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慕姐4208626
您可以通过Array.prototype.reduce()构建一个复杂的对象来遍历源数组,以跟踪每个对象的出现、到目前为止最常看到的对象以及实际看到最常见对象的次数。所以,一旦一些数组项超过maxCount它就变成了mostOften。这样你就可以只通过所有项目来找到获胜者: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)
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慕码人2483693
这里重要的是为您认为“相同”的对象提供可靠的密钥。我会建议JSON.stringify由两个属性组成的数组: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是带有任何外来字符的字符串,或布尔值,或null, 或原语(的数组)数组,......这仍然有效。
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森林海
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);