RxJava:如何使用我自己的边界函数对任意大量的项进行分组

Woj*_*ela 5 java rx-java

我有一个可以发出字符串的observable,我想用第一个字符对它们进行分组.这样做很容易groupBy:

Observable<String> rows = Observable.just("aa", "ab", "ac", "bb", "bc", "cc");

Observable<List<String>> groupedRows = rows.groupBy(new Func1<String, Character>() {
  public Character call(String row) {
    return row.charAt(0);
  }
}).flatMap(new Func1<GroupedObservable<Character, String>, Observable<List<String>>>() {
  public Observable<List<String>> call(GroupedObservable<Character, String> group) {
    return group.toList();
  }
});

groupedRows.toBlocking().forEach(new Action1<List<String>>() {
  public void call(List<String> group) {
    System.out.println(group);
  }
});

// Output:
// [aa, ab, ac]
// [bb, bc]
// [cc]
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但它对我的目的并不好,因为groupBy只有在源可观察量发出时才完成每个组onComplete.因此,如果我有很多行,它们将完全聚集在内存中,并且只在最后一行"刷新"并写入输出.

我需要像buffer运算符这样的东西,但是我自己的函数表示每个组的边界.我实现了它(知道行总是按字母排序):

Observable<String> rows = Observable.just("aa", "ab", "ac", "bb", "bc", "cc");

ConnectableObservable<String> connectableRows = rows.publish();

Observable<String> boundarySelector = connectableRows.filter(new Func1<String, Boolean>() {
  private char lastChar = 0;

  public Boolean call(String row) {
    char currentChar = row.charAt(0);
    boolean isNewGroup = lastChar != 0 && (currentChar != lastChar);
    lastChar = currentChar;
    return isNewGroup;
  }
});

Observable<List<String>> groupedRows = connectableRows.buffer(boundarySelector);

connectableRows.connect();

groupedRows.toBlocking().forEach(new Action1<List<String>>() {
  public void call(List<String> group) {
    System.out.println(group);
  }
});

// Output:
// []
// []
// []
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它不起作用,因为它boundarySelector正在"吃掉"行,我认为这很奇怪,因为我特意用来ConnectableObservable表示在开始发射之前需要两个订阅者(boundarySelectorgroupedRows)rows.

好奇如果我延迟rows1秒,那么这段代码就可以了.

所以问题是:如何使用我自己的边界函数对任意数量的行进行分组?

aka*_*okd 2

Observable<Integer> source = Observable.range(0, 100);

source
.groupBy(k -> k / 10)
.publish(groups -> groups
        .map(g -> Pair.of(g.getKey(), g.takeUntil(groups)))
        .flatMap(kv -> 
            kv.second
            .doOnNext(v -> System.out.println(kv.first + " value " + v))
            .doOnCompleted(() -> System.out.println(kv.first + " done"))
        ))
.subscribe()
;
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