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如果在rxjs中没有连续重复3次,如何跳过0值

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Reza · 技术社区 · 4 年前

1 > 4 > 0 > 3 > 1 > 0 > 0 > 0 > 1 > 2 > 0 > 0 > 0 > 0 > 0 > 0

我想要 0 仅当它连续重复3次或更多次时才会发出,因此预期结果如下

1 > 4 > 3 > 1 > 0 > 1 > 2 > 0

在rxjs中如何实现这一点?

0 回复 | 直到 4 年前

Rafi Henig 4 年前

您可能希望按照以下示例实现它:

const { from } = rxjs;
const { scan, filter, map} = rxjs.operators;


const input = [
    1, 4, 0, 3, 1, 0, 0, 0, 
    1, 2, 0, 0, 0, 0, 0, 0,
    2, 0, 2, 4, 0, 0, 3, 9,
    4, 3, 0, 0, 0, 0, 0, 0,
];

from(input).pipe(
    scan(({ counter }, current) => {

        if (current === 0) return { current, counter: ++counter, emit: counter == 3 }
        else return { current, counter: 0, emit: true };

    }, { emit: false, counter: 0, current: undefined }),

    filter(x => x.emit), 

    map(x => x.current) 
)
.subscribe(console.log)

<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.6.2/rxjs.umd.min.js"></script>

Guilhermevrs 4 年前

实际上,您可以使用scan/filter/map。

const {
  Observable,
  from
} = rxjs;

function groupRepeatedNumbers(opts = {
  minLimit: 3,
  desiredNumber: 0
}) {
  return function(source) {
    let numberCounter = 0;
    return new Observable(subscriber => {
      source.subscribe({
        next(value) {
          if (value !== opts.desiredNumber) {
            numberCounter = 0;
            subscriber.next(value);
          } else if (++numberCounter === opts.minLimit) {
            subscriber.next(value);
          }
        },
        error(error) {
          subscriber.error(error);
        },
        complete() {
          subscriber.complete();
        }
      })
    });
  }
}

const input = [
  1, 4, 0, 3, 1, 0, 0, 0,
  1, 2, 0, 0, 0, 0, 0, 0,
  2, 0, 2, 4, 0, 0, 3, 9,
  4, 3, 0, 0, 0, 0, 0, 0,
];

from(input).pipe(
    groupRepeatedNumbers()
  )
  .subscribe(console.log)

<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.6.2/rxjs.umd.min.js"></script>

您可以看到,您甚至可以将操作员配置为按需操作,如:

groupRepeatedNumbers({
  minLimit: 5,
  desiredNumber: 3
}) // Will group the number 3, and only emit when it gets 5 or more occasions

frido 4 年前

defer 跟踪自定义运算符中的变量(如计数器),同时为每个订户提供自己的新变量(如果您有多个订户,这一点很重要)。

import { defer, MonoTypeOperatorFunction, Observable } from 'rxjs'; 
import { filter } from 'rxjs/operators';

function sequentialLimit<T>(
  condition: (value: T) => boolean, 
  limit: number
): MonoTypeOperatorFunction<T> {
  return (source: Observable<T>) => defer(() => {
    let counter = 0;
    return source.pipe(
      filter(v => {
        condition(v) ? counter++ : counter = 0;
        return counter === 0 || counter === limit;
      })
    );
  });
}

用法:

const input = [
  1, 4, 0, 3, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0,
  2, 0, 2, 4, 0, 0, 3, 9, 4, 3, 0, 0, 0, 0, 0, 0,
];

from(input).pipe(
  sequentialLimit(v => v === 0, 3)
).subscribe(console.log)

https://stackblitz.com/edit/rxjs-zpwnxq?file=index.ts