  {id: 1, val: 60},
  {id: 2, val: 2},
  {id: 3, val: 89},
  {id: 4, val: 78}
];
const array2 = [1,6,8,79,45,21,65,85,32,654];
const max = array1.reduce((acc, item) => acc = acc > item.val ? acc : item.val, 0);
const max2 = array2.reduce((acc, item) => acc = acc > item ? acc : item, 0);

console.log(max);
console.log(max2);