如何用单个reduce()优化代码，基于日志属性筛选正确开关门人员？

我家有三个人：['John', 'Jane', 'Jack']，我们记录了人员开关门的日志数据：

const logs = [
  { name: "John", status: "opened" },
  { name: "Jane", status: "opened" },
  { name: "Jack", status: "opened" },
  { name: "Jane", status: "closed" },
  { name: "Jack", status: "closed" },
];

从日志能看出，只有['Jane', 'Jack']完成了完整的开门+关门操作，John只开了门没关门。我现在用三步代码能识别出John：

let openers = logs.reduce((acc, log) => {
  if (log.status === "opened") {
    acc.push(log.name);
  }
  return acc;
}, []);

console.log(openers);

let closers = logs.reduce((acc, log) => {
  if (log.status === "closed") {
    acc.push(log.name);
  }
  return acc;
}, []);

console.log(closers);

let result = [];
closers.forEach((closer) => {
  if (openers.includes(closer)) {
    result.push(closer);
  }
});

console.log(result);

现在想把这三步逻辑用单个reduce()一次性完成，求代码优化方案。

优化方案1：状态标记统计法 #

通过reduce维护一个记录每个人操作状态的对象，最后筛选出只开门没关门的人：

const logs = [
  { name: "John", status: "opened" },
  { name: "Jane", status: "opened" },
  { name: "Jack", status: "opened" },
  { name: "Jane", status: "closed" },
  { name: "Jack", status: "closed" },
];

const unclosedPersons = Object.entries(
  logs.reduce((acc, { name, status }) => {
    // 初始化未记录人员的操作状态
    if (!acc[name]) {
      acc[name] = { opened: false, closed: false };
    }
    // 更新对应操作的状态标记
    acc[name][status] = true;
    return acc;
  }, {})
)
// 筛选出仅完成开门、未完成关门的人
.filter(([_, { opened, closed }]) => opened && !closed)
.map(([name]) => name);

console.log(unclosedPersons); // 输出: ["John"]

优化方案2：计数简化法 #

通过计数记录操作次数：开门加1、关门减1，最终计数为1的就是只开门没关门的人：

const logs = [
  { name: "John", status: "opened" },
  { name: "Jane", status: "opened" },
  { name: "Jack", status: "opened" },
  { name: "Jane", status: "closed" },
  { name: "Jack", status: "closed" },
];

const unclosedPersons = Object.entries(
  logs.reduce((acc, { name, status }) => {
    // 开门操作+1，关门操作-1，初始值默认为0
    acc[name] = (acc[name] || 0) + (status === "opened" ? 1 : -1);
    return acc;
  }, {})
)
// 计数为1说明仅存在开门操作，无对应关门操作
.filter(([_, count]) => count === 1)
.map(([name]) => name);

console.log(unclosedPersons); // 输出: ["John"]

这两种方案都只遍历日志一次，比原代码的三次遍历（两次reduce+一次forEach）效率更高，逻辑也更紧凑。

内容的提问来源于stack exchange，提问作者code-8