是否有一种方法来迭代Java SparseArray (Android) ?我使用sparsearray很容易地按索引获取值。我找不到一个。


当前回答

看来我找到解决办法了。我没有正确地注意到keyAt(index)函数。

所以我会这样说:

for(int i = 0; i < sparseArray.size(); i++) {
   int key = sparseArray.keyAt(i);
   // get the object by the key.
   Object obj = sparseArray.get(key);
}

其他回答

答案是否定的,因为SparseArray没有提供。正如pst所说,这个东西不提供任何接口。

您可以从0 - size()开始循环,并跳过返回null的值,但仅此而已。

正如我在我的评论中所述,如果你需要迭代使用Map而不是SparseArray。例如,使用按键顺序迭代的TreeMap。

TreeMap<Integer, MyType>

或者你只需要创建你自己的ListIterator:

public final class SparseArrayIterator<E> implements ListIterator<E> {

private final SparseArray<E> array;
private int cursor;
private boolean cursorNowhere;

/**
 * @param array
 *            to iterate over.
 * @return A ListIterator on the elements of the SparseArray. The elements
 *         are iterated in the same order as they occur in the SparseArray.
 *         {@link #nextIndex()} and {@link #previousIndex()} return a
 *         SparseArray key, not an index! To get the index, call
 *         {@link android.util.SparseArray#indexOfKey(int)}.
 */
public static <E> ListIterator<E> iterate(SparseArray<E> array) {
    return iterateAt(array, -1);
}

/**
 * @param array
 *            to iterate over.
 * @param key
 *            to start the iteration at. {@link android.util.SparseArray#indexOfKey(int)}
 *            < 0 results in the same call as {@link #iterate(android.util.SparseArray)}.
 * @return A ListIterator on the elements of the SparseArray. The elements
 *         are iterated in the same order as they occur in the SparseArray.
 *         {@link #nextIndex()} and {@link #previousIndex()} return a
 *         SparseArray key, not an index! To get the index, call
 *         {@link android.util.SparseArray#indexOfKey(int)}.
 */
public static <E> ListIterator<E> iterateAtKey(SparseArray<E> array, int key) {
    return iterateAt(array, array.indexOfKey(key));
}

/**
 * @param array
 *            to iterate over.
 * @param location
 *            to start the iteration at. Value < 0 results in the same call
 *            as {@link #iterate(android.util.SparseArray)}. Value >
 *            {@link android.util.SparseArray#size()} set to that size.
 * @return A ListIterator on the elements of the SparseArray. The elements
 *         are iterated in the same order as they occur in the SparseArray.
 *         {@link #nextIndex()} and {@link #previousIndex()} return a
 *         SparseArray key, not an index! To get the index, call
 *         {@link android.util.SparseArray#indexOfKey(int)}.
 */
public static <E> ListIterator<E> iterateAt(SparseArray<E> array, int location) {
    return new SparseArrayIterator<E>(array, location);
}

private SparseArrayIterator(SparseArray<E> array, int location) {
    this.array = array;
    if (location < 0) {
        cursor = -1;
        cursorNowhere = true;
    } else if (location < array.size()) {
        cursor = location;
        cursorNowhere = false;
    } else {
        cursor = array.size() - 1;
        cursorNowhere = true;
    }
}

@Override
public boolean hasNext() {
    return cursor < array.size() - 1;
}

@Override
public boolean hasPrevious() {
    return cursorNowhere && cursor >= 0 || cursor > 0;
}

@Override
public int nextIndex() {
    if (hasNext()) {
        return array.keyAt(cursor + 1);
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public int previousIndex() {
    if (hasPrevious()) {
        if (cursorNowhere) {
            return array.keyAt(cursor);
        } else {
            return array.keyAt(cursor - 1);
        }
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public E next() {
    if (hasNext()) {
        if (cursorNowhere) {
            cursorNowhere = false;
        }
        cursor++;
        return array.valueAt(cursor);
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public E previous() {
    if (hasPrevious()) {
        if (cursorNowhere) {
            cursorNowhere = false;
        } else {
            cursor--;
        }
        return array.valueAt(cursor);
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public void add(E object) {
    throw new UnsupportedOperationException();
}

@Override
public void remove() {
    if (!cursorNowhere) {
        array.remove(array.keyAt(cursor));
        cursorNowhere = true;
        cursor--;
    } else {
        throw new IllegalStateException();
    }
}

@Override
public void set(E object) {
    if (!cursorNowhere) {
        array.setValueAt(cursor, object);
    } else {
        throw new IllegalStateException();
    }
}
}

对于使用Kotlin的人来说,老实说,到目前为止迭代SparseArray最简单的方法是:使用Anko或Android KTX的Kotlin扩展!(感谢Yazazzello指出Android KTX)

简单地调用forEach {i, item ->}

看来我找到解决办法了。我没有正确地注意到keyAt(index)函数。

所以我会这样说:

for(int i = 0; i < sparseArray.size(); i++) {
   int key = sparseArray.keyAt(i);
   // get the object by the key.
   Object obj = sparseArray.get(key);
}

公认的答案有一些漏洞。SparseArray的美妙之处在于它允许在折痕处有空隙。我们可以有两个地图,在SparseArray中。

(0,true)
(250,true)

注意这里的大小是2。如果我们遍历大小,我们将只得到映射到索引0和索引1的值。因此不能访问键值为250的映射。

for(int i = 0; i < sparseArray.size(); i++) {
   int key = sparseArray.keyAt(i);
   // get the object by the key.
   Object obj = sparseArray.get(key);
}

最好的方法是遍历数据集的大小,然后在数组上使用get()检查这些索引。这里有一个适配器的例子,我允许批量删除项目。

for (int index = 0; index < mAdapter.getItemCount(); index++) {
     if (toDelete.get(index) == true) {
        long idOfItemToDelete = (allItems.get(index).getId());
        mDbManager.markItemForDeletion(idOfItemToDelete);
        }
    }

我认为理想情况下SparseArray家族应该有一个getKeys()方法,但可惜它没有。