[英]RxJs zip operator equivalent in xstream?
您好,我想弄清楚xstream中是否有与RxJs运算符zip等效的东西,或者至少是一种获得相同行为的方式。 如果有人需要澄清区别,下面的大理石图将显示。
zip in rxjs
|---1---2---3-----------5->
|-a------b------c---d----->
"zip"
|-1a----2b------3c-----5d->
whereas 'combineLatest' aka 'combine' in xstream does
|---1---2----------4---5->
|----a---b---c---d------->
"combine"
|-1a----2a-2b-2c-2d-4d-5d>
感谢任何帮助,因为我是流编程的新手。 先感谢您!
对于xstream,我还需要一个zip运算符。 因此,我从现有的运营商创建了自己的公司。 压缩需要任意数量的流。
function zip(...streams) {
// Wrap the events on each stream with a label
// so that we can seperate them into buckets later.
const streamsLabeled = streams
.map((stream$, idx) => stream$.map(event => ({label: idx + 1, event: event})));
return (event$) => {
// Wrap the events on each stream with a label
// so that we can seperate them into buckets later.
const eventLabeled$ = event$.map(event => ({label: 0, event: event}));
const labeledStreams = [eventLabeled$, ...streamsLabeled];
// Create the buckets used to store stream events
const buckets = labeledStreams.map((stream, idx) => idx)
.reduce((buckets, label) => ({...buckets, [label]: []}), {});
// Initial value for the fold operation
const accumulator = {buckets, tuple: []};
// Merge all the streams together and accumulate them
return xs.merge(...labeledStreams).fold((acc, event) => {
// Buffer the events into seperate buckets
acc.buckets[event.label].push(event);
// Does the first value of all the buckets have something in it?
// If so, then there is a complete tuple.
const tupleComplete = Object.keys(acc.buckets)
.map(key => acc.buckets[key][0])
.reduce((hadValue, value) => value !== undefined
? true && hadValue
: false && hadValue,
true);
// Save completed tuple and remove it from the buckets
if (tupleComplete) {
acc.tuple = [...Object.keys(acc.buckets).map(key => acc.buckets[key][0].event)];
Object.keys(acc.buckets).map(key => acc.buckets[key].shift());
} else {
// Clear tuple since all columns weren't filled
acc.tuple = [];
}
return {...acc};
}, accumulator)
// Only emit when we have a complete tuple
.filter(buffer => buffer.tuple.length !== 0)
// Just return the complete tuple
.map(buffer => buffer.tuple);
};
}
可以与compose一起使用。
foo$.compose(zip(bar$)).map(([foo, bar]) => doSomething(foo, bar))
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