我有一系列對象，例如：const arr1 = [ {"from":"Monica","to":"Rachel"}, {"from":"Monica","to":"Rachel"}, {"from":"Monica","to":"Chandler"}, {"from":"Monica","to":"Chandler"}, {"from":"Ross","to":"Chandler"}, {"from":"Ross","to":"Monica"},];我想對兩個(gè)鍵（“from”和“to”）相同的所有唯一實(shí)例進(jìn)行排序和計(jì)數(shù)。我可以成功地計(jì)算每個(gè)案例，只比較一個(gè)鍵（'from' 或 'to'），但我找不到關(guān)于如何比較兩者的解決方案。這是我的示例代碼：let arr2 = Object.values(arr1.reduce((c, { from, to }) => { c[from] = c[from] || { from, to, count: 0 }; c[from].count++; return c;}, {}));console.log(arr2);這是我現(xiàn)在得到的結(jié)果（因?yàn)榇a只比較“來自”鍵）：console.log(arr2);// Array(2)// 0: {from: "Monica", to: "Rachel", count: 4}// 1: {from: "Ross", to: "Chandler", count: 2}// length: 2這是我想要達(dá)到的結(jié)果：console.log(arr2);// Array(2)// 0: {from: "Monica", to: "Rachel", count: 2}// 1: {from: "Monica", to: "Chandler", count: 2}// 2: {from: "Ross", to: "Chandler", count: 1}// 3: {from: "Ross", to: "Monica", count: 1}// length: 4