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尋找一種方法來比較和計算數組中所有嵌套對象的鍵/值對

JavaScript

尚方寶劍之說 2023-05-19 15:06:56

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"},];我想對兩個鍵（“from”和“to”）相同的所有唯一實例進行排序和計數。我可以成功地計算每個案例，只比較一個鍵（'from' 或 'to'），但我找不到關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);這是我現在得到的結果（因為代碼只比較“來自”鍵）：console.log(arr2);// Array(2)// 0: {from: "Monica", to: "Rachel", count: 4}// 1: {from: "Ross", to: "Chandler", count: 2}// length: 2這是我想要達到的結果：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

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4 回答

慕的地10843

TA貢獻1785條經驗獲得超8個贊

使用Array.reduce，您可以通過from和to變量對鍵對輸入數組進行分組。
對于重復，計數將增加，最后，結果將存儲在groupedBy對象的每個鍵的值上。
您只能使用提取值Object.values。

const arr1 = [

{"from":"Monica","to":"Rachel"},

{"from":"Monica","to":"Chandler"},

{"from":"Ross","to":"Chandler"},

{"from":"Ross","to":"Monica"},

];

const groupedBy = arr1.reduce((acc, cur) => {

const key = `${cur.from}_${cur.to}`;

acc[key] ? acc[key].count ++ : acc[key] = {

...cur,

count: 1

};

return acc;

}, {});

const result = Object.values(groupedBy);

console.log(result);

反對回復 2023-05-19

慕容3067478

TA貢獻1773條經驗獲得超3個贊

您可以使用帶有分隔符的組合鍵from和值。to

const

array = [{ from: "Monica", to: "Rachel" }, { from: "Monica", to: "Rachel" }, { from: "Monica", to: "Chandler" }, { from: "Monica", to: "Chandler" }, { from: "Ross", to: "Chandler" }, { from: "Ross", to: "Monica" }],

result = Object.values(array.reduce((r, { from, to }) => {

const key = [from, to].join('|');

r[key] ??= { from, to, count: 0 };

r[key].count++;

return r;

}, {}));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

反對回復 2023-05-19

繁星點點滴滴

TA貢獻1803條經驗獲得超3個贊

替代：

from通過and組合對數組進行排序to，在此之后，重復項將在排序后的數組中相鄰。
遍歷排序的數組以刪除重復項和計數。

const arr1 = [

{"from":"Monica","to":"Rachel"},

{"from":"Monica","to":"Chandler"},

{"from":"Ross","to":"Chandler"},

{"from":"Ross","to":"Monica"},

];

var arr2 = arr1.sort((obj1, obj2) => (obj1.from + obj1.to).localeCompare(obj2.from + obj2.to));

for(var i = arr2.length - 1; i >= 0; i--){

if(i > 0 && arr2[i].from + arr2[i].to == arr2[i-1].from + arr2[i-1].to){

arr2[i-1].count = arr2[i].count ? arr2[i].count + 1 : 2;

arr2.splice(i, 1);

}else if(!arr2[i].count){

arr2[i].count = 1;

}

console.log(arr2);

反對回復 2023-05-19

侃侃無極

TA貢獻2051條經驗獲得超10個贊

循環每個對象，獲取它的值from并to進行比較。

const arr = [

{"from": "A", "to": "B"},

{"from": "A", "to": "C"},

{"from": "A", "to": "A"},

];

for (var i = 0; i < arr.length; i++) {

if (arr[i].from == arr[i].to) {

console.log('Matches!');

}

else {

console.log('No match');

}

反對回復 2023-05-19

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尋找一種方法來比較和計算數組中所有嵌套對象的鍵/值對

尋找一種方法來比較和計算數組中所有嵌套對象的鍵/值對

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