在对象数组中的某个属性中查找所有具有最大值的对象，并从同一对象返回其他属性的值

Question

我知道这里有类似的问题，但使用这些方法只返回一个最大值。我需要确定数组的哪些对象在给定属性中具有该最大值，并在给定属性中具有最大值的那些对象中返回某个 (other) 属性的值。

我有一个名为 week 的对象数组，它有两个属性 "name" 和 "traffic"：

[
 { name: "Saturday", traffic: 12 },
 { name: "Sunday", traffic: 12 },
 { name: "Monday", traffic: 13 },
 { name: "Tuesday", traffic: 9 },
 { name: "Wednesday", traffic: 10 },
 { name: "Thursday", traffic: 8 },
 { name: "Friday", traffic: 13 },
]

​
在这种情况下，Monday 和 Frid​​ay 具有属性 “Traffic”的最大值 这是 13 我需要一种方法来返回一个 string 包含最高的一天的名称strong>"Traffic" value 如果只有一天，以及包含名称 （作为字符串） 的数组具有最高“流量” 价值的天，如果有超过一天的最高“流量” value，因为在这种情况下会返回一个包含 Monday 和 Frid​​ay 的数组。

我试过这个：

function getMaxTr() {
    return week.reduce((max, p) => p.traffic > max ? 
      p.traffic : max, week[0].traffic); 
}

但是这样我只得到了属性 “traffic” 的一个最大值，即 13。

还有这个：

let max = week [week.length - 1];

通过最后一个，我得到一个具有最大流量值的对象，如下所示：

Object { name: "Friday", traffic: 13 }

Answer 1

如果你想返回最大traffic值的对象的name，你可以使用Array#filter()、Array#reduce()的组合和 Array#map() 这样的方法：

let maxTraffic = arr.reduce(function(a, b) {
    return a.traffic > b.traffic ? a.traffic : b.traffic;
});
var result = arr.filter(a => a.traffic == maxTraffic).map(a => a.name);

这将返回一个array，其中包含具有最大traffic 值的元素的名称。

演示：

这是一个工作演示：

var arr = [
 { name: "Saturday", traffic: 12 },
 { name: "Sunday", traffic: 12 },
 { name: "Monday", traffic: 13 },
 { name: "Tuesday", traffic: 9 },
 { name: "Wednesday", traffic: 10 },
 { name: "Thursday", traffic: 8 },
 { name: "Friday", traffic: 13 },
];

let maxTraffic = arr.reduce(function(a, b) {
    return a.traffic > b.traffic ? a.traffic : b.traffic;
});
var result = arr.filter(a => a.traffic == maxTraffic).map(a => a.name);

console.log(result);

Answer 2

如果reduce数组只包含一个元素，您可以使用函数reduce对日期和函数pop进行分组。

var array = [ { name: "Saturday", traffic: 12 }, { name: "Sunday", traffic: 12 }, { name: "Monday", traffic: 13 }, { name: "Tuesday", traffic: 9 }, { name: "Wednesday", traffic: 10 }, { name: "Thursday", traffic: 8 }, { name: "Friday", traffic: 13 }],
    reduced = array.reduce((a, {name, traffic}) => {
      if (traffic > a.highest) {
        a.current = [name];    
        a.highest = traffic;    
      } else if (traffic === a.highest) a.current.push(name);
  
      return a;
    }, {highest: 0, current: []}).current,
    result = reduced.length === 1 ? reduced.pop() : reduced;

console.log(result);

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当样本仅包含一个流量值最高的对象时，此代码 sn-p 将结果显示为字符串：

var array = [ { name: "Saturday", traffic: 12 }, { name: "Sunday", traffic: 12 }, { name: "Monday", traffic: 1 }, { name: "Tuesday", traffic: 9 }, { name: "Wednesday", traffic: 10 }, { name: "Thursday", traffic: 8 }, { name: "Friday", traffic: 13 }],
    reduced = array.reduce((a, {name, traffic}) => {
      if (traffic > a.highest) {
        a.current = [name];    
        a.highest = traffic;    
      } else if (traffic === a.highest) a.current.push(name);
  
      return a;
    }, {highest: 0, current: []}).current,
    result = reduced.length === 1 ? reduced.pop() : reduced;

console.log(result);

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Answer 3

你可以使用reduce。在每次迭代中，检查结果中是否存在具有较低或相等 traffic 属性的元素，如果是，则替换前一种情况的整个内容，或将相等元素添加到结果中。如果以上都没有返回 true，则只需再次返回最后一次迭代的元素。

const arr = [
    { name: "Saturday", traffic: 12 },
    { name: "Sunday", traffic: 12 },
    { name: "Monday", traffic: 13 },
    { name: "Tuesday", traffic: 9 },
    { name: "Wednesday", traffic: 10 },
    { name: "Thursday", traffic: 8 },
    { name: "Friday", traffic: 13 },
];

let res = arr.reduce((a, b) => {
    let now = a.pop();
    if (now.traffic < b.traffic) return [b];
    if (now.traffic === b.traffic) return [...a, now, b];
    return [...a, now];
}, [arr[0]]).map(e => e.name);

res = res.length > 1 ? res : res[0];

console.log(res);