Javascript RxJS中的分支和(重新)连接值对
我想创建一个Javascript RxJS中的分支和(重新)连接值对,javascript,rxjs,rxjs5,Javascript,Rxjs,Rxjs5,我想创建一个 拆分值并在单独的流上处理每个部分 每个流都将转换数据,我无法控制应用的转换 (re-)将部分值与其对应的计数器部分连接起来 我这样做的原因是为了确保值的完整性。或者至少在某种程度上 因为每个流可能有一些异步操作,所以它们在加入流时不会按顺序出现。使用某种concat()也不起作用,因为它会阻止所有传入值。处理应该并行进行 举例说明: o |
concat()
也不起作用,因为它会阻止所有传入值。处理应该并行进行
举例说明:
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)
但是,(a)这将导致始终为每个值设置/删除async
。我希望部分流只初始化一次。此外,(b)这只适用于一个异步流
我还尝试使用窗口*
,但无法想出如何再次重新连接值。还试图使用goupBy
,但运气不佳
编辑: 这是我目前的尝试。它有上述问题(a)<代码>初始化…和
完成…
只应记录一次
const doSomethignAsync=data$=>{
console.log('Init…')//应该发生一次。
返回数据$
.mergeMap(val=>Rx.Observable.of(val.data).delay(val.delay))
.finally(()=>console.log('Completed…');//不应该发生
};
const input$=新的Rx.Subject();
const out$=输入$
.合并地图(
({a,b})=>Rx.Observable.of(b).let(doSomethignAsync),
({a},asyncResult)=>({a,b:asyncResult})
)
.subscribe({a,b})=>{
如果(a==b){
log(`rejoined[${a},${b}]正确。`);
}否则{
log(`Joined[${a},${b}]…`);//不应该发生
}
});
输入$.next({a:1,b:{data:1,delay:2000});
输入$.next({a:2,b:{data:2,delay:1000});
输入$.next({a:3,b:{data:3,delay:3000});
输入$.next({a:4,b:{data:4,delay:0})代码>
这是一个需要解决的相当复杂的问题,具体如何解决将取决于未包含的用例的非常具体的细节
这就是说,这里有一种可能的方法可以做出一系列假设。它有点通用,就像要与let
一起使用的自定义运算符
(旁注:我将其命名为“collate”,但这是一个糟糕且误导性极强的名称,但没有时间命名……)
下面是一个用法示例:
const result$ = input$.let(
collate({
key: 'a',
work: a => {
// do stuff with "a"
return Observable.of(a).map(d => d + '-processed-A');
}
}, {
key: 'b',
work: b => {
// do stuff with "b"
return Observable.of(b).map(d => d + '-processed-B');
}
})
);
给定输入{a:'1',b:'1}
它将输出{a:'1-processed-a',b:'1-processed-b'}
等等,在尽可能多地并发执行时正确分组--它所做的唯一缓冲是将特定输入的所有段匹配在一起
这是一个正在运行的演示
崩溃
可能有更清晰/更简单的方法来实现这一点,特别是如果您可以硬编码一些东西,而不是使它们通用。但让我们来分析一下我所做的
const collate = (...segments) => source$ =>
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$.reduce(
(obj, result) => Object.assign(obj, result),
{}
)
);
constcollate=(…段)=>source$=>
来源$
//对于每个输入对象,我们使用索引作为ID
//(由Rx提供为自动递增)
.mergeMap((对象,索引)=>{
//分段是我们应该如何进行的配置
//将数据分块到并发处理通道中。
//所以我们返回一个数组,mergeMap将使用它
//好像它是一个可观察的,或者我们可以使用
//可观察。从(arr)到更清晰
返回segments.map({key,work})=>{
常量输入=对象[键];
//“work”函数将返回
//可以观察到的东西,如
常量输出$=工作(输入);
返回可观察的.from(输出$).map(输出=>({
//把我们结束的指数放在后面,我们就可以
//将每个部分重新缝合在一起
指数
结果:{[键]:输出}
}))
})
})
//我已在mergeMap中返回数组
//所以我们需要再展平一层。这是
//相当令人困惑……可能更清晰,但#约洛
.mergeAll()
//现在我们有了每个片段的所有结果流
//没有保证的顺序,所以我们需要将它们组合在一起
.群比(
obj=>obj.index,
obj=>obj.result,
//这很难解释。它被用作通知程序
//要说何时完成组$,我们需要完成它
//在我们收到该组的每个片段之后
//我们跳过所有的通知,除了我们期望的最后一个
//但请记住,这不会跳过下游的元素!
//仅作为durationSelector通知程序的一部分
组$=>组$.skip(segments.length-1)
)
.mergeMap(组$=>
//合并返回到一个对象中的每个分段对象
//所以它的形状和它进来时一样,但结果是什么
组$.reduce(
(obj,result)=>Object.assign(obj,result),
{}
)
);
我没有考虑或担心错误处理/传播可能如何工作,因为这在很大程度上取决于您的用例。如果您无法控制每个段的处理,那么还包括某种类型的超时和。建议采取(1)
,否则您可能会泄露订阅。我想您可以使用pairwise()
然后merge()
这两条流,但我不确定您想要完成什么,因此我可以给出更准确的建议。我添加了更多信息。你不明白哪一部分?不清楚你到底需要完成什么。还不清楚你所说的平行是什么意思——它在不同的环境中对不同的人意味着不同的事情。e、 g.当您仍在等待匹配的“a”或“b”时,缓冲策略是什么?你只是同时做这对,而不是每对的倍数吗
const result$ = input$.let(
collate({
key: 'a',
work: a => {
// do stuff with "a"
return Observable.of(a).map(d => d + '-processed-A');
}
}, {
key: 'b',
work: b => {
// do stuff with "b"
return Observable.of(b).map(d => d + '-processed-B');
}
})
);
const collate = (...segments) => source$ =>
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),
{}
)
);