Question

我正在玩https://beta.golang.org/pkg/net/http/#Server.Shutdown。更好地学习rxjs。

我正在处理的数组如下。我只是试图将ID从数组中删除。

我能够得到工作的答案，但觉得答案可以写得更好。

var movieLists = [
        {
            name: "New Releases",
            videos: [
                {
                    "id": 70111470,
                    "title": "Die Hard",
                    "boxart": "http://cdn-0.nflximg.com/images/2891/DieHard.jpg",
                    "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                    "rating": 4.0,
                    "bookmark": []
                },
                {
                    "id": 654356453,
                    "title": "Bad Boys",
                    "boxart": "http://cdn-0.nflximg.com/images/2891/BadBoys.jpg",
                    "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                    "rating": 5.0,
                    "bookmark": [{ id: 432534, time: 65876586 }]
                }
            ]
        },
        {
            name: "Dramas",
            videos: [
                {
                    "id": 65432445,
                    "title": "The Chamber",
                    "boxart": "http://cdn-0.nflximg.com/images/2891/TheChamber.jpg",
                    "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                    "rating": 4.0,
                    "bookmark": []
                },
                {
                    "id": 675465,
                    "title": "Fracture",
                    "boxart": "http://cdn-0.nflximg.com/images/2891/Fracture.jpg",
                    "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                    "rating": 5.0,
                    "bookmark": [{ id: 432534, time: 65876586 }]
                }
            ]
        }
    ];

以下是我提出的答案

          movieLists
                  .map( movieList => movieList.videos
                  .map(video => video.id)  )   // don't like this part
                  .concatAll()

我基本上将地图嵌套在另一个地图中，然后调用concat all。

是否有可能重构第二个地图胖箭头，以便它可以位于第一个地图之外？

.map()
.map()

Answer 1

只需更改运营商的顺序即可。

Observable.from(movieLists)
  .map(movie => movie.videos)
  .concatAll()
  .map(movie => movie.id);

Answer 2

映射到电影后可以使用concatAll()：

var movieLists = [
    {
        name: "New Releases",
        videos: [
            {
                "id": 70111470,
                "title": "Die Hard",
                "boxart": "http://cdn-0.nflximg.com/images/2891/DieHard.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 4.0,
                "bookmark": []
            },
            {
                "id": 654356453,
                "title": "Bad Boys",
                "boxart": "http://cdn-0.nflximg.com/images/2891/BadBoys.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 5.0,
                "bookmark": [{ id: 432534, time: 65876586 }]
            }
        ]
    },
    {
        name: "Dramas",
        videos: [
            {
                "id": 65432445,
                "title": "The Chamber",
                "boxart": "http://cdn-0.nflximg.com/images/2891/TheChamber.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 4.0,
                "bookmark": []
            },
            {
                "id": 675465,
                "title": "Fracture",
                "boxart": "http://cdn-0.nflximg.com/images/2891/Fracture.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 5.0,
                "bookmark": [{ id: 432534, time: 65876586 }]
            }
        ]
    }
];

Rx.Observable.from(movieLists)
    .map(movieList => movieList.videos)
    .concatAll()
    .map(video => video.id)
    .do(console.log)
    .subscribe();

<script src="https://unpkg.com/@reactivex/rxjs@5.0.3/dist/global/Rx.js"></script>

<强>但是：基本上你想要展平数组并提取id - 这里根本不需要RxJS，使用原生方法也应该更快：

var movieLists = [
    {
        name: "New Releases",
        videos: [
            {
                "id": 70111470,
                "title": "Die Hard",
                "boxart": "http://cdn-0.nflximg.com/images/2891/DieHard.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 4.0,
                "bookmark": []
            },
            {
                "id": 654356453,
                "title": "Bad Boys",
                "boxart": "http://cdn-0.nflximg.com/images/2891/BadBoys.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 5.0,
                "bookmark": [{ id: 432534, time: 65876586 }]
            }
        ]
    },
    {
        name: "Dramas",
        videos: [
            {
                "id": 65432445,
                "title": "The Chamber",
                "boxart": "http://cdn-0.nflximg.com/images/2891/TheChamber.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 4.0,
                "bookmark": []
            },
            {
                "id": 675465,
                "title": "Fracture",
                "boxart": "http://cdn-0.nflximg.com/images/2891/Fracture.jpg",
                "uri": "http://api.netflix.com/catalog/titles/movies/70111470",
                "rating": 5.0,
                "bookmark": [{ id: 432534, time: 65876586 }]
            }
        ]
    }
];

// Step 1: get all the videos as an multi-dimensional array
const videos = movieLists.map(movie => movie.videos);
// Step 2: flatten the array
const flatVideos = [].concat.apply([], videos);
// Step 3: Map to ids
const videoIds = flatVideos.map(video => video.id);
console.log(videoIds);

我已经用3个步骤编写了这个，但这些也可以连接起来。

Answer 3

您可以在一个操作员中执行此操作：

Observable.from(movieLists)
  .concatMap(
    movie => movie.videos, 
    (_, video) => video.id
  )

refactor fat arrow嵌套rxjs流

3 个答案: