如何比较存在于两种不同 React 状态中的数组对象

Question

我想比较两种不同 React 状态下的对象数组。我的状态值如下所示：

review1 = [{name: John, Title: Manager},{name: Peter, Title: Engineer}, {name: Serena, Title: Supervisor}]
review2 = [{personName: John, ismanager: yes}, {name: Peter, ismanager: no}, {personName: Serena, ismanager: yes},{personName: John, ismanager:yes}]

现在我想统计一下经理的人数。如果倍数，则计数应增加。在这种情况下，约翰是两个位置的经理，那么约翰的计数是2，彼得是0和塞雷娜1

我正在尝试下面的功能，但它没有按预期工作。

let filter = this.state.review1.map(elem=>elem.personName)
count = this.state.review2.map(data=> data.name).reduce((n,x) =>n +(x === filter.personName && data.ismanager === "yes"),null)
<div> John: {ismanger.count}, Serena: {ismanager.count}, Peter: {ismanager.count} </div> 

Answer 1

您可以直接计算有条件的经理。

var review2 = [{ personName: 'John', ismanager: 'yes' }, { personName: 'Peter', ismanager: 'no' }, { personName: 'Serena', ismanager: 'yes' }, { personName: 'John', ismanager: 'yes' }],
    count = Array.from(
        review2.reduce((m, { personName, ismanager }) => m.set(personName, (m.get(personName) || 0) + (ismanager === 'yes')), new Map),
        ([name, count]) => ({ name, count })
    );
console.log(count);

Answer 2

我想

我们只需要在数组review2做吧？试试这个。

演示位置：播放代码

review2 = [{personName: 'John', ismanager: 'yes'}, {name: 'Peter', ismanager: 'no'}, {personName: 'Serena', ismanager: 'yes'},{personName: 'John', ismanager:'yes'}]
const countMap = review2.reduce((accum, person) => {
  const isPersonName = !!person.personName;
  const name = isPersonName ? person.personName : person.name;
  if (accum[name] && person.ismanager === 'yes') {
    accum[name]++;
  } else if (person.ismanager === 'yes') {
    accum[name] = 1;
  }
  return accum;
}, {})
console.log(countMap);