javascript - 如何在 typescript 中使用map-reduce过滤具有最大值属性的列表？

Question

我有一个列表如下:

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}];

我想过滤在 typescript 中使用 Map-Reduce 获得最大技能值的列表。

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

我尝试了如下并得到了预期的输出:

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Map-Reduce寻找答案。

Answer 1

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。否则检查技能是否相等，然后压入实际值并返回累加器。

要从一定数量的属性中获取最大对象，您可以存储每个属性的最大值并过滤部分结果集。

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);

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