TA貢獻1852條經驗 獲得超7個贊
您可以通過Array.prototype.reduce()構建一個復雜的對象來遍歷源數組，以跟蹤每個對象的出現、到目前為止最?？吹降膶ο笠约皩嶋H看到最常見對象的次數。
所以，一旦一些數組項超過maxCount它就變成了mostOften。
這樣你就可以只通過所有項目來找到獲勝者：
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)
TA貢獻1860條經驗 獲得超9個贊
這里重要的是為您認為“相同”的對象提供可靠的密鑰。我會建議JSON.stringify由兩個屬性組成的數組：
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是帶有任何外來字符的字符串，或布爾值，或null, 或原語（的數組）數組，......這仍然有效。
TA貢獻2011條經驗 獲得超2個贊
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);
舉報