如何使用类型记录中的map-约简来筛选属性最大值的列表？

Question

我有一个清单如下：

let scores = [{name: "A", skills: 50, result: 80},
              {name: "B", skills: 40, result: 90},
              {name: "C", skills: 60, result: 60},
              {name: "D", skills: 60, result: 70}];

我想过滤的名单谁获得最大的技能值使用地图-减少在打字稿。

output = [{name: "C", skills: 60, result: 60},
          {name: "D", skills: 60, result: 70}];

我尝试了如下，并获得了预期的产出：

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let scores = [{name: "A", skills: 50, result: 80},
              {name: "B", skills: 40, result: 90},
              {name: "C", skills: 60, result: 60},
              {name: "D", skills: 60, result: 70}];

let maxSkillValue = Math.max.apply(Math, scores.map(function (o) { return o.skills; }));
let output = scores.filter(i => i.skills === maxSkillValue);

console.log(output);

​

使用地图寻找答案-减少。

Answer 1

谢谢您的回答。如果任何对象在“技能”或“结果”中都有最大值，我如何筛选第一个列表？我的预期输出:输出={ name："B"，技能: 40，结果: 90 }，{ name："C"，技能: 60，结果: 60 }，{ name："D"，技能: 60，结果: 70 }- ImtiazNur

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let scores = [{name: "A", skills: 500, result: 80},
              {name: "B", skills: 6000, result: 90},
              {name: "C", skills: 6000, result: 60},
              {name: "D", skills: 20, result: 60},
              {name: "E", skills: 10, result: 90}];

const maxSkill = scores.reduce((acc, curr) => curr.skills > acc ? curr.skills : acc, 0);
const maxResult = scores.reduce((acc, curr) => curr.result > acc ? curr.result : acc, 0);
const maxSkillsAndRes = scores.reduce((r, o) => (o.skills === maxSkill || o.result === maxResult) ? [...r,o] : r, []);

console.log(maxSkillsAndRes);

​

sry，大约第三行，你最好写：

const maxSkillsAndRes = scores.filter( o => o.skills === maxSkill || o.result === maxResult );

输出相同，但更具可读性。

Answer 2

您可以通过检查结果集或最后存储的最大技能值来减少数组，然后将此对象作为数组中的新结果集。其他人检查技能是否相等，然后推送实际值并返回累加器。

要从一定数量的属性中获取max对象，可以存储每个属性的最大值，并筛选部件结果集。

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let scores = [{ name: "A", skills: 50, result: 80 }, { name: "B", skills: 40, result: 90 }, { name: "C", skills: 60, result: 60 }, { name: "D", skills: 60, result: 70 }],
    maxSkills = scores.reduce((r, o) => {
        if (!r || r[0].skills < o.skills) return [o];
        if (r[0].skills === o.skills) r.push(o);
        return r;
    }, undefined),
    keys = ['skills', 'result'],
    max = scores.reduce((r, o) => {
        keys.forEach(k => r[k] = Math.max(k in r ? r[k] : -Infinity, o[k]));
        r._ = [...r._, o].filter(q => keys.some(k => r[k] === q[k]));
        return r;
    }, { _: [] })._;

console.log(maxSkills);
console.log(max);

.as-console-wrapper { max-height: 100% !important; top: 0; }

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