用RX Java或首先获取过滤器索引的绝佳方法 [英] Elegant way to get index of filter or first with RX Java

问题描述

我只是在练习RX Java，想获取数组中与过滤器匹配的项的位置.我看不到任何明显的方法.我当时在寻找压缩范围和可迭代的可观察范围之类的东西，但是它很快变得比for循环更冗长和复杂.

I'm just practicing RX Java and wanted to get the position in an array for items which match a filter. I can't see any obvious way to do it. I was looking at maybe zipping a range and iterable observable or something but it was quickly getting way more verbose and complicated than for loops.

推荐答案

RxJava中曾经有mapWithIndex和zipWithIndex运算符，但是它们已被删除，请参见

There used to be mapWithIndex and zipWithIndex operators in RxJava, but they were removed, see here why.

因此，您必须编写一些库样板:

So you have to write some library boilerplate once:

class Indexed<T> {
    final int index;
    final T value;
    public Indexed(T value, int index) {
        this.index = index;
        this.value = value;
    }
    @Override
    public String toString() {
        return index + ") " + value;
    }
}

Iterable<Integer> naturals = IntStream.iterate(0, i -> i + 1)::iterator;

但是随后，您可以使其简洁明了:

But then, you can get it reasonably concise:

Observable<String> obs = Observable.just("zero", "one", "two", "three");
obs.zipWith(naturals, (s, i) -> new Indexed<String>(s, i))
   .filter(e -> e.value.length() > 4)
   .subscribe(System.out::println);

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