正确过滤具有多个条件的对象数组和一个数组

我正在创建一个Todo App，作为学习PostgreSQL和使用JavaScript进行功能编程的练习。这个想法是，可以有多个容器，每个容器都有单独的过滤器设置，每个容器从Todos的主列表中读取数据，然后每个容器根据其过滤器设置显示其结果。这实际上是在React-Redux和Nodejs中完成的，但是我在下面的示例中做了更好的说明。

数据：

// Example todos
let a = {
  added: "2021-02-01T05:00:00.000Z",
  completed: null,
  description: "This is just a test.",
  name: "testing",
  tags: [],
  todo_id: 3,
}
let b = {
  added: "2021-02-01T05:00:00.000Z",
  completed: null,
  description: "This is just a test.",
  name: "testing",
  tags: ["one", "two"],
  todo_id: 4
}
let c = {
  added: "2021-02-01T05:00:00.000Z",
  completed: null,
  description: "This is just a test.",
  name: "tesin",
  tags: ["one"],
  todo_id: 5
}
let d = {
  added: "2021-02-01T05:00:00.000Z",
  completed: '2021-03-01T05:00:00.000Z',
  description: "This is just a test.",
  name: "testing",
  tags: ["two", "testing"],
  todo_id: 6
}

// Example filter_group 
// This is just to show what the initial state looks like in my React app
const filter_groups = {
  "main": {
    byTag: 'two',
    byName: '',
    byCompleted: true
  }
}

// Object.keys(filter_groups).map() passes the filter settings down as props
// Just stuck it in a variable for the example
const filter = filter_groups['main']

// Putting example Todos in array
const todo_list = [a,b,c,d]

当前的解决方案非常简单，丑陋且效率低下。它一次应用每个过滤器：

// ------- CURRENT SOLUTION ------- 

// if user is trying to filter byCompleted, then filter todo_list
// else do nothing todo_list and pass it to the next filter 
// (similar thing applies down the chain)
const byCompleted = filter.byCompleted === true ?
  todo_list.filter ( (t) => {
    return t.completed !== null
  })
  :
  todo_list


//IF user is trying to filter byTag...
const byTag = filter.byTag !== '' ?
  // THEN filter the byCompleted array above
  byCompleted.filter( (t) => {
    // IF there are tags on the Todo...
    let matches = t.tags ?
      // THEN filter for matches
      t.tags.filter( (tag) => {
        return tag === filter.byTag
      }) 
      : 
      // ELSE make this an empty array
      []

    // return any Todos with match
    return matches.length > 0
  })
  :
  byCompleted

// IF user is trying to filter byName...
const byName = filter.byName !== '' ?
  // THEN filter byTag Todos
  byTag.filter( (t) => {
    return t.name === P.filter.byName
  })
  :
  byTag

console.log(byName);

我真的很想知道是否有一种方法可以一次全部应用整个过滤器。在示例中，您将看到过滤器设置为byTag: 'two', byComplete: true使控制台仅输出一个结果的过滤器。我过去尝试通过一次过滤器传递使事情生效的尝试将返回带有two标签的所有内容。

我尝试查看了文档，发现过滤器功能还有更多功能，但是并没有点击，并且找不到适合我的用例的示例。

编辑：这是codeandbox上的全部内容：https ://codesandbox.io/s/recursing-almeida-cdz65 ? file =/ src/index.js