在我的应用程序中,我使用第三方库(Spring Data for MongoDB准确地说)。

这个库的方法返回Iterable<T>,而我的其余代码期望Collection<T>。

有没有什么实用的方法可以让我快速地把一个转换成另一个?为了这么简单的事情,我希望避免在代码中创建一堆foreach循环。


当前回答

你也可以编写自己的实用方法:

public static <E> Collection<E> makeCollection(Iterable<E> iter) {
    Collection<E> list = new ArrayList<E>();
    for (E item : iter) {
        list.add(item);
    }
    return list;
}

其他回答

从CollectionUtils:

List<T> targetCollection = new ArrayList<T>();
CollectionUtils.addAll(targetCollection, iterable.iterator())

下面是这个实用方法的完整源代码:

public static <T> void addAll(Collection<T> collection, Iterator<T> iterator) {
    while (iterator.hasNext()) {
        collection.add(iterator.next());
    }
}

试试Cactoos的StickyList:

List<String> list = new StickyList<>(iterable);

两个评论

There is no need to convert Iterable to Collection to use foreach loop - Iterable may be used in such loop directly, there is no syntactical difference, so I hardly understand why the original question was asked at all. Suggested way to convert Iterable to Collection is unsafe (the same relates to CollectionUtils) - there is no guarantee that subsequent calls to the next() method return different object instances. Moreover, this concern is not pure theoretical. E.g. Iterable implementation used to pass values to a reduce method of Hadoop Reducer always returns the same value instance, just with different field values. So if you apply makeCollection from above (or CollectionUtils.addAll(Iterator)) you will end up with a collection with all identical elements.

With Guava you can use Lists.newArrayList(Iterable) or Sets.newHashSet(Iterable), among other similar methods. This will of course copy all the elements in to memory. If that isn't acceptable, I think your code that works with these ought to take Iterable rather than Collection. Guava also happens to provide convenient methods for doing things you can do on a Collection using an Iterable (such as Iterables.isEmpty(Iterable) or Iterables.contains(Iterable, Object)), but the performance implications are more obvious.

因为RxJava是一个锤子,而这个看起来像钉子,你可以这样做

Observable.from(iterable).toList().toBlocking().single();