寻找一种方法来比较和计算数组中所有嵌套对象的键/值对

Question

我有一系列对象，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"},

];

我想对两个键（“from”和“to”）相同的所有唯一实例进行排序和计数。

我可以成功地计算每个案例，只比较一个键（'from' 或 'to'），但我找不到关于如何比较两者的解决方案。

这是我的示例代码：

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);

这是我现在得到的结果（因为代码只比较“来自”键）：

console.log(arr2);

// Array(2)

// 0: {from: "Monica", to: "Rachel", count: 4}

// 1: {from: "Ross", to: "Chandler", count: 2}

// length: 2

这是我想要达到的结果：

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

Answer 1

使用Array.reduce，您可以通过from和to变量对键对输入数组进行分组。对于重复，计数将增加，最后，结果将存储在groupedBy对象的每个键的值上。您只能使用 提取值Object.values。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"},];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);

Answer 2

您可以使用带有分隔符的组合键from和值。toconst    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; }

Answer 3

替代：from通过and组合对数组进行排序to，在此之后，重复项将在排序后的数组中相邻。遍历排序的数组以删除重复项和计数。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"},]; 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);

Answer 4

循环每个对象，获取它的值from并to进行比较。const arr = [&nbsp;{"from": "A", "to": "B"},&nbsp;{"from": "A", "to": "C"},&nbsp;{"from": "A", "to": "A"},];for (var i = 0; i < arr.length; i++) {&nbsp; if (arr[i].from == arr[i].to) {&nbsp; &nbsp; console.log('Matches!');&nbsp; }&nbsp; else {&nbsp; &nbsp; console.log('No match');&nbsp; }}