te')); return $arr; } /* 遍历用户所有主题 * @param $uid 用户ID * @param int $page 页数 * @param int $pagesize 每页记录条数 * @param bool $desc 排序方式 TRUE降序 FALSE升序 * @param string $key 返回的数组用那一列的值作为 key * @param array $col 查询哪些列 */ function thread_tid_find_by_uid($uid, $page = 1, $pagesize = 1000, $desc = TRUE, $key = 'tid', $col = array()) { if (empty($uid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('uid' => $uid), array('tid' => $orderby), $page, $pagesize, $key, $col); return $arr; } // 遍历栏目下tid 支持数组 $fid = array(1,2,3) function thread_tid_find_by_fid($fid, $page = 1, $pagesize = 1000, $desc = TRUE) { if (empty($fid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('fid' => $fid), array('tid' => $orderby), $page, $pagesize, 'tid', array('tid', 'verify_date')); return $arr; } function thread_tid_delete($tid) { if (empty($tid)) return FALSE; $r = thread_tid__delete(array('tid' => $tid)); return $r; } function thread_tid_count() { $n = thread_tid__count(); return $n; } // 统计用户主题数 大数量下严谨使用非主键统计 function thread_uid_count($uid) { $n = thread_tid__count(array('uid' => $uid)); return $n; } // 统计栏目主题数 大数量下严谨使用非主键统计 function thread_fid_count($fid) { $n = thread_tid__count(array('fid' => $fid)); return $n; } ?>javascript - How to emit every n-th value? - Stack Overflow
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javascript - How to emit every n-th value? - Stack Overflow

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I'm using mousemove event to create an observable.

Observable.fromEvent(document, 'mousemove')

I need to emit every 10-th event. What do I do?

I'm using mousemove event to create an observable.

Observable.fromEvent(document, 'mousemove')

I need to emit every 10-th event. What do I do?

Share Improve this question edited Mar 20, 2017 at 13:30 martin 96.9k26 gold badges203 silver badges235 bronze badges asked Mar 20, 2017 at 12:59 manidosmanidos 3,4647 gold badges39 silver badges74 bronze badges
Add a ment  | 

2 Answers 2

Reset to default 10

I can think of four different ways to do it:

bufferCount()

Observable.range(1, 55)
  .bufferCount(10)
  .map(arr => arr[arr.length - 1])
  .subscribe(val => console.log(val));

windowCount()

Observable.range(1, 55)
  .windowCount(10)
  .switchMap(window => window.takeLast(1))
  .subscribe(val => console.log(val));

debounce()

let source = Observable.range(1, 55).publish();

source
  .debounce(val => debounceNotifier)
  .subscribe(val => console.log(val));

let debounceNotifier = source
  .bufferCount(10)
  .publish();
debounceNotifier.connect();

source.connect();

scan()

Observable.range(1, 55)
  .scan((acc, val) => {
    if (acc.length === 10) {
      acc = [];
    }
    acc.push(val);
    return acc;
  }, [])
  .filter(acc => acc.length === 10)
  .map(acc => acc[acc.length - 1])
  .subscribe(val => console.log(val));

However, when using scan() it'll will discard the last value 55.

See demo for all of them: https://jsbin./yagayot/14/edit?js,console

Here's an even simpler and considerably faster approach, which I tested with RxJS 6:

range(1, 10000000)
  .pipe(
    filter(function(value, index) { 
      return index % 10 === 0; 
    }),
  );

This code is twice as fast as the bufferCount and windowCount approaches in the other answer: https://jsperf./observable-nth/1

This is likely because the filter operator uses a simple counter, instead of having to keep a buffer of the last n elements. I assume this is even faster when n is larger or the elements themselves are bigger. With RxJS 6, you can also easily make this into your own custom operator:

const takeEveryNth = (n: number) => filter((value, index) => index % n === 0);
// usage: rxjs.range(1, 10000000).pipe(takeEveryNth(10));

It is also the code used in the official docs explaining how to create custom operators: https://github./ReactiveX/rxjs/blob/6.2.2/doc/pipeable-operators.md

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