我想创建一个
的流- 在单独的流上分配值并处理每个部分
- 每个流将转换数据,我无法控制应用转换
- (重新 - (加入部分值及其相应的计数器部分
我要这样做的原因是要确保值的完整性。或至少在其中的某些部分。
因为每个流可以在加入流时都不会出现一些异步操作。使用某种concat()
也无效,因为它会阻止所有传入值。处理应并行进行。
说明:
o
|
| [{a1,b1}, {a2,b2}, ...]
|
+
/
{a<x>} / {b<x>}
/
| |
| + async(b<x>) -> b'<x>
| |
/
/
/
/
+ join(a<x>, b'<x>)
|
| [{a1,b'1}, {a2,b'2}, ...]
|
(subscribe)
我知道这可以通过result selector
函数完成。例如
input$.mergeMap(
({a, b}) => Rx.Observable.of(b).let(async),
({a}, processedB) => ({a, b:processedB})
);
但是,(a(这将导致async
始终为每个值设置/拆卸。我希望部分流只能初始化一次。另外,(b(这仅适用于一个异步流。
我也尝试使用window*
,但无法弄清楚如何再次重新加入值。还尝试使用goupBy
没有运气。
编辑:
这是我目前的尝试。它具有上述问题(a(。Init...
和Completed...
只能记录一次。
const doSomethignAsync = data$ => {
console.log('Init...') // Should happen once.
return data$
.mergeMap(val => Rx.Observable.of(val.data).delay(val.delay))
.finally(() => console.log('Completed...')); // Should never happen
};
const input$ = new Rx.Subject();
const out$ = input$
.mergeMap(
({ a, b }) => Rx.Observable.of(b).let(doSomethignAsync),
({ a }, asyncResult ) => ({ a, b:asyncResult })
)
.subscribe(({a, b}) => {
if (a === b) {
console.log(`Re-joined [${a}, ${b}] correctly.`);
} else {
console.log(`Joined [${a}, ${b}]...`); // Should never happen
}
});
input$.next({ a: 1, b: { data: 1, delay: 2000 } });
input$.next({ a: 2, b: { data: 2, delay: 1000 } });
input$.next({ a: 3, b: { data: 3, delay: 3000 } });
input$.next({ a: 4, b: { data: 4, delay: 0 } });
<script src="https://unpkg.com/rxjs/bundles/Rx.min.js"></script>
这是一个相当复杂的问题,确切的工作方式将取决于未包括的用例非常具体的细节。
也就是说,这是一种可能的假设的一种可能的方法。它是有点通用的,例如与let
一起使用的自定义操作员。
(旁注:我将其命名为"整理",但这是一个不好且高度误导的名称,但是没有时间来命名事物...(
const collate = (...segments) => source$ =>
source$
.mergeMap((obj, index) => {
return segments.map(({ key, work }) => {
const input = obj[key];
const output$ = work(input);
return Observable.from(output$).map(output => ({
index,
result: { [key]: output }
}))
})
})
.mergeAll()
.groupBy(
obj => obj.index,
obj => obj.result,
group$ => group$.skip(segments.length - 1)
)
.mergeMap(group$ =>
group$.reduce(
(obj, result) => Object.assign(obj, result),
{}
)
);
这是一个用法示例:
const result$ = input$.let(
collate({
key: 'a',
work: a => {
// do stuff with "a"
return Observable.of(a).map(d => d + '-processed-A');
}
}, {
key: 'b',
work: b => {
// do stuff with "b"
return Observable.of(b).map(d => d + '-processed-B');
}
})
);
给定的输入{ a: '1', b: '1 }
它将输出{ a: '1-processed-A', b: '1-processed-B' }
等,依此类推,在尽可能多地进行同时进行的同时进行分组 - 唯一的缓冲是将所有段匹配在一起的特定输入。
这是一个正在运行的演示https://jsbin.com/yuruvar/edit?js ,, output
故障
可能会有更清晰/更简单的方法来执行此操作,尤其是如果您可以进行硬编码而不是使它们成为通用的情况。但是让我们分解我做了什么。
const collate = (...segments) => source$ =>
source$
// for every input obj we use the index as an ID
// (which is provided by Rx as autoincrementing)
.mergeMap((obj, index) => {
// segments is the configuration of how we should
// chunk our data into concurrent processing channels.
// So we're returning an array, which mergeMap will consume
// as if it were an Observable, or we could have used
// Observable.from(arr) to be even more clear
return segments.map(({ key, work }) => {
const input = obj[key];
// the `work` function is expected to return
// something Observable-like
const output$ = work(input);
return Observable.from(output$).map(output => ({
// Placing the index we closed over lets us later
// stitch each segment back to together
index,
result: { [key]: output }
}))
})
})
// I had returned Array<Observable> in mergeMap
// so we need to flatten one more level. This is
// rather confusing...prolly clearer ways but #YOLO
.mergeAll()
// now we have a stream of all results for each segment
// in no guaranteed order so we need to group them together
.groupBy(
obj => obj.index,
obj => obj.result,
// this is tough to explain. this is used as a notifier
// to say when to complete() the group$, we want complete() it
// after we've received every segment for that group, so in the
// notifier we skip all except the last one we expect
// but remember this doesn't skip the elements downstream!
// only as part of the durationSelector notifier
group$ => group$.skip(segments.length - 1)
)
.mergeMap(group$ =>
// merge every segment object that comes back into one object
// so it has the same shape as it came in, but which the results
group$.reduce(
(obj, result) => Object.assign(obj, result),
{}
)
);
我不考虑或不必担心错误处理/传播可能会如何工作,因为这高度取决于您的用例。如果您无法控制每个细分市场的处理,则建议包括某种超时和.take(1)
,否则您可能会泄漏订阅。