在对象数组中找到最常见的项目

Question

我需要帮助制作一个函数来搜索，使用vanilla或例如lodash来循环一个对象数组，并获得数组中出现次数最多的对象，这是我的对象：

[{"s":97,"p":75},{"s":99,"p":93},{"s":97,"p":75},{"s":97,"p":76},{"s":97,"p":75},{"s":97,"p":75},{"s":97,"p":74},{"s":86,"p":80},{"s":97,"p":73},{"s":97,"p":71},{"s":97,"p":71}]

结果应该是：

{"s":97,"p":75}

提前感谢

Answer 1

您可以使用Array.prototype.reduce()遍历源数组，构建一个复杂的对象，以跟踪每个对象的出现、迄今为止最常看到的对象以及最常看到对象的实际次数。

因此，一旦一些数组项超过maxCount，它就变成mostOften。

有了它，你可以只对所有项目进行一次遍历，就可以找到获胜者：

const src = [{"s":97,"p":75},{"s":99,"p":93},{"s":97,"p":75},{"s":97,"p":76},{"s":97,"p":75},{"s":97,"p":75},{"s":97,"p":74},{"s":86,"p":80},{"s":97,"p":73},{"s":97,"p":71},{"s":97,"p":71}],
{mostOften} = src.reduce((r,{s,p}) => {
const hash = s+'ud8ff'+p
r.hashCount[hash] = (r.hashCount[hash]||0) + 1
r.hashCount[hash] > r.maxCount &&
(r.mostOften = {s,p}, r.maxCount = r.hashCount[hash])
return r
}, {hashCount: {}, mostOften: null, maxCount: 0})

console.log(mostOften)      

Answer 2

这里重要的是为您认为的"相同的"；对象。我建议在由两个属性组成的阵列上使用JSON.stringify：

let data = [{"s":97,"p":75},{"s":99,"p":93},{"s":97,"p":75},{"s":97,"p":76},{"s":97,"p":75},{"s":97,"p":75},{"s":97,"p":74},{"s":86,"p":80},{"s":97,"p":73},{"s":97,"p":71},{"s":97,"p":71}];
let keys = Object.fromEntries(data.map(o => [ JSON.stringify([o.s, o.p, "s" in o, "p" in o]), o ]));
let counter = {};
for (let key in keys) counter[key] = (counter[key] || 0) + 1;
let bestKey = Object.entries(counter).reduce((max, [key, count]) =>
count > max[1] ? [key, count] : max, ["", 0])[0];

let result = keys[bestKey];
console.log(result);

即使s或p是具有任何外来字符的字符串，或布尔值，或null，或基元的数组。。。这仍然有效。

Answer 3

let arr = [{"s":97,"p":75},{"s":99,"p":93},{"s":97,"p":75},{"s":97,"p":76},{"s":97,"p":75},{"s":97,"p":75},{"s":97,"p":74},{"s":86,"p":80},{"s":97,"p":73},{"s":97,"p":71},{"s":97,"p":71}]
let obj = {};
let maxCount = 0;
let result;
arr.forEach(e => {
let key = `s:${e.s}:p:${e.p}`;
obj[key] = obj[key] || 0;
obj[key] += 1;
if(obj[key] > maxCount){
maxCount = obj[key];
result = e;
}
});
console.log(result);